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oracle 的lob & long

tacy lee — Mon, 23 Jun 2008 17:18:00 GMT

一直认为lob�c�d��的性能要好�q�long�Q�但是之前只了解到l(f��)ong的种�U�限�Ӟ��oracle也是不推荐��用long�c�d��Q�这几天�׃��一个项目问题，产品里面一个表字段用了long�c�d��Q�分析下来操作long的时候，性能有所影响�Q�想把它�Ҏ(gu��)��lob�Q�就��单验证了一�?/p>

首先创徏两个��试表：(x��)

create table test_long (a int primary key,b long);
create table test_clob (a int primary key,b clob);

用附件java代码�Q�往两个表里面各插入100条数据，保证插入数据是一��L(f��ng)��Q�lob字段长度�?0k�Q�如果小�?k�Q�oracle可以把它保存到到表内�Q�不�?x��)存储在表外�Q�性能没有问题�Q�这个我基本��定�Q�而且我们应用中这个字�D늻��怼�(x��)��过4k�Q��?

做一个简单查询对比一下：(x��)

SQL> set autotrace traceonly;
SQL> select * from test_clob where a=1;

�l�计信息
----------------------------------------------------------
        331 recursive calls
          0 db block gets
         69 consistent gets
          4 physical reads
          0 redo size
       1278 bytes sent via SQL*Net to client
        837 bytes received via SQL*Net from client
          5 SQL*Net roundtrips to/from client
         12 sorts (memory)
          0 sorts (disk)
          1 rows processed

SQL> select * from test_long where a=1;

�l�计信息
----------------------------------------------------------
        236 recursive calls
          0 db block gets
         43 consistent gets
          0 physical reads
          0 redo size
        675 bytes sent via SQL*Net to client
        531 bytes received via SQL*Net from client
          3 SQL*Net roundtrips to/from client
          5 sorts (memory)
          0 sorts (disk)
          1 rows processed

�Ҏ(gu��)��一下，long开销比lob��，当然你可以把lob字段启用�~�存�Q�把4�ơ物理读��L��Q�但�q�是多了�Q?3-43�Q�次逻辑读，update也试了一下，lob产生的redo比long大，��׃��列出来了�Q�有兴趣的可以自��p��?

��试下来�Q�看来之前的认识不对�Q�不��定的东西最好还是动手试试，当然对于新应用，�q�是不徏议用long�Q�毕竟oracle已经废弃它了�?

testClobLong.java

tacy lee 2008-06-24 01:18 发表评论

tacy lee — Wed, 28 May 2008 15:05:00 GMT

有时候，我们可能希望看到l(f��)r的出错页面：(x��)比如lr出错�Q�但是后台服务器没有错误日志�Q�这时候，我们希望能看到错误页面的内容来判断问题出在什么地方，但是lr没有提供�c�M��的功�?

我们可以通过一�U�变通的办法来实玎ͼ�(x��)

首先扑ֈ�你出错的��面�Q�保存该��面到参数里面：(x��)

web_set_max_html_param_len(“2048”);

web_reg_save_param(“FILED”,”LB=”,”RB=”,”Search=Body”,LAST);

然后输出到日志里面：(x��) lr_output_message(”#######################################%s”,lr_eval_string(”{FILED}”));

修改lr run-time的几个设�|�：(x��)

1、Always send messages

2、continue on error �Q�这��h��能保证运行lr_output_message)

�q�样lr�?x��)把所有的lr_output_message输出保存到日志文�?

当然你不要下载资源文�Ӟ��否则保存到的��׃��是html��面了，可能是一个gif :(

最后，�l�合lr controller的错误信息，定位到出错的vuser id�Q�查看该vuser的log文�g��p��看到错误��面�?

非常有效的一个小技巧，用它解决了一个难�~�的问题�?

tacy lee 2008-05-28 23:05 发表评论

ibm jdk 1.5�~�省用的gc�{�略性能很差

tacy lee — Mon, 14 Apr 2008 12:38:00 GMT

摘要: �q�几天测试一个引擎的性能�Q�用一个单表查询的case�Q�测试出来的�l�果�?10tps�Q�cpu也正常，�?5%左右�Q�也没怀疑�?

后面再重新测试的时候，加上了gc log�Q�用gc分析工具分析了一下gc的吞吐量�Q�发现吞吐量奇低�Q�竟然只�?7%左右�Q�很是奇怪，看了一下gc日志�Q�所有都是global gc�Q?怀疑gc�{�略有问题，查了一下资料，参考了下面一��文章：(x��) 阅读全文

tacy lee 2008-04-14 20:38 发表评论

�q�发是啥

tacy lee — Tue, 18 Mar 2008 07:33:00 GMT

一个用��L(f��ng)��d��是一个用戯��求，一个webservice�c�M��的调用也��一个请求，�{�等

一个用户在某个旉��点上当然只能发�v一个用戯��求，一个用戯��求就是一个�ƈ�?br>

我们一般纠�~�在同一事物�q�发�q�是不同事务�q�发�?br>

可能在一个时间点上，�?00个用户在发送浏览，查询动作�Q?0个用户在下订单，5个用户在做付?g��u)Ƒ֊�作，你说�q�个旉��点上有多��个�q�发��h��Q�当然是115个了

衡量一个系�l�性能主要靠的��是�q�个吞吐量（tps�Q?br>

当然我们也非常关心同�?00个用户�ƈ发下订单的时候系�l�是否能支撑�Q�这是通常我们大部分�h理解的�ƈ发）�Q�我们会(x��)说这是核心业务，我们要得出数据（是否要考虑背景业务呢，呵呵�Q�很难说的清楚，我一般就不考虑�Q?/p>

tacy lee 2008-03-18 15:33 发表评论

工作日志-OOM事�g

tacy lee — Sun, 16 Mar 2008 14:38:00 GMT

某项目，�q�前开始报OOM�Q�频率保持在一月一�ơ，发生OOM的时候，heap free size�q�有7�?00M�Q�比较奇怪，�q�后�pȝ��上集��，�pȝ��发生OOM的频率开始变得频�J�，基本�?-5天，�׃��用的是sun jdk 1.4.2_08�Q�无法获取到heap dump�Q�徏议用户升�U�到1.4.2_14�Q�该版本以后sun��d��了HeapDumpOnOutOfMemoryError参数�Q�便于获取dump帮助诊断该类问题�Q�，4天之后，我们获取��C��heapdump文�g�Q�通过对dump的分析，基本上排除了对象泄漏�?/p>

�Ҏ(gu��)��环境�Q?4bit Solaris + 32bit JDK�Q�，客户把Heap最大设�|��ؓ(f��)2G�Q�开始怀�?2bit JDK无法分配�q�么大的Heap�Q�经�q�验证，不存在这��L(f��ng)��问题�Q�sun�|�站也有相关说明�Q�在solaris 64bit�pȝ��上，32bit jdk最大可以设�|�到4G�Q?/p>

但是从dump看到application classes loader大小已经��C��60M以上�Q�有�Ҏ(gu��)��疑Perm��|�太��导��_(d��)��查了一下sun的文档，Perm区缺省大��ؓ(f��)64M�Q�估计是应用加蝲太多classes��D��Perm区溢出，

我们也简单模拟了一下Perm溢出�Q�强制设�|�max perm大小�?2M�Q��ƈ对GC�q�行了监控，�l�果和我们预想的一��_(d��)��看下面的gc log�Q?

151.836: [Full GC 151.836: [Tenured: 25735K->25736K(1048576K), 0.8380858 secs] 25911K->25736K(1557568K), [Perm : 32767K->32767K(32768K)], 0.8382804 secs]
152.676: [Full GC 152.676: [Tenured: 25736K->25722K(1048576K), 0.8464782 secs] 25752K->25722K(1557568K), [Perm : 32767K->32766K(32768K)], 0.8466638 secs]
153.525: [Full GC 153.525: [Tenured: 25722K->25724K(1048576K), 0.8419056 secs] 25738K->25724K(1557568K), [Perm : 32767K->32767K(32768K)], 0.8420986 secs]
154.368: [Full GC 154.368: [Tenured: 25724K->25724K(1048576K), 0.8398816 secs] 25724K->25724K(1557568K), [Perm : 32767K->32767K(32768K)], 0.8400498 secs]
155.212: [Full GC 155.212: [Tenured: 25724K->25725K(1048576K), 0.8365448 secs] 25788K->25725K(1557568K), [Perm : 32767K->32767K(32768K)], 0.8367370 secs]
156.050: [Full GC 156.050: [Tenured: 25725K->25722K(1048576K), 0.8422488 secs] 25725K->25722K(1557568K), [Perm : 32767K->32766K(32768K)], 0.8424328 secs]
156.895: [Full GC 156.895: [Tenured: 25722K->25724K(1048576K), 0.8443532 secs] 25738K->25724K(1557568K), [Perm : 32767K->32767K(32768K)], 0.8445450 secs]
157.740: [Full GC 157.741: [Tenured: 25724K->25724K(1048576K), 0.8427754 secs] 25740K->25724K(1557568K), [Perm : 32767K->32767K(32768K)], 0.8429634 secs]
158.587: [Full GC 158.588: [Tenured: 25724K->25726K(1048576K), 0.8352290 secs] 25820K->25726K(1557568K), [Perm : 32767K->32767K(32768K)], 0.8354212 secs]
159.424: [Full GC 159.424: [Tenured: 25726K->25723K(1048576K), 0.8435336 secs] 25726K->25723K(1557568K), [Perm : 32767K->32766K(32768K)], 0.8437092 secs]
160.270: [Full GC 160.270: [Tenured: 25723K->25725K(1048576K), 0.8477722 secs] 25739K->25725K(1557568K), [Perm : 32767K->32767K(32768K)], 0.8479596 secs]
161.119: [Full GC 161.119: [Tenured: 25725K->25725K(1048576K), 0.8543338 secs] 25725K->25725K(1557568K), [Perm : 32767K->32767K(32768K)], 0.8545040 secs

从日志看�Q�和我们现场的状况非常相��|��heap�I�间充��Q�但是perm已经��C��32M�Q�无法再�q�一步分配空��_(d��)��直接��D��jvm频繁做Full GC�Q�控制台也开始抛出OOM�Q�Perm引�v的回攉��是full gc�Q�，�q�样看基本我们判断是Perm太小�Q�导致无法加载classes��D��?

和客��h��通之后，我们本来打算�q�一步验证（在生产环节打开PrintGCDetail�Q�获取详�l�的GC log�Q�，后面仔细��查nohup.out,发现里面已经抛出�?OutOfMemoryError:PermGen Space�Q�至此我们确定是Perm讄��不合理导致了本次事故�Q�和客户��认之后�Q�我们在启动参数中加上了MaxPermSize

后面惛_��中间上了集群之后�Q�eos加蝲了大量的jboss cache class�Q�这也直接解释了��Z��么这�D�|��间OOM出现的频率比之前更频�J�的原因

�q�里�ȝ��一下，希望对碰到类似问题的tx有借鉴意义�Q�强烈徏议用sun jdk 1.4.2的同学升�U�到>=1.4.2_12�Q�便于对OOM问题的诊断，�q�加上GC log协助验证�?

�q�里再介�l�一下JVM发生OOM的几�U�情况：(x��)

1、java.lang.OutOfMemoryError: Java heap space

�q�是我们�q�_��理解的OOM�Q�是�׃��heap space��实没有�I�间分配�Q�这�U�一般是�׃��内存泄漏��D��Q�也有可能是heap space讄��太小。需要具体分�?

2、java.lang.OutOfMemoryError: PermGen space

jvm规范里面有定义一个method space�Q�这里主要放classes和method list和一个string pool�Q�string有一个intern�Ҏ(gu��)��Q�通过�q�个�Ҏ(gu��)��定义的string都放在这里（好像不常用）�Q�这里设�|�不太小�?x��)导致OOM�Q�缺�?4M�Q�主要由于现在应用依赖的�W�三方类��来��多�Q�导致这�c�问题频�J�发生，需要引起重�?

3、Requested array size exceeds VM limit
�q�种是由于申��L(f��ng)��array size��出了heap space大小�Q�比如在一�?56M的heap space中申请一�?12M的array�Q�这�U�基本都是应用bug��D��

4、request bytes for . Out of swap space?
�q�种是由于heap size讄��相对于系�l�物理内存太大，��D��pȝ��swap space不��Q�这�U�的解决办法��是减小heap size大小

5�?lt;reason> (Native method)
�q�种估计是最�ȝ��的了�Q�也是最��碰到的�Q�是�׃��jni或native method��D��Q�如果自己没有写�q�类的东西，基本可以说是jdk问题

tacy lee 2008-03-16 22:38 发表评论

关于oracle中的timestamp和date�c�d��

tacy lee — Tue, 25 Dec 2007 16:42:00 GMT

之前一直认为类��|��(x��)where timestamp>date �q�种子句是不走烦引的

下面��单做一个验证：(x��)

c:>sqlplus / as sysdba
sys@EOS >create table test as select table_name,to_timestamp(last_analyzed) date_test from dba_tables;

表已创徏�?br />
sys@EOS> create index idx_test_date on test (date_test);

索引已创建�?br />
sys@EOS> desc test
名称                                                  是否为空? �c�d��
----------------------------------------------------- -------- ----------------
--------------------
TABLE_NAME                                            NOT NULL VARCHAR2(30)
DATE_TEST                                                      TIMESTAMP(0)

sys@EOS> select date_test from test where date_test > TO_DATE('2007-11-5 00:00:00','yyyy-MM-dd HH24:mi:ss');

执行计划
----------------------------------------------------------
Plan hash value: 944171586

-------------------------------------------------------------------------------- --
| Id | Operation        | Name          | Rows | Bytes | Cost (%CPU)| Time |
-------------------------------------------------------------------------------- --
|   0 | SELECT STATEMENT |               |     1 |    22 |     1   (0)| 00:00:01 |
|* 1 | INDEX RANGE SCAN| IDX_TEST_DATE |     1 |    22 |     1   (0)| 00:00:01 |
-------------------------------------------------------------------------------- --

Predicate Information (identified by operation id):
---------------------------------------------------

   1 - access("DATE_TEST">TIMESTAMP'2007-11-05 00:00:00')

Note
-----
   - dynamic sampling used for this statement

�l�计信息
----------------------------------------------------------
          7 recursive calls
          0 db block gets
         18 consistent gets
          0 physical reads
          0 redo size
        280 bytes sent via SQL*Net to client
        374 bytes received via SQL*Net from client
          1 SQL*Net roundtrips to/from client
          0 sorts (memory)
          0 sorts (disk)
          0 rows processed

从上面可以清楚看刎ͼ�timestamp>date情况下，走烦�?br />
�U�正我之前的认识�?br />
另外再补充一下，date�q�个数据�c�d��一般情况下很少用，��产品里面所有的date数据�c�d��全部改�ؓ(f��)timestamp

tacy lee 2007-12-26 00:42 发表评论

Websphere到底是否需要配�|�IHS

tacy lee — Thu, 13 Dec 2007 06:19:00 GMT

作者：(x��)tacy lee

有用Websphere做过��目的�h可能都知道，ibm一般都��在Websphere前面加一个IHS来做webserver�Q�据说这��h��能�?x��)提�?0%左右�Q�这栯��是否有道理呢�Q�下面我做了一个简单的��试来验证：(x��)

��试环境�Q?/p>

��g�Q?/p>

应用服务器：(x��)Dell6600

压力��试客户端：(x��)自用�W�记本（T2050 1.6G)

软�g�Q?/p>

�pȝ��Q�CentOS 4.4

Websphere 6.0.2.17+IHS6.0.2.17�Q�部�|�在同一台机器上�Q?/p>

首先配置好Websphere和IHS�Q�发布一个简单的��试应用�Q�用loadrunner来测试一下不同的�l�合看看�Q�录制一个打开首页��可以了�Q�，下面是我的测试数据：(x��)

��试�Ҏ(gu��)��	每秒处理��h��?/td>	响应旉��	服务器CPU
直接��h��Websphere	4600/s	0.013s	28%
通过IHS转发��h��	6800/s	0.009s	26%

数据昄��Q�这�q�不是一点点提升�Q�竟然快接近50%�Q�把静态资源放�|�到IHS中测试了一把，基本和通过IHS转发差不多，�E�微有些提升�Q�不�q�放到IHS中可以方便Cache�Q�Edge Server��包括了Caching Proxy component�Q?/p>

下面记录一下如何放�|�静态资源文件到IHS中：(x��)

1、打开Plugins中的plugin-cfg.xml�Q�修改如下内容：(x��)

"default_host_eos_URIs">
   "JSESSIONID" AffinityURLIdentifier="jsessionid" Name="/*.jsp"/>
   "JSESSIONID" AffinityURLIdentifier="jsessionid" Name="/*.do"/>
   "JSESSIONID" AffinityURLIdentifier="jsessionid" Name="/eosmgr/*"/>
   "JSESSIONID" AffinityURLIdentifier="jsessionid" Name="/axis/*"/>
   "JSESSIONID" AffinityURLIdentifier="jsessionid" Name="/axis2/*"/>
   "JSESSIONID" AffinityURLIdentifier="jsessionid" Name="/eoshome_deploy/*"/>

也可以通过修改WEB-INF下ibm-web-ext.xmi中的fileServingEnabled为false�Q�然后重新生成plugin-cfg.xml�Q�但是我试了一下好像不好用�?/p>

另外Websphere�Q�fixpacks 5.1.1.17, 6.0.2.25 and 6.1.0.15�Q�之后的版本�l�Webcontainer增加了一个自定义参数

com.ibm.ws.webcontainer.disallowAllFileServing

讑֮�它�ؓ(f��)true产生同样的效果（而且他会(x��)覆盖ibm-web-ext.xmi中的讄��Q��?/p>

2、拷贝你的所有资源文件到IHS的Root Directory�?/p>

3、重启IHS

del.icio.us Tags: websphere,ihs,configuration,tuning

tacy lee 2007-12-13 14:19 发表评论

db2诊断�p�d��?--捕获sql执行情况

tacy lee — Sun, 25 Nov 2007 06:51:00 GMT

作者：(x��)tacy lee

在应用��用过�E�中�Q�我们经�怼�(x��)��到应用响应旉��很慢�Q�甚��x��有响应，但是应用服务器可能�ƈ不是很繁忙，cpu利用率也非常低，引�v�q�种状况的原因有很多�U�，比如环境问题�Q�应用资源泄漏，数据库原因等�{�，本文主要是从一�ơ应用性能诊断�q�程来谈谈如何通过数据库诊断应用性能问题�?/p>

问题�Q?/p>

��试�q�程中发现应用中某个跌��{��面执行旉��比较长，�pȝ��压力不大�Q�cpu利用很低�Q�该��面需要从cache中取数据�Q�第一�ơ的时候加载cache�Q�从数据库中查询回数据�ƈcache�Q��?/p>

诊断�Q?/p>

��面逻辑比较��单，我们先用loadrunner模拟�q�发��试一下这个页面，然后再数据库端捕获sql执行情况�?/p>

1、打开db2监控开�?/p>

#db2 connect to eos

#db2 update monitor switches using statement on

#db2 reset monitor all

2、几分钟之后�Q�我们收集sql�l�计快照

#db2 get snapshot for dynamic sql on eos > dysqlstatus.out

现在�l�计信息已经存放在dysqlstatus.out中，你可以��用�Q意方便的文本处理工具查看�Q�我一般用windows上的gvim来处理，打开dysqlstatus.out

Number of executions               = 1

Number of compilations             = 1

Worst preparation time (ms)        = 2

Best preparation time (ms)         = 2

Internal rows deleted              = 0

Internal rows inserted             = 0

Rows read                          = 2

Internal rows updated              = 0

Rows written                       = 0

Statement sorts                    = 0

Statement sort overflows           = 0

Total sort time                    = 0

Buffer pool data logical reads     = Not Collected

Buffer pool data physical reads    = Not Collected

Buffer pool temporary data logical reads   = Not Collected

Buffer pool temporary data physical reads  = Not Collected

Buffer pool index logical reads    = Not Collected

Buffer pool index physical reads   = Not Collected

Buffer pool temporary index logical reads  = Not Collected

Buffer pool temporary index physical reads = Not Collected

Total execution time (sec.ms)      = 0.000377

Total user cpu time (sec.ms)       = 0.010000

Total system cpu time (sec.ms)     = 0.000000

Statement text                     = select ACTIVITYDEFID,ACTIVITYINSTID from wfworkitem  where 
PROCESSINSTID=104199 and CURRENTSTATE = 4


......

��单说一下vi中的处理

:g!/Total execution time/d

只保留文本中的sql执行旉����Q�我们要按照执行旉���来排�?br>


通过vim的visual功能选择执行旉���块（�{�号后面的数字）�Q�然后排�?br>
Total execution time (sec.ms)      = 0.050590

Total execution time (sec.ms)      = 0.000170

Total execution time (sec.ms)      = 0.000247

Total execution time (sec.ms)      = 0.000292

Total execution time (sec.ms)      = 0.000474

Total execution time (sec.ms)      = 0.000330

Total execution time (sec.ms)      = 0.000348

Total execution time (sec.ms)      = 0.000279

Total execution time (sec.ms)      = 0.000385

Total execution time (sec.ms)      = 0.000296

Total execution time (sec.ms)      = 0.000261

Total execution time (sec.ms)      = 0.000195

Total execution time (sec.ms)      = 0.000226

Total execution time (sec.ms)      = 0.000227

Total execution time (sec.ms)      = 0.000193

......

:'<,'>!sort



排序后的�l�果�Q�部分）

Total execution time (sec.ms)      = 2.027776

Total execution time (sec.ms)      = 2.203624

Total execution time (sec.ms)      = 2.504677

Total execution time (sec.ms)      = 2.951256

Total execution time (sec.ms)      = 3.119875

Total execution time (sec.ms)      = 3.303277

Total execution time (sec.ms)      = 3.303517

Total execution time (sec.ms)      = 4.017133

Total execution time (sec.ms)      = 4.043329

Total execution time (sec.ms)      = 4.252125

Total execution time (sec.ms)      = 4.400952

Total execution time (sec.ms)      = 4.606765

Total execution time (sec.ms)      = 5.208087

Total execution time (sec.ms)      = 5.778598

Total execution time (sec.ms)      = 8.117470

Total execution time (sec.ms)      = 9797.905136

可以看到最长时间的sql total执行旉��耗费�?797.905123s.

现在我们到dysqlstatus.out中去找这条语�?/p>

Number of executions               = 4602

Number of compilations             = 4294967295

Worst preparation time (ms)        = 2

Best preparation time (ms)         = 2

Internal rows deleted              = 0

Internal rows inserted             = 0

Rows read                          = 2963688

Internal rows updated              = 0

Rows written                       = 0

Statement sorts                    = 0

Statement sort overflows           = 0

Total sort time                    = 0

Buffer pool data logical reads     = Not Collected

Buffer pool data physical reads    = Not Collected

Buffer pool temporary data logical reads   = Not Collected

Buffer pool temporary data physical reads  = Not Collected

Buffer pool index logical reads    = Not Collected

Buffer pool index physical reads   = Not Collected

Buffer pool temporary index logical reads  = Not Collected

Buffer pool temporary index physical reads = Not Collected

Total execution time (sec.ms)      = 9797.905136

Total user cpu time (sec.ms)       = 9.290000

Total system cpu time (sec.ms)     = 1.230000

Statement text                     = select * from XXXX_T_CNFACTIVITYDEF

�q�条语句��d��执行�?602�ơ，�q�_��每次的执行时�?S�Q�而且�q�些数据应该是被cache��h��?nbsp; �Q�）

�ȝ��Q?/p>

上面的方法简单�ȝ��了从数据库层面对应用的性能问题诊断�Q�希望对大家有所帮助�Q�对于数据库快照诊断问题的思�\对于��L��数据库通用

补充一个unix上脚本处理方式：(x��)

sqlsort.sh

awk 'BEGIN{RS="";FS="\n";ORS="\n"};/Statement text/{print $1, $21, $24}' $1 | awk '$5 > 0 {print "AvgTime:", $11/$5, "\t", $0}'| sort -n | head -n $2|awk '{print $0, "\n"}'

使用�Q?sqlsort.sh dysqlstate.out 10�Q�显�C�Top ten�Q?/div>

del.icio.us Tags: db2,java,tuning,diagnostic

tacy lee 2007-11-25 14:51 发表评论

tacy lee — Sat, 24 Nov 2007 13:18:00 GMT

作者：(x��)tacy lee

在应用中�Q�我们经�怼�(x��)��到sql执行很慢�Q�但是数据库cpu和内存��用率又不高的情况�Q�类似的问题基本上由于锁�Q�排序等原因造成�Q�本文主要描�q�如何去定位锁等待问题，谁在锁等待？�{�待谁持有的锁？锁在那个表？

一、测试准�?/p>

1、先在session1执行如下操作�Q�创建测试表

#db2 connect to eos
#export DB2OPTIONS=+C
#db2 "create table tacy_test (a int not null primary key,b varchar(10))"
#db2 "insert into tacy_test values(1,'a')"
#db2 "insert into tacy_test values(2,'a')"
#db2 "insert into tacy_test values(3,'a')"
#db2 "insert into tacy_test values(4,'a')"
#db2 commit

2、在session2执行如下操作

#db2 connect to eos
#export DB2OPTIONS=+C

二、��生一个lock wait

在session1做一个表更新�Q?/p>

#db2 "update tacy_test set b='b' where a=4"

sql执行成功

在session2做同��h��新操作：(x��)

#db2 "update tacy_test set b='c' where a=4"

�q�程被挂��L(f��ng)��?/p>

三、定位锁�{�待

1、先来看看应用的情况�Q?/p>

#db2pd -db eos -applications

Database Partition 0 -- Database EOS -- Active -- Up 0 days 07:37:37

Applications:
Address    AppHandl [nod-index] NumAgents  CoorPid    Status                  C-AnchID C-StmtUID  L-AnchID L-StmtUID  Appid                           
0x10140040 8        [000-00008] 1          8425       Lock-wait               80       2          66       1          *LOCAL.db2inst1.071124043739    
0x100CE540 7        [000-00007] 1          8358       UOW-Waiting             0        0          80       2          *LOCAL.db2inst1.071124043708

可以看到有一个应用的状态处于Lock-wait

2、现在我们来看看应用在等什�?/p>

#db2pd -db eos -locks showlock wait

Database Partition 0 -- Database EOS -- Active -- Up 0 days 07:42:56

Locks:
Address    TranHdl    Lockname                   Type       Mode Sts Owner      Dur HldCnt     Att Rlse
0x2C8E0760 3          02001806078066020000000052 Row        ..X  W   2          1   0          0   0x0  TbspaceID 2 TableID 1560 RecordID 0x2668007

锁的�c�d��为Row�Q�行锁）�Q�X锁（排他锁）�Q�下面是我们最兛_��的锁的位�|?/p>

TbspaceID 2 TableID 1560 RecordID 0x2668007

其中TbspaceID��I�间ID�Q�TableID��的ID�Q�RecordID代表具体位置�Q�全部应该是0x0266807,其中前面三个字节为page number�Q��ؓ(f��)0x02668�Q�后面一个字节代表solt identifier�Q��ؓ(f��)0x07

3、找到相应的�?/p>

#db2 "select tbspace,tabschema,tabname,tableid,tbspaceid from syscat.tables where tbspaceid=2 and tableid=1560"

TBSPACE       TABSCHEMA   TABNAME    TABLEID TBSPACEID
------------  ----------- ---------- ------- ---------
USERSPACE1    DB2INST1    TACY_TEST     1560         2

  1 record(s) selected.

4、根据RecordID扑ֈ�锁在哪行

db2提供了一个强大的数据分析工具db2dart�Q�可以dump出相应的page数据

#db2dart eos /dd /tsi 2 /oi 1560 /ps 157312p /np 1 /v y

Warning: The database state is not consistent.

Warning: Reorg rows MAY be due to the inconsistent state of the database.
                  DB2DART Processing completed with warning(s)!
                        Complete DB2DART report found in:
/home/db2inst1/sqllib/db2dump/DART0000/EOS.RPT

其中tsi��I�间id(2)�Q�oi��id(1560)�Q�ps为page number(0x0266807)�Q�需要�{换�ؓ(f��)十进�Ӟ��在结��ֿ��d��p�Q�np代表你要获取的页敎ͼ�v为是否详�l�输�?/p>

现在我们来看看EOS.RPT

______________________________________________________________________________

        _______                    DART                   _______ 

   D a t a b a s e   A n a l y s i s   a n d   R e p o r t i n g   T o o l

                           IBM    DB2    6000
______________________________________________________________________________

DART (V8.1.0)  Report:
2007-11-24-20.59.51.355893

            Database Name: EOS
            Report name: EOS.RPT
            Old report back-up: EOS.BAK
            Database Subdirectory: /opt/db2/db2inst1/NODE0000/SQL00001
            Operational Mode: Database Inspection Only (INSPECT)

______________________________________________________________________________
------------------------------------------------------------------------------


Action option: DD 
Table-object-ID: 1560; Tablespace-ID: 2; First-page: 157312p; Number-pages: 1; Verbose: y

Warning: The database state is not consistent.

Warning: Reorg rows MAY be due to the inconsistent state of the database.
Connecting to Buffer Pool Services...

   Table object report phase start.
   Dump format is verbose.

                           ______________________________________

         Page 0 of object 1560 from table space 2.

         BPS Page Header:

                     Page Data Offset = 48
                     Page Data Length = 4048
                             Page LSN = 0000 AE97 AE41
                   Object Page Number = 0
                     Pool Page Number = 157312
                            Object ID = 1560
                          Object Type = Data Object

               Data Page Header:

                           Slot Count = 8
                     Total Free Space = 2784
                  Total Reserve Space = 0
               Youngest Reserve Space = n/a
                         Youngest TID = n/a
                    Free Space Offset = 2799
                  Maximum Record Size = 23

               Data Records:


            Slot 0:

               Offset Location = 3996  (xF9C)
               Record Length = 32  (x20)

               Record Type = Data Object Header Control Record

                  Page count = 1
         Object Creation LSN = 0000 AE97 800C
                Object State = x0000
          UDI Since Runstats = 0
                  DART Field = x00000000

            Slot 1:

               Offset Location = 2992  (xBB0)
               Record Length = 1004  (x3EC)

               Record Type = Free Space Control Record

               Free space entries:
                 0:  2884 (x0B44),  4028 (x0FBC),  4028 (x0FBC),  4028 (x0FBC)
                 4:  4028 (x0FBC),  4028 (x0FBC),  4028 (x0FBC),  4028 (x0FBC)
                 8:  4028 (x0FBC),  4028 (x0FBC),  4028 (x0FBC),  4028 (x0FBC)
               省略。。�?
                  492:  4028 (x0FBC),  4028 (x0FBC),  4028 (x0FBC),  4028 (x0FBC)
                  496:  4028 (x0FBC),  4028 (x0FBC),  4028 (x0FBC),  4028 (x0FBC)

            Slot 2:

               Offset Location = 2916  (xB64)
               Record Length = 76  (x4C)

               Record Type = Table Directory Record

                  MetaIndex Root Page = 157377
                  Index Type = 2
                  Table Descriptor Pointer  --  Page 157312  Slot 3
                  Max Insert Search = 0
                  Flags = x02000200
                     bit representation = 00000010 00000000 00000010 00000000
                  Check pending info:
                     Constraint status    = x00
                     Constraint RID       = Page 0 Slot 0
                     last BID          = x00000000

            Slot 3:

               Offset Location = 2892  (xB4C)
               Record Length = 24  (x18)

               Record Type = Table Description Record

                  Number of Columns = 2


                  Column 1:
                  Type is Long Integer
                  Length = 4
                  Prohibits NULLs
                  Prohibits Default
            Fixed offset: 0

                  Column 2:
                  Type is Fixed Length Character String
                  Length = 10
                  Allows NULLs
                  Prohibits Default
            Fixed offset: 4

            Slot 4:

               Offset Location = 2869  (xB35)
               Record Length = 23  (x17)

               Record Type = Table Data Record (FIXEDVAR)

               Fixed part length value = 15

                  Column 1:
            Fixed offset: 0
                  Type is Long Integer
                  Value = 1

                  Column 2:
            Fixed offset: 4
                  Type is Fixed Length Character String
                      61202020 20202020 2020                 a               


            Slot 5:

               Offset Location = 2846  (xB1E)
               Record Length = 23  (x17)

               Record Type = Table Data Record (FIXEDVAR)

               Fixed part length value = 15

                  Column 1:
            Fixed offset: 0
                  Type is Long Integer
                  Value = 2

                  Column 2:
            Fixed offset: 4
                  Type is Fixed Length Character String
                      61202020 20202020 2020                 a               


            Slot 6:

               Offset Location = 2823  (xB07)
               Record Length = 23  (x17)

               Record Type = Table Data Record (FIXEDVAR)

               Fixed part length value = 15

                  Column 1:
            Fixed offset: 0
                  Type is Long Integer
                  Value = 3

                  Column 2:
            Fixed offset: 4
                  Type is Fixed Length Character String
                      61202020 20202020 2020                 a               


            Slot 7:

               Offset Location = 2800  (xAF0)
               Record Length = 23  (x17)

               Record Type = Table Data Record (FIXEDVAR)

               Fixed part length value = 15

                  Column 1:
            Fixed offset: 0
                  Type is Long Integer
                  Value = 4

                  Column 2:
            Fixed offset: 4
                  Type is Fixed Length Character String
                      61202020 20202020 2020                 a               


         Slots Summary:  Total=8,  In-use=8,  Deleted=0.

      
   Table object report phase end.
                     ______________________________________

                  DB2DART Processing completed with warning(s)!
                     Warning(s) detected during processing.
                     ______________________________________

                        Complete DB2DART report found in:
/home/db2inst1/sqllib/db2dump/DART0000/EOS.RPT
    _______    D A R T    P R O C E S S I N G    C O M P L E T E    _______

扑ֈ�Solt 7 (0x07)�Q�ok�Q�你现在可以清楚的知道应用等待的Row�?4,a)

�ȝ��

通过上面的方法，我们��单描�q�C��一个db2锁问题的定位�Ҏ(gu��)��Q�希望能�l�大家在分析和定位应用性能问题的时候�v��C��定的帮助

del.icio.us Tags: db2,lock,tuning,diagnostic

tacy lee 2007-11-24 21:18 发表评论

深入理解Loadrunner中的Browser Emulation

tacy lee — Mon, 05 Nov 2007 16:19:00 GMT

作者：(x��)tacy lee

一�Q�基本介�l?/strong>

在Loadrunner的��用中�Q�对于Run-time Settings下的browser emulation讄��是比较容易让��Z�生困惑的地方。下面我们结合sniffer来具体看看每个选项的用途，以及(qi��ng)�Ҏ(gu��)��试的影响�?

Browser Emulation �?/p>
二：(x��)案例和工�?/strong>

1. ��试案例�Q?/b>
打开�|�站首页两次�Q�对比不同Browser Emulation讄��下loadrunner的行为，脚本如下�?
Action() { web_url("www.primeton.com", "URL=http://www.primeton.com/", "Resource=0", "RecContentType=text/html", "Referer=", "Snapshot=t2.inf", "Mode=HTML", LAST); web_url("www.primeton.com", "URL=http://www.primeton.com/", "Resource=0", "RecContentType=text/html", "Referer=", "Snapshot=t2.inf", "Mode=HTML", LAST); return 0; }

2. sniffer工具
开源工��P��(x��)Wireshark(前��n是ethereal)�Q�www.wireshark.org�Q?
三：(x��)��试�q�程

��Z��方便描述�Q�我们约定用�Q?
A代表Simulate
B代表Cache
C代表Check
D代表Download
E代表Simulate
F代表Clear
首先讄��Run
1. ��L��所有选项
�l�果�Q�共获取数据
1 0.000000 2 0.036053 203.81.29.137 92 1.415887 94 1.449960 203.81.29.137

在这�U�情况下�Q�数
�q�时�Q�你即��选
2. 选择E(F)
�l�果�Q�共获取数据
1 0.000000 2 0.037013 203.81.29.137 51 0.651919 203.81.29.137 53 0.676377 203.81.29.137 101 1.347949 203.81.29.137

在这�U�情况下�Q�数
3. 选择DE(F)
�l�果�Q�共获取数据
1 0.000000 2 0.037911 203.81.29.137 9 0.141816 203.81.29.137 426 3.334889 428 3.372253 203.81.29.137 448 4.395488 457 4.439604 203.81.29.137 859 7.593610 870 7.659680 203.81.29.137 888 8.511308 890 8.549451 203.81.29.137 892 8.566246 893 8.601893 203.81.29.137 899 8.702628 904 8.741807 203.81.29.137 1298 11.809456 1310 11.878665 203.81.29.137 1341 12.771707 1348 12.813950 203.81.29.137 1759 16.032952 1761 16.068296 203.81.29.137 1779 16.983042 1781 17.016836 203.81.29.137

在这�U�情况下�Q�数
4. 选择ADE
�l�果�Q�共获取数据

在这�U�情况下�Q�由
6. 关于BC选项
C的解释（Check
C比较�Ҏ(gu��)��理
B的解释（Cache
LR对于资源文�g的cache�
当然如果你想把GIF也cac
四：(x��) �l�论 browser cache URLs requiring content(HTMLs) for newer versions of stored pages every visit to the page non-HTML resources a new user on each iteratioin cache on each iteration Logic中的iteration�?。让Action�q�行两次�Q�看看��@环运行脚本两�ơ，数据包和�q�接数的变化�? �?5个，建立�q�接1个（�U�色标识�Q�，断开�q�接1个（蓝色标识�Q? rpcode-wrapper"> gray 1px solid; padding-right: 4px; border-top: gray 1px solid; padding-left: 4px; font-size: 8pt; padding-bottom: 4px; margin: 20px 0px 10px; overflow: auto; border-left: gray 1px solid; width: 97.5%; cursor: text; line-height: 12pt; padding-top: 4px; border-bottom: gray 1px solid; font-family: consolas, 'Courier New', courier, monospace; background-color: #f4f4f4; max-height: 200px">
No. Time Source Destination Protocol Info 192.168.1.61 203.81.29.137 TCP 13835 > http [SYN] Seq=0 Len=0 MSS=1460 WS=2 192.168.1.61 TCP http > 13835 [SYN, ACK] Seq=0 Ack=1 Win=17280 Len=0 MSS=1440 WS=0 192.168.1.61 203.81.29.137 TCP 13835 > http [FIN, ACK] Seq=817 Ack=71762 Win=257760 Len=0 192.168.1.61 TCP http > 13835 [FIN, ACK] Seq=71762 Ack=818 Win=16464 Len=0 据包非常��（没有选择下蝲资源文�g入css,js,gif�{�）�Q�而且你可以看刎ͼ�打开4�ơ首��，只徏立了一个tcp�q�接�? 择A�Q�发现数据包的数量量��|��有变化，因�ؓ(f��)cache主要�q�是针对资源文�g �?02个，建立�q�接2个（�U�色标识�Q�，断开�q�接2个（蓝色标识�Q?/p> gray 1px solid; padding-right: 4px; border-top: gray 1px solid; padding-left: 4px; font-size: 8pt; padding-bottom: 4px; margin: 20px 0px 10px; overflow: auto; border-left: gray 1px solid; width: 97.5%; cursor: text; line-height: 12pt; padding-top: 4px; border-bottom: gray 1px solid; font-family: consolas, 'Courier New', courier, monospace; background-color: #f4f4f4; max-height: 200px">
No. Time Source Destination Protocol Info 192.168.1.61 203.81.29.137 TCP 13886 > http [SYN] Seq=0 Len=0 MSS=1460 WS=2 192.168.1.61 TCP http > 13886 [SYN, ACK] Seq=0 Ack=1 Win=17280 Len=0 MSS=1440 WS=0 48 0.618117 192.168.1.61 203.81.29.137 TCP 13886 > http [FIN, ACK] Seq=409 Ack=35882 Win=257760 Len=0 49 0.644106 192.168.1.61 203.81.29.137 TCP 13887 > http [SYN] Seq=0 Len=0 MSS=1460 WS=2 192.168.1.61 TCP http > 13886 [FIN, ACK] Seq=35882 Ack=410 Win=16872 Len=0 192.168.1.61 TCP http > 13887 [SYN, ACK] Seq=0 Ack=1 Win=17280 Len=0 MSS=1440 WS=0 99 1.310379 192.168.1.61 203.81.29.137 TCP 13887 > http [FIN, ACK] Seq=409 Ack=35882 Win=257760 Len=0 192.168.1.61 TCP http > 13887 [FIN, ACK] Seq=35882 Ack=410 Win=16872 Len=0 据包非常��（没有选择下蝲资源文�g入css,js,gif�{�）�Q�对比第一�U�情况，你会(x��)发现它徏立了两个�q�接�Q�这��是E的作用，它对于每�ơ�P代都当成一个新的用��P��需要重新徏立连接�? �?782个，建立�q�接6个（�U�色标识�Q�，断开�q�接6个（蓝色标识�Q?/p> gray 1px solid; padding-right: 4px; border-top: gray 1px solid; padding-left: 4px; font-size: 8pt; padding-bottom: 4px; margin: 20px 0px 10px; overflow: auto; border-left: gray 1px solid; width: 97.5%; cursor: text; line-height: 12pt; padding-top: 4px; border-bottom: gray 1px solid; font-family: consolas, 'Courier New', courier, monospace; background-color: #f4f4f4; max-height: 200px">
No. Time Source Destination Protocol Info 192.168.1.61 203.81.29.137 TCP 14016 > http [SYN] Seq=0 Len=0 MSS=1460 WS=2 192.168.1.61 TCP http > 14016 [SYN, ACK] Seq=0 Ack=1 Win=17280 Len=0 MSS=1440 WS=0 6 0.107432 192.168.1.61 203.81.29.137 TCP 14017 > http [SYN] Seq=0 Len=0 MSS=1460 WS=2 192.168.1.61 TCP http > 14017 [SYN, ACK] Seq=0 Ack=1 Win=17280 Len=0 MSS=1440 WS=0 192.168.1.61 203.81.29.137 TCP 14017 > http [FIN, ACK] Seq=1852 Ack=150284 Win=257484 Len=0 192.168.1.61 TCP http > 14017 [FIN, ACK] Seq=150284 Ack=1853 Win=16998 Len=0 192.168.1.61 203.81.29.137 TCP 14020 > http [SYN] Seq=0 Len=0 MSS=1460 WS=2 192.168.1.61 TCP http > 14020 [SYN, ACK] Seq=0 Ack=1 Win=17280 Len=0 MSS=1440 WS=0 192.168.1.61 203.81.29.137 TCP 14016 > http [FIN, ACK] Seq=2849 Ack=377404 Win=257484 Len=0 192.168.1.61 TCP http > 14016 [FIN, ACK] Seq=377404 Ack=2850 Win=15935 Len=0 192.168.1.61 203.81.29.137 TCP 14020 > http [FIN, ACK] Seq=1602 Ack=208150 Win=257760 Len=0 192.168.1.61 TCP http > 14020 [FIN, ACK] Seq=208150 Ack=1603 Win=17280 Len=0 192.168.1.61 203.81.29.137 TCP 14022 > http [SYN] Seq=0 Len=0 MSS=1460 WS=2 192.168.1.61 TCP http > 14022 [SYN, ACK] Seq=0 Ack=1 Win=17280 Len=0 MSS=1440 WS=0 192.168.1.61 203.81.29.137 TCP 14023 > http [SYN] Seq=0 Len=0 MSS=1460 WS=2 192.168.1.61 TCP http > 14023 [SYN, ACK] Seq=0 Ack=1 Win=17280 Len=0 MSS=1440 WS=0 192.168.1.61 203.81.29.137 TCP 14022 > http [FIN, ACK] Seq=1550 Ack=159770 Win=257484 Len=0 192.168.1.61 TCP http > 14022 [FIN, ACK] Seq=159770 Ack=1551 Win=17280 Len=0 192.168.1.61 203.81.29.137 TCP 14026 > http [SYN] Seq=0 Len=0 MSS=1460 WS=2 192.168.1.61 TCP http > 14026 [SYN, ACK] Seq=0 Ack=1 Win=17280 Len=0 MSS=1440 WS=0 192.168.1.61 203.81.29.137 TCP 14023 > http [FIN, ACK] Seq=3151 Ack=367918 Win=257484 Len=0 192.168.1.61 TCP http > 14023 [FIN, ACK] Seq=367918 Ack=3152 Win=17280 Len=0 192.168.1.61 203.81.29.137 TCP 14026 > http [FIN, ACK] Seq=1602 Ack=208150 Win=257760 Len=0 192.168.1.61 TCP http > 14026 [FIN, ACK] Seq=208150 Ack=1603 Win=17280 Len=0 据包的数量非常大�Q�连接也很多�Q�由于没有cache功能�Q�每�ơ打开��面都需要重��C��载所有的资源文�g�? �?25个，建立�q�接3个，断开�q�接3�?不再标识了，syn即�ؓ(f��)�q�接��h��Q�fin即�ؓ(f��)断开��h��Q?/p> gray 1px solid; padding-right: 4px; border-top: gray 1px solid; padding-left: 4px; font-size: 8pt; padding-bottom: 4px; margin: 20px 0px 10px; overflow: auto; border-left: gray 1px solid; width: 97.5%; cursor: text; line-height: 12pt; padding-top: 4px; border-bottom: gray 1px solid; font-family: consolas, 'Courier New', courier, monospace; background-color: #f4f4f4; max-height: 200px">
No. Time Source Destination Protocol Info 203.81.29.137 TCP 14189 > http [SYN] Seq=0 Len=0 MSS=1460 WS=2 192.168.1.61 TCP http > 14189 [SYN, ACK] Seq=0 Ack=1 Win=17280 Len=0 MSS=1440 WS=0 203.81.29.137 TCP 14190 > http [SYN] Seq=0 Len=0 MSS=1460 WS=2 192.168.1.61 TCP http > 14190 [SYN, ACK] Seq=0 Ack=1 Win=17280 Len=0 MSS=1440 WS=0 203.81.29.137 TCP 14190 > http [FIN, ACK] Seq=1852 Ack=150284 Win=257484 Len=0 192.168.1.61 TCP http > 14190 [FIN, ACK] Seq=150284 Ack=1853 Win=16998 Len=0 203.81.29.137 TCP 14189 > http [FIN, ACK] Seq=1504 Ack=235576 Win=257760 Len=0 203.81.29.137 TCP 14192 > http [SYN] Seq=0 Len=0 MSS=1460 WS=2 192.168.1.61 TCP http > 14189 [FIN, ACK] Seq=235576 Ack=1505 Win=17280 Len=0 192.168.1.61 TCP http > 14192 [SYN, ACK] Seq=0 Ack=1 Win=17280 Len=0 MSS=1440 WS=0 203.81.29.137 TCP 14192 > http [FIN, ACK] Seq=409 Ack=35882 Win=257760 Len=0 192.168.1.61 TCP http > 14192 [FIN, ACK] Seq=35882 Ack=410 Win=16872 Len=0 ache发挥作用�Q�数据包�Ҏ(gu��)��W�三�U�情况大大减��，几乎�{�于打开一�ơ首��늚�数据量（449个数据包�Q�，只有�W�一�ơ打开��面需要完整下载页面（包括资源文�g�Q�，后面的三�ơ打开��面都只要下载HTML��面�Q�不包括资源文�g�Q��? 果�Q�共获取数据�?42个，建立�q�接4个，断开�q�接4�?/p> gray 1px solid; padding-right: 4px; border-top: gray 1px solid; padding-left: 4px; font-size: 8pt; padding-bottom: 4px; margin: 20px 0px 10px; overflow: auto; border-left: gray 1px solid; width: 97.5%; cursor: text; line-height: 12pt; padding-top: 4px; border-bottom: gray 1px solid; font-family: consolas, 'Courier New', courier, monospace; background-color: #f4f4f4; max-height: 200px">
No. Time Source Destination Protocol Info 203.81.29.137 TCP 14292 > http [SYN] Seq=0 Len=0 MSS=1460 WS=2 192.168.1.61 TCP http > 14292 [SYN, ACK] Seq=0 Ack=1 Win=17280 Len=0 MSS=1440 WS=0 203.81.29.137 TCP 14294 > http [SYN] Seq=0 Len=0 MSS=1460 WS=2 192.168.1.61 TCP http > 14294 [SYN, ACK] Seq=0 Ack=1 Win=17280 Len=0 MSS=1440 WS=0 203.81.29.137 TCP 14294 > http [FIN, ACK] Seq=1852 Ack=150284 Win=257484 Len=0 192.168.1.61 TCP http > 14294 [FIN, ACK] Seq=150284 Ack=1853 Win=16998 Len=0 203.81.29.137 TCP 14292 > http [FIN, ACK] Seq=1504 Ack=235576 Win=257760 Len=0 203.81.29.137 TCP 14297 > http [SYN] Seq=0 Len=0 MSS=1460 WS=2 192.168.1.61 TCP http > 14292 [FIN, ACK] Seq=235576 Ack=1505 Win=17280 Len=0 192.168.1.61 TCP http > 14297 [SYN, ACK] Seq=0 Ack=1 Win=17280 Len=0 MSS=1440 WS=0 203.81.29.137 TCP 14298 > http [SYN] Seq=0 Len=0 MSS=1460 WS=2 192.168.1.61 TCP http > 14298 [SYN, ACK] Seq=0 Ack=1 Win=17280 Len=0 MSS=1440 WS=0 203.81.29.137 TCP 14297 > http [FIN, ACK] Seq=1550 Ack=159770 Win=257484 Len=0 192.168.1.61 TCP http > 14297 [FIN, ACK] Seq=159770 Ack=1551 Win=17280 Len=0 203.81.29.137 TCP 14298 > http [FIN, ACK] Seq=1806 Ack=226090 Win=257760 Len=0 192.168.1.61 TCP http > 14298 [FIN, ACK] Seq=226090 Ack=1807 Win=17076 Len=0 于选择了F�Q�在�q�代的时候清除了cache�Q�所以每�ơ�P代都需要重��C��载资源文件。数据包差不多等于第三种情况的一半，�U�等于打开两次首页的数据量�Q?49×2个数据包�Q��? for newer versions of stored pages every visit to the page�Q?/i> 解�Q�类似IE讄��中的每次��查，如果不设�|�C�Q�LR对于已经cache的文件就不会(x��)重新向服务器��h��Q�如果选择C�Q�你��可以在数据包中发现很多304信息�? URLs requiring content(HTMLs)�Q?/i> q�不�?x��)真正cache在内存中或者在��盘上，�q�个选项表示�Q�对于一些需要用到的兌��Q�校验，��面解析内容真正cache在内存中�Q�减��客��L(f��ng)��的重复工作�? he到内存中�Q�你可以在Advanced中设�|�，选择Specify URL requiring content in addition to HTML pages�Q�加入条目image/gif�Q��ƈ��N��。当Vuser�q�行的时候，你可以对比一下mmdrv.exe�q�程的内存消耗（内存占用�?x��)更多）�?

通过上面的测试分析，我们大概知道了每个选项的真正含义，你需要根据你的测试目的来选择合适的讄��Q?
1�?对于一个具体的应用��试�Q�对于前端Web Server不可忽略�Q�缺省设�|�非常合适，不需要调��_(d��)��有时候需要考虑把C选上�Q?
注意�Q�很多�h在录制脚本的时候，�?f��n)惯把登入操作放到vuser_init中，�q�时候缺省设�|�可能会(x��)抛错�Q�徏议把�q�类的操作都攑օ�到action�?
2�?如果你更��x��后端应用服务器的性能或者说做一些架构的验证分析�Q�那你缺省设�|�对于你来说��׃��合适了�Q�你需要选择取消所有的讄��V�?
当然你也可以�Ҏ(gu��)��自己的具体情况做不同调整�Q�但是一定要真正理解�q�些选项的具体含义才能做��C��犯错�?/p>
del.icio.us Tags: performance , testing , loadrunner

tacy lee 2007-11-06 00:19 发表评论

tacy lee — Tue, 30 Oct 2007 05:09:00 GMT

故障现象�Q?/p>
��试一�D�|��间后应用无响应，�q�接池不能放大，jvm crash�Q�日志报对象分配��p�|

问题诊断�Q?/p>
�W�一个阶�D�|��websphere问题

到现��Z��后，回放脚本��试几分钟，应用开始无法响应，后台也没有异常，update jdk之后�Q�系�l�能正常响应了，但是发现新的问题�Q�db2�q�接池始�l�无法放大，最大只能到30�Q�而且�pȝ��也会(x��)抛OOM�Q�导致系�l�异常推出，从系�l�日志看�Q�是因�ؓ(f��)应用中的大对象分配导致的�Q?M大小�Q?/p>
期间�Q�关于连接池无法攑֤�问题想了很多办法�Q�包括修改db2 maxappls,maxagents�q�些参数�Q�更新数据库驱动�Q�而且��定不是db2的问题（在创�?0之后�Q�我们依然可以通过其他方式�q�接到db2�Q�说明db2的连接限制确实放大了�Q�，当然我们productdatasource�q�个池子大小我已�l�放大到100了�?/p>
中间�q�发现测试脚本没有正常启动流�E�，排查后发现是loadrunner的问题，用我机器上的lr录制正常�Q�错误代码提�C�是字段长度限制�Q�莫名其妙）�?/p>
关于jvm crach我们也调整了heap讄��Q?xms256m,-xmx1536m

但是问题依然存在。后面我们重新安装了应用�Q�所有的讄��采用�~�省配置�Q�没有打��M��补丁�Q�系�l�这个时候竟然可以正常跑了，只是响应很慢�Q�而且旉��曲线一直往上抛�Q�测试一�D�|��间系�l�无响应�Q?/p>
查了一下配�|�，发现productdatasource的缺省设�|�竟然就�?0�Q?strong>�q�个时候基本判断是之前的websphere的设�|�修�Ҏ(gu��)��有生�?/strong>

重新修改jvm和连接池配置�Q�这时候系�l�正常，数据�q�接也达�?3个，然后开始测试大�q�发�?/p>

阶段二就是调整数据库配置

1、第一个是db2 default buffer pool�Q�缺省配�|�buffer�?m�Q�这个一定要注意修改

2、第二个是db2的lock数量�Q�在�~�省基础上好像放大了100�?/p>
3、sort heap�Q�排序区�Q�防止排序溢出）

�q�些调整都要通过db2的状态来调整�Q�可以通过get snapshot指��o(h��)来获得数据库状态，buffer不够�?x��)出现大量的逻辑读，lock不够�?x��)抛lock溢出(�?x��)导致锁升��Q�，sort heap不够�?x��)提�C�排序溢出（�q�时候排序会(x��)在硬盘做�Q?/p>

回头看看�q�次支持�Q?/p>
1、websphere配置修改不生效，我后面仔�l�想了想�Q�这个websphere很多公司用可能大安��乱改了一通，另一问题是我们的使用�?f��n)惯�Q�websphere强烈不徏议用kill直接杀�q�程方式停服务器�Q�websphere不但是一个java�q�程�Q�还有很多的附属�q�程�Q�直接kill也很�Ҏ(gu��)��D��websphere不正�?/p>
2、jvm crach问题�Q�这个我��大家看看�q�篇文档http://www-1.ibm.com/support/docview.wss?rs=180&context=SSEQTP&q1=fragmentation

&uid=swg21176363&loc=en_US&cs=utf-8&lang=en

如何��d��位jvm问题�Q�首先看nativerr.log日志�Q�如果出现OOM�Q�这里会(x��)有记录，当发现OOM的时候，可以打开jvm的verbosegc�Q�分析verbosegc和jvm dump file�Q�上面文档里提到一个很重要的东西就是pinned对象�Q�这也是ibm为啥不徏议设�|�ms=mx的原因�?/p>

tacy lee 2007-10-30 13:09 发表评论

tacy lee — Tue, 30 Oct 2007 04:59:00 GMT

先徏立一个监控器
db2 "create event monitor SQLCOST for statements write to file '/home/db2inst1'"
再设�|�事务状态�ؓ(f��)打开
db2 "set event monitor SQLCOST state=1"
注：(x��)1为打开�Q?为关闭，攉��数据之后�Q�记得关闭你的监控器�Q�否则。。�?/font>
跑你的测试后�Q�在你的/home/db2inst1目录下会(x��)生成一些evm文�g
用下面指令获取诊断信息：(x��)
db2evmon -db eos51 -evm SQLCOST>sqlcost1.txt
完成之后删除你的监控�?
db2 "drop event monitor SQLCOST"
生成的采样例子，从下面的例子中，你可以清除的看到SQL执行的时��_(d��)��CPU消耗情况，排序是否溢出�Q�BufferPool的��用情况，�Ҏ(gu��)��q�些信息�Q�SQL的执行效率一目了�Ӟ��(x��)
26) Statement Event ...
Appl Handle: 336
Appl Id: C0A80421.O905.0ABDA5065446
Appl Seq number: 0657
Record is the result of a flush: FALSE
-------------------------------------------
Type : Dynamic
Operation: Execute
Section : 7
Creator : NULLID
Package : SYSSN300
Consistency Token : SYSLVL01
Package Version ID :
Cursor : SQL_CURSN300C7
Cursor was blocking: FALSE
Text : update WFProcessInst set relateData=? where processInstID= ?
-------------------------------------------
Start Time: 04/25/2007 14:57:19.402248
Stop Time: 04/25/2007 14:57:19.409622
Exec Time: 0.007374 seconds
Number of Agents created: 1
User CPU: 0.000000 seconds
System CPU: 0.000000 seconds    [licl1]
Fetch Count: 0
Sorts: 0
Total sort time: 0
Sort overflows: 0    [licl2]
Rows read: 1
Rows written: 1
Internal rows deleted: 0
Internal rows updated: 0
Internal rows inserted: 0
Bufferpool data logical reads: 9
Bufferpool data physical reads: 0
Bufferpool temporary data logical reads: 0
Bufferpool temporary data physical reads: 0
Bufferpool index logical reads: 3
Bufferpool index physical reads: 0
Bufferpool temporary index logical reads: 0
Bufferpool temporary index physical reads: 0    [licl3]
SQLCA:
sqlcode: 0
sqlstate: 00000

[licl1]SQL执行旉��和CPU消耗情�?

[licl2]SQL的排序情况，可以看到�q�个SQL没有排序�Q�当然也没有排序溢出

[licl3]Bufferpool的��用情况，逻辑��d��物理�ȝ��Ҏ(gu��)��

del.icio.us Tags: database , db2 , tuning

tacy lee 2007-10-30 12:59 发表评论

Loadrunner Analysis之Web Page Diagnostics

tacy lee — Tue, 23 Oct 2007 11:04:00 GMT

作者：(x��)tacy lee

��单介�l�一下Loadrunner Analysis中的Web Page Diagnostics模块的��用，很多人对于测试之后的�l�果数据分析�怸�着头脑�Q�其实loadrunner Analysis�l�你提供了很好的文档�Q�大家没事可以多��ȝ��Q�多��d��遍对于性能��试你就入门�?�Q�）

Web Page Diagnostics �Q�以下简�U�WPD�Q�，�q�是LR Analysis中非帔R��要的一块，搞清楚这部分的内容会(x��)让你��走很多弯�\�Q�很多环境问题都可以通过它来定位�Q�比如客��L(f��ng)��Q�网�l�。通过它可以你可以比较好的来定位是环境的问题还是应用本�w�的问题�Q�当然更重要的是Web��面本��n的问题�?/p>
WPD包括下面几个图表�Q?/p>
Web Page Diagnostics     �q�是张��d��Q�包括下面几张Over Time囄��内容

Page Component Breakdown     ��面中每个元素的�q�_��响应旉��占整个页面响应时间的癑ֈ��?/p>
Page Component Breakdown(Over Time)     在整个测试过�E�中�Q��Q意一�U�内��面中每个元素的响应旉��Q�例如在runtime中设�|�了browser cache�Q�页面中的资源文件就只会(x��)在第一�ơ下载，后面的页面响应时间也��׃��包括�q�些元素的时��_(d��)��q�在Page Component Breakdown中是看不出来的，因�ؓ(f��)Page Component Breakdown是整个测试期间内的��^均时间。当�Ӟ��是否启用了cache�Q�通过over time囑ְ�能看出来�Q?/p>
Page Download Time Breakdown    ��面中每个元素的响应旉��分割图，响应旉��被分割�ؓ(f��)以下几个部分�Q�DNS Resolution,Connection,First Buffer,SSL Handshaking,Receive,FTP Authentication,Client,Error

Page Download Time Breakdown(Over Time)      在整个测试过�E�中�Q��Q意一�U�内��面中每个元素的响应旉��分割�?/p>
Time to First Buffer Breakdown      First Buffer Time旉��分割为Network Time和Server Time�Q�客��L(f��ng)��http��h��发送到接收到服务器端的应答包（ACK�Q��ؓ(f��)Network Time�Q�从接收到ACK到完成First Buffer接受为Server Time

Time to First Buffer Breakdown(Over Time)      基本同上�Q��Q意一�U�内�?/p>
Downloaded Component Size(KB)      ��面中每个元素的大小�Q�KB�Q?/p>
介绍了这么多�Q�具体如何分析呢�Q?/p>
首先打开Web Page Diagnostics图，来看看下面一个例子Download Time图：(x��)

上图存在两个问题�Q?/p>
1、receive旉��很长

�q�个一般是�|�络问题�Q�当然如果你��认�|�络不存在问题，那么你就要看看是不是客户端的问题�Q�客��L(f��ng)��也可能会(x��)造成Receive�q�长�Q�这个千万要注意�Q?/p>
2、页面问�?/p>
��面上包括了非常多的囄��Q�而且囄��g��都没有优化，最大的竟然�?63K�Q�记下来�Q�这可是�|�证�?�Q�）

很多时候，你可以根据DNS,Connection,Receive来看出是否存在网�l�问题，�Ҏ(gu��)��Client来判断是否存在客��L(f��ng)��问题�?/p>
看看�Q�挺��单的吧！ ^_^

换个囄��看，Page Component Breakdown(Over Time)

很清楚吧�Q�页面元素都被cache了，说明场景启用了browser cache�Q�页面的响应旉��只包括红�U�和蓝线�?/p>
Time to First Buffer Breakdown(Over Time) �Q�图��׃��贴了�Q�这个图非常重要�Q�也最复杂�Q�这里的��g��l�对�Q�当�|�络状况不好的时候，server time很可能包括网�l�时��_(d��)��因�ؓ(f��)很多��面元素比较?y��u)��（��?k的样子）�Q�在First Buffer��完成传输，所以一定要注意分析�?/p>
��唠叨到�q�里吧，�Ƣ迎拍砖

del.icio.us Tags: loadrunner , performance , testing

tacy lee 2007-10-23 19:04 发表评论

tacy lee — Fri, 19 Oct 2007 08:22:00 GMT

作者：(x��)tacy lee
weblogic.xml

servlet-reload-check-secs
The element defines whether a WebLogic Server will check to see if a servlet has been modified, and if it has been modified, reloads it.

The value -1 means never check the servlets. This is the default value in a production environment.

The value 0 means always check the servlets.

The value 1 means check the servlets every second. This is the default value in a development environment.

A value specified in the console will always take precedence over a manually specified value.
resource-reload-check-secs
The element is used to perform metadata caching for cached resources that are found in the resource path in the Web application scope. This parameter identifies how often WebLogic Server checks whether a resource has been modified and if so, it reloads it.

The value -1 means metadata is cached but never checked against the disk for changes. In a production environment, this value is recommended for better performance.

The value 0 indicates not to do any metadata caching. Customers who keep changing their files must set this parameter to a value greater than or equal to 0.

The value 1 means reload every second. This is the default value in a development environment.

Values specified for this parameter using the Admin Console are given precedence.
native-io-enabled
To use native I/O while serving static files with weblogic.servlet.FileServlet, which is implicitly registered as the default servlet, set native-io-enabled to true. (The default value is false.) native-io-enabled element applies only on Windows.

page-check-seconds
Sets the interval, in seconds, at which WebLogic Server checks to see if JSP files have changed and need recompiling. Dependencies are also checked and recursively reloaded if changed.

The value -1 means never check the pages. This is the default value in a production environment.

The value 0 means always check the pages.

The value 1 means check the pages every second. This is the default value in a development environment.

In a production environment where changes to a JSP are rare, consider changing the value of pageCheckSeconds to 60 or greater, according to your tuning requirements.
JDBC

讄��Initial Capacity�{�于Maximum Capacity

讄��Statement cache�Q�注意，对于每个打开的statement�Q�DBMS都会(x��)�l�护一个cursor�Q�这个��D��|�过大会(x��)��D�� java.sql.SQLException: ORA-01000: maximum open cursors exceeded�c�M��的错误。当�Ӟ��你要清楚�Q�statement cache的大��是指每个连接能cache的statement敎ͼ�例如你设�|�connection pool size = 100 ,讄��Statement Cache = 10�Q�那�pȝ��最大维持的cursor�?00*10�Q?/p>

Network connection

Enable Native IO �Q�注意，不是java的NIO�Q�采用Java muxer方式处理�q�接�Q�对于大�q�发的系�l�媄响巨大，java需要�ؓ(f��)每个�q�接��h��起一个线�E�来处理�Q?/p>

修改Accept Backlog�Q�当应用服务器出现拒�l�连接的时�?/p>

启动脚本

使用productmode启动weblogic

讄��-xms�{�于-xmx

��量使用jrockit

work manager

�?版本以后�Q�weblogic用work manager取代了thread queue�Q�默认情况下�Q�weblogic有一个default work manager�Q�采用fair share方式�q�_��׃�n�U�程

一般你不需要自己创建work manager�Q�除非你有如下需求：(x��)

你的应用有优先��

你需要满��SLA定义的响应时�?/p>

需要指定最��线�E�约束来避免服务器死�?/p>

del.icio.us Tags: weblogic , tuning , tips

tacy lee 2007-10-19 16:22 发表评论

��单的java io��试

tacy lee — Thu, 20 Sep 2007 06:27:00 GMT
     摘要: 作者：(x��)tacy.lee

一个简单的java io��试�Q�不同的实现�Ҏ(gu��)��看一个zip包的解压速度  阅读全文

tacy lee 2007-09-20 14:27 发表评论

Ajax profiler tools

tacy lee — Tue, 18 Sep 2007 15:08:00 GMT

一个简单的ajax性能分析工具�Q�比较实用�?/p>
1、下�?a >Ajax View�q�安装运行，它会(x��)监听�?888端口

2、打开IE或者Firefox�Q�设�|�代理：(x��)localhost:8888

3、打开你的ajax��面�Q�执�?/p>
4、另外开一个窗口，打开http://fakeurl.com/?&AJAXVIEWREQUEST=GET=main.html

5、选择左边的链接JS Performance Statistics�Q�你��可以看到具体的执行旉��

延��阅读�Q?a title="http://www.fasterj.com/articles/javascript.shtml" >Javascript Performance

del.icio.us Tags: ajax , profiler , performance , java

tacy lee 2007-09-18 23:08 发表评论

Http协议的几个概念和web browser优化

tacy lee — Fri, 07 Sep 2007 09:28:00 GMT
     摘要: 作者：(x��)tacy lee

写本文的目的��是��x��清除http1.1�?.0的具体区别，看过很多别�h写的关于他们的概念，可能是没动手��L��试，��L��模模�p�糊  阅读全文

tacy lee 2007-09-07 17:28 发表评论

亚洲精品国产综合久久一线,国产亚洲综合一区二区三区,亚洲国产精品日韩在线观看

oracle 的lob & long

ibm jdk 1.5�~�省用的gc�{�略性能很差

�q�发是啥

工作日志-OOM事�g

关于oracle中的timestamp和date�c�d��

Websphere到底是否需要配�|�IHS

db2诊断�p�d���?--捕获sql执行情况

深入理解Loadrunner中的Browser Emulation

作者：(x��)tacy lee

Loadrunner Analysis之Web Page Diagnostics

weblogic.xml

JDBC

Network connection

启动脚本

work manager

���单的java io���试

Ajax profiler tools

Http协议的几个概念和web browser优化

db2诊断�p�d��?--捕获sql执行情况

��单的java io��试