??xml version="1.0" encoding="utf-8" standalone="yes"?>
]]>System.out.println("-------------2----------"+uuid);
if(flag == 1){
String url_success="wangdian_extend_update.jsp?dwdm="+dwdm+"&wxyid="+wxyid+"&uuid="+uuid;
out.println("<script>window.setTimeout(\"window.location='"+url_success+"'\",10)</script>");
}else{
out.println("dp|!");
String url_fail="wangdian_extend.jsp?dwdm="+dwdm+"&wxyid="+wxyid;
out.println("<script>window.setTimeout(\"window.location='"+url_fail+"'\",10)</script>");
}
]]><table align="center" width="640">
<tr >
<td>
对应|点:
<select name="wangdian" id="wangdian" style="width:80pt;" onchange="window.location='/safety-wxy/ent/wangdian_extend_update.jsp?dwdm=<%=dwdm%>&wxyid=<%=wxyid%>&uuid='+this.value;">
<option value="0"></option>
<%
String oid = "";
String wdname = "";
Connection connection;
try
{
connection = DataFactory.getConnection("wxyinfor");
}
catch(Exception e)
{
throw new SQLException(e.toString());
}
Statement statement=null;
ResultSet resultset=null;
//System.out.println("--------"+uuid);
String sql = "select oid,wdname,wdaddress,wdfzr,wdlxdh from T_xzwdext where dwdm = '"+dwdm+"' and wxyid = '"+wxyid+"'";
System.out.println("----test sql ---"+sql);
try
{
statement = connection.createStatement();
//resultset = statement.executeUpdate(sql);
resultset=statement.executeQuery(sql);
while (resultset.next())
{
//oid = resultset.getString("oid");
%>
<option value="<%=resultset.getString("oid") %>"><%=resultset.getString("wdname") %></option>
<%
//wdname = resultset.getString("wdname");
}
}
catch(Exception e)
{
System.out.println(e.getMessage());
}
finally
{
try
{
if(null!=resultset)
{
resultset.close();
resultset = null;
}
}catch(SQLException e1){}
try
{
if(null!=statement)
{
statement.close();
statement = null;
}
}
catch(SQLException e2){}
try
{
if(null!=connection)
{
connection.close();
connection = null;
}
}
catch(SQLException e3){}
}
%>
<%
System.out.println(usertype);
if(usertype!=null & !"adm".equals(usertype)){
%>
<input name="wadd" type="button" id="wadd"
value="d|点" onclick="window.location='/safety-wxy/ent/wangdian_extend.jsp?dwdm=<%=dwdm%>&wxyid=<%=wxyid%>'">
<%
}
%>
</td>
</tr>
</table>
]]><select name="wangdian" id="wangdian" style="width:80pt;" onchange="window.location='/safety-wxy/ent/wangdian_extend_update.jsp?dwdm=<%=dwdm%>&wxyid=<%=wxyid%>&uuid='+this.value;">
解决onchange时?获得value的?
]]>
取得当前旉的时分秒
可以试试看这玩意
不知道是否有问题Q?br />
16. 查询除每个雇员的姓名Q工资,部门名称Q工资在公司的等U?及其领导的姓名以及工资所在公司的{
----->先查出来每个雇员的姓名,工资Q部门名Uͼ工资在公司的{
select e.ename 雇员名称,e.sal 雇员薪水,d.dname 部门名称,s.grade 工资{
from emp e,salgrade s,dept d
where e.sal between losal and hisal and e.deptno = d.deptno
----->再查 每个雇员?领导名字
select e.ename 雇员名称,e.sal 雇员薪水,
d.dname 部门名称,s.grade 工资{,
p.ename 雇员领导
from emp e,salgrade s,dept d,emp p
where e.sal between losal and hisal
and e.deptno = d.deptno
and e.mgr = p.empno
---->再查 每个领导的工资等U?br />
select e.ename 雇员名称,e.sal 雇员薪水,
d.dname 部门名称,s.grade 工资{,
p.ename 雇员领导,ls.grade 领导工资{
from emp e,salgrade s,dept d,emp p,salgrade ls
where e.sal between s.losal and s.hisal
and e.deptno = d.deptno
and e.mgr = p.empno
and p.sal between ls.losal and ls.hisal
---->优化
select e.ename 雇员名称,e.sal 雇员薪水,
d.dname 部门名称,
decode(s.grade,1,'W五{',2,'W四{',3,'W三{',4,'W二{',5,'W一{') 工资{,
p.ename 雇员领导,
decode(ls.grade,1,'W五{',2,'W四{',3,'W三{',4,'W二{',5,'W一{') 领导工资{
from emp e,salgrade s,dept d,emp p,salgrade ls
where e.sal between s.losal and s.hisal
and e.deptno = d.deptno
and e.mgr = p.empno
and p.sal between ls.losal and ls.hisal
17. 左右链接
左连接表C?左边表中所有的字段都会列出?但是双没有的以I显C?br />
双接表C?双表中所有的字段都会列出?但是左边没有的以I显C?br />
18. sql1999语法
19. l函?br />
count(),max(),min(),avg(),sum()
分组l计------------->
group by ?having 可以q用?br />
select e.deptno,d.dname,avg(e.sal)
from emp e,dept d
where e.deptno = d.deptno
group by e.deptno,d.dname having avg(e.sal)>2000
20. 昄出非销售h员工作名UC及从事同一工作雇员的月工资的d。ƈ且要满
从事同一工作的雇员的月工资合计大?000Q输出结果按月工资的合计l计
---->先显C全部的非销售h?br />
select ename,job from emp where job<>'SALESMAN'
---->按工作分l?同时求出工资的d
select job,sum(sal) from emp where job<>'SALESMAN' group by job
---->月工?gt;5000
select job,sum(sal) from emp where job<>'SALESMAN' group by job having sum(sal)>5000
---->排序
select job,sum(sal) s from emp where job<>'SALESMAN' group by job having sum(sal)>5000 order by s
21. 求出q_工资最高的部门
select deptno,max(avg(sal)) from emp group by deptno-------错误
select max(avg(sal)) from emp group by deptno-----正确
22. 子查?br />
---->查询?654工资高的雇员信息
select * from emp where sal > (select sal from emp where empno=7654)
---->查询?654工资高的雇员信息,?788从事相同工作的全部雇员信?br />
-->先查?654的工资select sal from emp where empno = 7654
-->再查?同时7788工作select job from emp where empno = 7788
-->l合select * from emp where sal > (select sal from emp where empno = 7654)
and job = (select job from emp where empno = 7788)
---->查询工资最低的雇员的姓?工作工资
-->先查出最低工资select min(sal) from emp
-->select ename,job,sal from emp where sal = (select min(sal) from emp)
---->查询部门名称Q部门的员工敎ͼ部门的^均工资,部门的最低收入雇员的姓名(q个语句只是适合一个最低工资不一L)
select e.deptno,count(sal),avg(sal),d.dname from emp e ,dept d where e.deptno = d.deptno
group by e.deptno,d.dname
-->select deptno,count(sal) c,avg(sal) from emp group by deptno
-->select d.dname,ed.c,ed.a from dept d,(select deptno,count(sal) c,avg(sal) a from emp group by deptno) ed where d.deptno = ed.deptno
-->select d.dname,ed.c,ed.a ,ed.min,e.ename from
dept d,
(select deptno,count(sal) c,avg(sal) a,min(sal) min from emp group by deptno) ed ,
emp e
where d.deptno = ed.deptno
and e.sal = ed.min
23. 子查询中的IN
->查询每个部门的最低工资的雇员信息
---->select * from emp where sal in(select min(sal) from emp group by deptno)
24. create table myemp as select * from emp
q个语句保存了emp表的原始信息会将表结构和数据完整的复制出?br />
q个操作不会把约束带q来的?/font>
to_date('String','yyyy-mm-dd')
create table myemp as select * from emp where empno=10表示只是包含10部门雇员的一张时表
25. 在oracle中每一个链接到数据库的H口Q都会徏立与数据库的一个session
一个session如果更新了数据库中的记录Q其他session是无法更新的要等待对Ҏ交之后才允许更新。一U死锁的概念
select * from tab;
select * from tab where tname = '';
26. select * from (select rownum rn,ename,job,sal,hiredate from emp where rownum<=10) e where e.rn>5分页
27. a.创徏用户:create user test identified by test(要创建用户必M用管理员用户sys)
没有session的权限无法登陆的ora-01045,解决办法授权grant
grant create session to test--->创建session的权限给test用户
****************************************q个时候用test用户登陆成功的话Q表C创Z一个session
b.一个新的用h有的权限都要分别赋予Q一个角色包含很多权?br />
c.Oracle中connectQresource
grant connect,resource to test--->q个时?可以?br />
d.理员可以修Ҏ通管理员密码alter user 用户?identified by 密码
e.alter user 用户?pasword expire 让用L一ơ密码失?强制修改
f.alter user 用户?account lock 用户锁?br />
g.alter user 用户?account unlock 解锁用户
28. 惌问其他用L表,需要授予其他用戯的权?br />
grant SELECT ,DELETE on scott.emp to test---scott.emp查询和删除的权限ltest用户
回收权限
revoke 权限 on 用户.表明 from 用户