commons-fileupload 1.2发布时的官方声明

leekiang — Sat, 27 Sep 2008 04:34:00 GMT

The Jakarta Commons team is glad to announce the availability of commons-fileupload 1.2. Commons Fileupload is a framework for handling HTTP file upload requests in servlets, portlets, and similar server side Java applications.
Compared to the previous version 1.1.1, the following notable changes have been made:
    * A streaming API has been added. The streaming API allows to handle arbitrarily large files without intermediary files while still keeping an extremely low memory profile.
    * The presence of a content-length header is no longer required.
    * Added support for progress listeners.
    * Added support for header continuation lines.
    * Added support for limiting the actual file size, as opposed to the request size.

leekiang 2008-09-27 12:34 发表评论

leekiang — Wed, 24 Sep 2008 06:01:00 GMT

1,以下是一些零��的记录�Q�不全�?br />

//字段为java.sql.Blob�c�d��
Fj fj = new Fj();
fj.setAttblob(Hibernate.createBlob(new byte[1]));//用empty_blob()替换?
session.save(fj);
session.flush();
session.refresh(fj, LockMode.UPGRADE);
org.hibernate.blob.SerializableBlob sb =
(org.hibernate.blob.SerializableBlob) fj.getAttblob();
oracle.sql.BLOB blob = (oracle.sql.BLOB) sb.getWrappedBlob();
OutputStream os = blob.getBinaryOutputStream();
//------

2,用jdbc��d��CLOB
http://hi.baidu.com/xh28025/blog/item/f61c2df1ef8130c47831aa70.html

String description = ""
   query = "select picstr from clobtest_table where id = '001'";
pstmt = con.prepareStatement(query);
ResultSet result = pstmt.executeQuery();
if(result.next()){
   oracle.jdbc.driver.OracleResultSet ors =
   (oracle.jdbc.driver.OracleResultSet)result;
   oracle.sql.CLOB clobtmp = (oracle.sql.CLOB) ors.getClob(1);

   if(clobtmp==null || clobtmp.length()==0){
   System.out.println("======CLOB对象为空 ");
   description = "";
   }else{
   description=clobtmp.getSubString((long)1,(int)clobtmp.length());//�?开�?
   System.out.println("======字符串�Ş式�?/span>"+description);
   }
}

leekiang 2008-09-24 14:01 发表评论

leekiang — Mon, 27 Aug 2007 03:58:00 GMT

我对表单上传的理�?可能有误):

最大上�?G.
通过 http 协议上传文�g(rfc1867协议概述�Q�jsp 应用举例�Q�客��L(f��ng)��发送内�Ҏ(gu��)��?
服务器接收到上传的流�Q�自己其实是不作��M��处理的，那个request�q�是原装�?谁来处理�q�个request�?一般采用第三方的工��P��q�里以commons fileupload��Z��.

DiskFileItemFactory factory = new DiskFileItemFactory();
factory.setSizeThreshold(4096);// 讄��~�冲,�q�个值决定了(ji��n)是fileinputstream�q�是bytearrayinputstream
factory.setRepository(new File("d:\\temp"));//讄��临时存放目录,默认是new File(System.getProperty("java.io.tmpdir"))
ServletFileUpload sfu = new ServletFileUpload(factory);
sfu.setSizeMax(100*1024*1024);//100M
List items = sfu.parseRequest(request);//传入的这个request�q�是原装�?/span>

见上面的代码�Q�commons fielupload通过ServletFileUpload�cȝ��parseRequest(request)�Ҏ(gu��)��处理�q�个原始��。�?/font>ServletFileUpload又会(x��)调用其爷��L(f��ng)��FileUploadBase的parseRequest(request)�Ҏ(gu��)��Q�然后又�?x��)调r(ji��)eturn parseRequest(new ServletRequestContext(request)),代码如下

        try {
            FileItemIterator iter = getItemIterator(ctx);
            List items = new ArrayList();
            FileItemFactory fac = getFileItemFactory();
            if (fac == null) {
                throw new NullPointerException(
                    "No FileItemFactory has been set.");
            }
            while (iter.hasNext()) {
                FileItemStream item = iter.next();
                FileItem fileItem = fac.createItem(item.getFieldName(),
                        item.getContentType(), item.isFormField(),
                        item.getName());
                try {
                    Streams.copy(item.openStream(), fileItem.getOutputStream(),
                            true);
                } catch (FileUploadIOException e) {
                    throw (FileUploadException) e.getCause();
                } catch (IOException e) {
                    throw new I(y��ng)OFileUploadException(//报错�l�常在这�?br />                            "Processing of " + MULTIPART_FORM_DATA
                            + " request failed. " + e.getMessage(), e);
                }
                if (fileItem instanceof FileItemHeadersSupport) {
                    final FileItemHeaders fih = item.getHeaders();
                    ((FileItemHeadersSupport) fileItem).setHeaders(fih);
                }
                items.add(fileItem);
            }
            return items;
        } catch (FileUploadIOException e) {
            throw (FileUploadException) e.getCause();
        } catch (IOException e) {
            throw new FileUploadException(e.getMessage(), e);
        }

�q�里注意,上传�?lt;input type=file>标记一定要有name,如果没有�Q�commons fielupload不作处理�?br /> commons fielupload�?x��)把上传的文件以��的方式写入到temp文�g�?临时文�g夹可以自��p��定，如果不手动设�|�，则是Servlet容器为web应用分配的��(f��)时目�?tomcat可能��是%TOMCAT_HOME%\temp�Q�我用weblogic�Ӟ��?font color="#ff0000">C:\DOCUME~1\yourname\LOCALS~1\Temp\。这些��(f��)时文件以"upload"开��_(d��)��格式�?tmp,例如"upload_47fdc3_11c9eb678b0__8000_00000043.tmp"
在上传过�E�中commons fielupload才知道上传的文�g的大��，如果你定义允许的最大附件�ؓ(f��)100M�Q�然后你上传一�?00M的文�?那么只有实际传输�?00M以后�Q�commons fielupload才知道超�q�了(ji��n)�Q�然后抛��Z��个异�?org.apache.commons.fileupload.FileUploadBase.SizeLimitExceededException)�?br /> commons fielupload处理完了(ji��n)以后(处理�q�程全在parseRequest(request)�Ҏ(gu��)��?/span>)�Q�返回一个List,里面的每一��已被封装�ؓ(f��)FileItem,你通过fileItem.isFormField()判断出是普通的表单属性呢�Q�还是一个文�Ӟ��如果是文�Ӟ��你可以从fileItem.getInputStream()获得输入��，�q�个输入��其实是一个FileInputStream.当然�Q�如果文件很��，则是ByteArrayInputStream.那么如何区分�Q�这两者的区别是由factory.setSizeThreshold(4096)��定的。大�?K,为FileInputStream,��于4K为ByteArrayInputStream

�q�样��可以解释struts用ActionForm的方式处理上传附件的一些问题了(ji��n)�Q�struts接收到enctype="multipart/form-data"的post��h��后，�?x��)看那个对应的action有没有配�|�a(ch��n)ctionform�Q�如果配�|�了(ji��n)�Q�就�?x��)作一些处理，所以你在action里得到的request已经不是一个普通的request�?ji��n)，而是一个被��装�q�的request。如果想得到原始的request�Q�就不要struts-config.xml里给action�c�配�|�a(ch��n)ctionform

tempDir指定的目录中可能�?x��)随着旉��推移出现很多后缀�?tmp"的垃圾文�Ӟ��commons-fileupload1.2提供�?ji��n)一个不错的解决�Ҏ(gu��)��Q�就是把下面的代码加入到web.xml中即可�?br />

org.apache.commons.fileupload.servlet.FileCleanerCleanup

来源:http://fratemity1314.spaces.live.com/Blog/cns!E3BCC13A1E72BB4F!222.entry

Q:I'm using FileUpload in an Action, but it's not working. Why?

A:Struts recognises multipart requests, and parses them automatically, presenting the request parameters to your code in the same manner as if they were regular request parameters. Since struts has already processed the request, and made it available in your form bean, the input stream is no longer available for parsing, so attempting to do so with FileUpload will fail.

Q:But I need to parse the request myself. How can I do that?

A:Struts parses multipart a request as a part of the process of populating your form bean from that request. If, for some reason, you need to have full control over the multipart parsing, you can do so by configuring your action mapping without an associated form bean. (A better way of doing this, however, is to replace the default multipart handler with your own. See the struts documentation for details.)

1�Q�直接用commons fileupload而不用struts的ActionForm�?表单里的属性��g��能用request.getParameter()获取�?ji��n)，而url里的queryString可以�?br />2, 获取form里的属性值的代码

                    String formname = fi.getFieldName();// 获取form中的名字
                    String formcontent = fi.getString();
                    if (formname.equals("id")) {
                        id = formcontent;
                    } else if (formname.equals("title")) {
                        title = formcontent;
                    } else if (formname.equals("memo")) {
                        memo = formcontent;
                    }

3, 表单里的file控�g的name不能为空�?br />4�Q�上传大文�g(190多M)时报异常
org.apache.commons.fileupload.FileUploadException: Processing of multipart/form-data request failed. EOF after reading only: "3567789 " of: "203323339 " promised bytes, out of which at least: "0 " were already buffered
http://forums.bea.com/thread.jspa?threadID=200033356
http://dev2dev.bea.com.cn/bbs/thread.jspa?forumID=121&threadID=20060&start=0&tstart=0
http://thisisxy.blogcn.com/diary,204014352.shtml
http://www.80diy.com/home/20050527/17/4040711.html

5�Q?br />org.apache.commons.fileupload.FileUploadException: Processing of multipart/form-data request failed. Read timed out
HTTP�z�d��时限制的时间太�?HTTP Keep-Alive Timeout.

6,InputStream is =formFile.getInputStream() ; //formFile是org.apache.struts.upload.FormFile
上传的文件小时这个输入流是java.io.ByteArrayInputStream�Q�上传比较大的文件时�q�个输入��是FileInputStream

7,讄��表单里的上传路径为只�?br />

onClick="uploadfile.click();tempfile.value=uploadfile.value;" value="��览..">

10,
http://eastpoint.javaeye.com/blog/99084
http://topic.csdn.net/u/20080131/13/649c57c7-204e-4bb6-9b09-49cca39f00b8.html
http://blog.niwota.com/nb/chenchuang?cat_self_id=291075

leekiang 2007-08-27 11:58 发表评论

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commons-fileupload 1.2发布时的官方声明