在文件上傳過程中碰到很多問題，首先是搞錯了類，剛開始時我用的是PostodMethod,以為一個
setrequestbody（）方法就可以搞定，結(jié)果改過來改過去也沒改出來什么名堂，最后改用的是MultipartPostMethod類，呵呵，
問題解決了，關(guān)鍵點是MultipartPostMethod類里的addParameter()和addPart()兩個方法都要用到，而且要注意順
序。不過馬上又出現(xiàn)了新的問題，httpclient不支持中文名的文件上傳，暈了。又在這上面浪費了一段時間。解決的途徑是。找到
httpclient3.0"rc"java"org"apache"commons"httpclient"util目錄下的
EncodingUtil.java,打開，找到文件里面這個地方：
public static byte[] getAsciiBytes(final String data) {
if (data == null) {
throw new IllegalArgumentException("Parameter may not be null"); }
try { return data.getBytes("US-ASCII"); }
catch (UnsupportedEncodingException e) {throw new HttpClientError("HttpClient requires ASCII support"); }
}
看
到了沒有，return
data.getBytes("US-ASCII");它的編碼方式是US-ASCII，問題就出在這里了，把這個取掉，換成"GBK"或者
"GB2312"保存以后編譯，重新運行程序，goooooooooooood。中文名文件現(xiàn)在可以上傳了，呵呵
Introducing FileUpload
The FileUpload component has the
capability of simplifying the handling of files uploaded to a server.
Note that the FileUpload component is meant for use on the server side;
in other words, it handles where the files are being uploaded to—not
the client side where the files are uploaded from. Uploading files from
an HTML form is pretty simple; however, handling these files when they
get to the server is not that simple. If you want to apply any rules
and store these files based on those rules, things get more difficult.
The FileUpload component remedies this situation, and in very few
lines of code you can easily manage the files uploaded and store them
in appropriate locations. You will now see an example where you upload
some files first using a standard HTML form and then using HttpClient
code.
Using HTML File Upload
The commonly used methodology to upload
files is to have an HTML form where you define the files you want to
upload. A common example of this HTML interface is the Web page you
encounter when you want to attach files to an email while using any of
the popular Web mail services.
In this example, you will create a simple HTML page where you
provide for three files to be uploaded. Listing 1-1 shows the HTML for
this page. Note that the enctype attribute for the form has the
multipart/form-data, and the input tag used is of type file. Based on
the of the action attribute, on form submission, the data is sent to
ProcessFileUpload.jsp.
Listing 1-1. UploadFiles.html
<HTML>
<HEAD>
< HTTP-EQUIV="Content-Type" CONTENT="text/html; charset=windows-1252"/>
<TITLE>File Upload Page</TITLE>
</HEAD>
<BODY>Upload Files
<FORM name="filesForm" action="ProcessFileUpload.jsp"
method="post" enctype="multipart/form-data">
File 1:<input type="file" name="file1"/><br/>
File 2:<input type="file" name="file2"/><br/>
File 3:<input type="file" name="file3"/><br/>
<input type="submit" name="Submit" ="Upload Files"/>
</FORM>
</BODY>
</HTML>
You can use a servlet to handle the file upload. I have used JSP to
minimize the code you need to write. The task that the JSP has to
accomplish is to pick up the files that are sent as part of the request
and store these files on the server. In the JSP, instead of displaying
the result of the upload in the Web browser, I have chosen to print
messages on the server console so that you can use this same JSP when
it is not invoked through an HTML form but by using HttpClient-based
code.
Listing 1-2 shows the JSP code. Note the code that checks whether
the item is a form field. This check is required because the Submit
button contents are also sent as part of the request, and you want to
distinguish between this data and the files that are part of the
request. You have set the maximum file size to 1,000,000 bytes using
the setSizeMax method.
Listing 1-2. ProcessFileUpload.jsp
<%@ page contentType="text/html;charset=windows-1252"%>
<%@ page import="org.apache.commons.fileupload.DiskFileUpload"%>
<%@ page import="org.apache.commons.fileupload.FileItem"%>
<%@ page import="jsp servlet ejb .util.List"%>
<%@ page import="jsp servlet ejb .util.Iterator"%>
<%@ page import="jsp servlet ejb .io.File"%>
html>
<head>
< http-equiv="Content-Type" content="text/html; charset=windows-1252">
<title>Process File Upload</title>
</head>
<%
System.out.println("Content Type ="+request.getContentType());
DiskFileUpload fu = new DiskFileUpload();
// If file size exceeds, a FileUploadException will be thrown
fu.setSizeMax(1000000);
List fileItems = fu.parseRequest(request);
Iterator itr = fileItems.iterator();
while(itr.hasNext()) {
FileItem fi = (FileItem)itr.next();
//Check if not form field so as to only handle the file inputs
//else condition handles the submit button input
if(!fi.isFormField()) {
System.out.println(""nNAME: "+fi.getName());
System.out.println("SIZE: "+fi.getSize());
//System.out.println(fi.getOutputStream().toString());
File fNew= new File(application.getRealPath("/"), fi.getName());
System.out.println(fNew.getAbsolutePath());
fi.write(fNew);
}
else {
System.out.println("Field ="+fi.getFieldName());
}
}
%>
<body>
Upload Successful!!
</body>
</html>
CAUTION With FileUpload 1.0 I found that when the form was submitted
using Opera version 7.11, the getName method of the class FileItem
returns just the name of the file. However, if the form is submitted
using Internet Explorer 5.5, the filename along with its entire path is
returned by the same method. This can cause some problems.
To run this example, you can use any three files, as the contents of
the files are not important. Upon submitting the form using Opera and
uploading three random XML files, the output I got on the Tomcat server
console was as follows:
Content Type =multipart/form-data; boundary=----------rz7ZNYDVpN1To8L73sZ6OE
NAME: academy.xml
SIZE: 951
D:"javaGizmos"jakarta-tomcat-4.0.1"webapps"HttpServerSideApp"academy.xml
NAME: academyRules.xml
SIZE: 1211
D:"javaGizmos"jakarta-tomcat-4.0.1"webapps"HttpServerSideApp"academyRules.xml
NAME: students.xml
SIZE: 279
D:"javaGizmos"jakarta-tomcat-4.0.1"webapps"HttpServerSideApp"students.xml
Field =Submit
However, when submitting this same form using Internet Explorer 5.5, the output on the server console was as follows:
Content Type =multipart/form-data; boundary=---------------------------7d3bb1de0
2e4
NAME: D:"temp"academy.xml
SIZE: 951
D:"javaGizmos"jakarta-tomcat-4.0.1"webapps"HttpServerSideApp"D:"temp"academy.xml
The browser displayed the following message: “The requested resource
(D:"javaGizmos"jakarta-tomcat-4.0.1"webapps"HttpServerSideApp"D:"temp"academy.xml
(The filename, directory name, or volume label syntax is incorrect)) is
not available.”
This contrasting behavior on different browsers can cause problems. One workaround that I found in an article at http://www.onjava.com/pub/a/onjava/2003/06/25/commons.html
is to first create a file reference with whatever is supplied by the
getName method and then create a new file reference using the name
returned by the earlier file reference. Therefore, you can insert the
following code to have your code work with both browsers (I wonder who
the guilty party is…blaming Microsoft is always the easy way out)
File tempFileRef = new File(fi.getName());
File fNew = new File(application.getRealPath("/"),tempFileRef.getName());
In this section, you uploaded files using a standard HTML form
mechanism. However, often a need arises to be able to upload files from
within your jsp servlet ejb code, without any browser or form coming
into the picture. In the next section, you will look at
HttpClient-based file upload.
Using HttpClient-Based FileUpload
Earlier in the article you saw
some of the capabilities of the HttpClient component. One capability I
did not cover was its ability to send multipart requests. In this
section, you will use this capability to upload a few files to the same
JSP that you used for uploads using HTML.
The class org.apache.commons.httpclient.methods.MultipartPostMethod
provides the multipart method capability to send multipart-encoded
forms, and the package org.apache.commons.httpclient.methods.multipart
has the support classes required. Sending a multipart form using
HttpClient is quite simple. In the code in Listing 1-3, you send three
files to ProcessFileUpload.jsp.
Listing 1-3. HttpMultiPartFileUpload.java
package com.commonsbook.chap9;
import jsp servlet ejb .io.File;
import jsp servlet ejb .io.IOException;
import org.apache.commons.httpclient.HttpClient;
import org.apache.commons.httpclient.methods.MultipartPostMethod;
public class HttpMultiPartFileUpload {
private static String url =
"http://localhost/yaoliang/ProcessFileUpload.jsp";
public static void main(String[] args) throws IOException {
HttpClient client = new HttpClient();
MultipartPostMethod mPost = new MultipartPostMethod(url);
client.setConnectionTimeout(8000);
// Send any XML file as the body of the POST request
File f1 = new File("D:/students.xml");
File f2 = new File("D:/demy.xml");
File f3 = new File("D:/demyRules.xml");
System.out.println("File1 Length = " + f1.length());
System.out.println("File2 Length = " + f2.length());
System.out.println("File3 Length = " + f3.length());
mPost.addParameter(f1.getName(),f1.getName(), f1);
mPost.addParameter(f2.getName(), f2.getName(), f2);
mPost.addParameter(f3.getName(), f3.getName(),f3);
FilePart part1 = new FilePart("file1",file);
FilePart part2 = new FilePart("file2",file);
FilePart part3 = new FilePart("file3",file);
mPost.addPart(part1);
mPost.addPart(part2);
mPost.addPart(part3);
int statusCode1 = client.executeMethod(mPost);
System.out.println("statusLine>>>" + mPost.getStatusLine());
mPost.releaseConnection();
}
}
In this code, you just add the files as parameters and execute the
method. The ProcessFileUpload.jsp file gets invoked, and the output is
as follows:
Content Type =multipart/form-data; boundary=----------------31415926535897932384
6
NAME: students.xml
SIZE: 279
D:"javaGizmos"jakarta-tomcat-4.0.1"webapps"HttpServerSideApp"students.xml
NAME: academy.xml
SIZE: 951
D:"javaGizmos"jakarta-tomcat-4.0.1"webapps"HttpServerSideApp"academy.xml
NAME: academyRules.xml
SIZE: 1211
D:"javaGizmos"jakarta-tomcat-4.0.1"webapps"HttpServerSideApp"academyRules.xml
Thus, file uploads on the server side become quite a simple task if you are using the Commons FileUpload component.