首先,为什么不要求你直接将表格提交给该政策手册?

<form action="http://example.com/upload.php" method="post" enctype="multipart/form-data">
    <input type="file" name="file">
    <input type="submit">
</form>

如果这不是一种选择,而且你真的需要将表格提交服务器,那么首先在联合调查组中形成一种超文本形式:

<form action="upload" method="post" enctype="multipart/form-data">
    <input type="file" name="file">
    <input type="submit">
</form>

在关于<代码>/upload的的服务器中,根据PHP的文字处理申请有两种选择。

如果PHP的文字采用same参数,并且能够处理上载文件,其方式与超文本格式指示服务器做的相同(我仍然只是让表格直接提交给PHP的文字,但不管怎么说),那么你可以让该信封为透明的代理玩.,而该代理只是从吉卜赛人的请求中立即将 by转给PHP。 java.net.URLConnection 就此而言,APIC是有用的。

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    HttpURLConnection connection = (HttpURLConnection) new URL("http://example.com/upload.php").openConnection();
    connection.setDoOutput(true); // POST.
    connection.setRequestProperty("Content-Type", request.getHeader("Content-Type")); // This one is important! You may want to check other request headers and copy it as well.

    // Set streaming mode, else HttpURLConnection will buffer everything in Java s memory.
    int contentLength = request.getContentLength();
    if (contentLength > -1) {
        connection.setFixedLengthStreamingMode(contentLength);
     } else {
        connection.setChunkedStreamingMode(1024);
    }

    InputStream input = request.getInputStream();
    OutputStream output = connection.getOutputStream();
    byte[] buffer = new byte[1024]; // Uses only 1KB of memory!
    for (int length = 0; (length = input.read(buffer)) > 0;) {
        output.write(buffer, 0, length);
    }
    output.close();

    InputStream phpResponse = connection.getInputStream(); // Calling getInputStream() is important, it s lazily executed!
    // Do your thing with the PHP response.
}

如果PHP的文字包含 different或more参数(ain,我只是改动超文本格式,以便它能够直接提交PHP的文字,那么,如果用户使用,Apache Commons 文档上载,以提取上载文档和

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    InputStream fileContent = null;
    String fileContentType = null;
    String fileName = null;

    try {
        List<FileItem> items = new ServletFileUpload(new DiskFileItemFactory()).parseRequest(request);
        for (FileItem item : items) {
            if (!item.isFormField() && item.getFieldName().equals("file")) { // <input type="file" name="file">
                fileContent = item.getInputStream();
                fileContentType = item.getContentType();
                fileName = FilenameUtils.getName(item.getName());
                break; // If there are no other fields?
            }            
        }
    } catch (FileUploadException e) {
        throw new ServletException("Parsing file upload failed.", e);
    }

    if (fileContent != null) {
        HttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost("http://example.com/upload.php");
        MultipartEntity entity = new MultipartEntity();
        entity.addPart("file", new InputStreamBody(fileContent, fileContentType, fileName));
        httpPost.setEntity(entity);
        HttpResponse phpResponse = httpClient.execute(httpPost);
        // Do your thing with the PHP response.
    }
}

• How to fire HTTP requests using URLConnection
• How to upload files in JSP/Servlet
• HttpClient tutorial

我头顶上的几种解决办法......

• Use HTTP POST to send the file. Probably not a good idea if the files are large.
• Use FTP. This easily gives you authentication, handling of large files, etc. Bonus points for using SFTP.
• Use a program like rsync [over ssh] to migrate the directory contents from one server to the other. Not a good solution if you have disk space concerns since you d be storing the same files twice, once per server.

此外,铭记着如何经常将图像推向你的保护伞。 你们不想试图把100个图像及其传输网络的袖珍藏在记忆中——在此情况下将图像保存在磁盘上。

在发送方面,您可使用。 http://hc.apache.org/client-3.x/docs/apache/file,http://clcl/methods/multi_FilePart. rel= nofolor/a>。