Question

我是java和servlet编程的新手。我不确定是否可以编写一个servlet，当从本地客户端机器传递URL时，将该文件上传到服务器。

基本上在客户端机器上我们有一个C＃程序，在服务器端我们安装了Apache-tomcat。我需要使用客户端计算机上的C＃程序将文件上传到服务器。

我应该提供更多信息（？）

先谢谢

Answer 1

请注意，此代码说明了一般概念，并且无法保证无需修改即可运行。 The C# file upload part

// this code shows you how the browsers wrap the file upload request, you still can fine a way simpler code to do the same thing.
   public void PostMultipleFiles(string url, string[] files)
{
    string boundary = "----------------------------" + DateTime.Now.Ticks.ToString("x");
    HttpWebRequest httpWebRequest = (HttpWebRequest)WebRequest.Create(url);
    httpWebRequest.ContentType = "multipart/form-data; boundary=" + boundary;
    httpWebRequest.Method = "POST";
    httpWebRequest.KeepAlive = true;
    httpWebRequest.Credentials = System.Net.CredentialCache.DefaultCredentials;
    Stream memStream = new System.IO.MemoryStream();
    byte[] boundarybytes =System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary     +"\r\n");
    string formdataTemplate = "\r\n--" + boundary + "\r\nContent-Disposition:  form-data; name=\"{0}\";\r\n\r\n{1}";
    string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\n Content-Type: application/octet-stream\r\n\r\n";
    memStream.Write(boundarybytes, 0, boundarybytes.Length);
    for (int i = 0; i < files.Length; i++)
    {
        string header = string.Format(headerTemplate, "file" + i, files[i]);
        //string header = string.Format(headerTemplate, "uplTheFile", files[i]);
        byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
        memStream.Write(headerbytes, 0, headerbytes.Length);
        FileStream fileStream = new FileStream(files[i], FileMode.Open,
        FileAccess.Read);
        byte[] buffer = new byte[1024];
        int bytesRead = 0;
        while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
        {
            memStream.Write(buffer, 0, bytesRead);
        }
        memStream.Write(boundarybytes, 0, boundarybytes.Length);
        fileStream.Close();
    }
    httpWebRequest.ContentLength = memStream.Length;
    Stream requestStream = httpWebRequest.GetRequestStream();
    memStream.Position = 0;
    byte[] tempBuffer = new byte[memStream.Length];
    memStream.Read(tempBuffer, 0, tempBuffer.Length);
    memStream.Close();
    requestStream.Write(tempBuffer, 0, tempBuffer.Length);
    requestStream.Close();
    try
    {
        WebResponse webResponse = httpWebRequest.GetResponse();
        Stream stream = webResponse.GetResponseStream();
        StreamReader reader = new StreamReader(stream);
        string var = reader.ReadToEnd();

    }
    catch (Exception ex)
    {
        response.InnerHtml = ex.Message;
    }
    httpWebRequest = null;
}

要了解上述代码是如何编写的，您可能需要查看How does HTTP file upload work?

POST /upload?upload_progress_id=12344 HTTP/1.1
Host: localhost:3000
Content-Length: 1325
Origin: http://localhost:3000
... other headers ...
Content-Type: multipart/form-data; boundary=----WebKitFormBoundaryePkpFF7tjBAqx29L

------WebKitFormBoundaryePkpFF7tjBAqx29L
Content-Disposition: form-data; name="MAX_FILE_SIZE"

100000
------WebKitFormBoundaryePkpFF7tjBAqx29L
Content-Disposition: form-data; name="uploadedfile"; filename="hello.o"
Content-Type: application/x-object

... contents of file goes here ...
------WebKitFormBoundaryePkpFF7tjBAqx29L--

最后你所要做的就是实现一个可以处理文件上传请求的servlet，然后你做任何你想对文件做的事情，take a look at this file upload tutorial

protected void processRequest(HttpServletRequest request,
        HttpServletResponse response)
        throws ServletException, IOException {
    response.setContentType("text/html;charset=UTF-8");

    // Create path components to save the file
    final String path = request.getParameter("destination");
    final Part filePart = request.getPart("file");
    final String fileName = getFileName(filePart);

    OutputStream out = null;
    InputStream filecontent = null;
    final PrintWriter writer = response.getWriter();

    try {
        out = new FileOutputStream(new File(path + File.separator
                + fileName));
        filecontent = filePart.getInputStream();

        int read = 0;
        final byte[] bytes = new byte[1024];

        while ((read = filecontent.read(bytes)) != -1) {
            out.write(bytes, 0, read);
        }
        writer.println("New file " + fileName + " created at " + path);
        LOGGER.log(Level.INFO, "File{0}being uploaded to {1}", 
                new Object[]{fileName, path});
    } catch (FileNotFoundException fne) {
        writer.println("You either did not specify a file to upload or are "
                + "trying to upload a file to a protected or nonexistent "
                + "location.");
        writer.println("<br/> ERROR: " + fne.getMessage());

        LOGGER.log(Level.SEVERE, "Problems during file upload. Error: {0}", 
                new Object[]{fne.getMessage()});
    } finally {
        if (out != null) {
            out.close();
        }
        if (filecontent != null) {
            filecontent.close();
        }
        if (writer != null) {
            writer.close();
        }
    }
}

private String getFileName(final Part part) {
    final String partHeader = part.getHeader("content-disposition");
    LOGGER.log(Level.INFO, "Part Header = {0}", partHeader);
    for (String content : part.getHeader("content-disposition").split(";")) {
        if (content.trim().startsWith("filename")) {
            return content.substring(
                    content.indexOf('=') + 1).trim().replace("\"", "");
        }
    }
    return null;
}

使用java servlet作为服务而无需Web浏览器的文件上载

1 个答案: