Httpclient 使用 multipart/form-data 进行 POST

Question

我需要上传图片。 为此，我必须将图像和 ID 传递给服务器。

dos.writeBytes("Content-Disposition: form-data; name=\"file_name\";filename=\""
            + fileName + "\"" + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"" + stringFieldName + "\""+ lineEnd);
dos.writeBytes(lineEnd);

fileName是图像路径。
stringFieldName是 user_id

我的代码如下。

public void webservicePhp(Long userId, Bitmap bmp) {
    String userIdParameter = String.valueOf(userId);
    String fileName = "temporary_holder.jpg";
    HttpURLConnection conn = null;
    DataOutputStream dos = null;
    DataInputStream inStream = null;
    String lineEnd = "\r\n";
    String twoHyphens = "--";
    String boundary = "*****";
    int bytesRead, bytesAvailable, bufferSize;
    byte[] buffer;
    int maxBufferSize = 1 * 1024 * 1024;

    String charset = "UTF-8";
    String responseFromServer = "";

    String stringFieldName = "user_id";

    String sourceFileUri = HomeScreen.get_path();
    String upLoadServerUri = "http://10.120.10.87:8080/ContactsManagerWeb/UploadImage";

    File sourceFile = new File(sourceFileUri);
    if (!sourceFile.isFile()) {
        Log.e("Huzza", "Source File Does not exist");
        return;
    }
    int serverResponseCode = 0;
    try { // open a URL connection to the Servlet
        FileInputStream fileInputStream = new FileInputStream(sourceFile);
        URL url = new URL(upLoadServerUri);
        conn = (HttpURLConnection) url.openConnection(); // Open a HTTP
                                     // connection to
                                     // the URL
        conn.setDoInput(true); // Allow Inputs
        conn.setDoOutput(true); // Allow Outputs
        conn.setUseCaches(false); // Don't use a Cached Copy
        conn.setRequestMethod("POST");
        conn.setRequestProperty("Connection", "Keep-Alive");
        conn.setRequestProperty("Accept-Charset", charset);

        conn.setRequestProperty("ENCTYPE", "multipart/form-data");
        conn.setRequestProperty("Content-Type",
                "multipart/form-data;boundary=" + boundary);
        conn.setRequestProperty("file_name", fileName);
        conn.setRequestProperty("user_id", userIdParameter);
        dos = new DataOutputStream(conn.getOutputStream());
        dos.write(query.getBytes(charset));
        dos.writeBytes(twoHyphens + boundary + lineEnd);
        dos.writeBytes("Content-Disposition: form-data; name=\"file_name\";filename=\""
                + fileName + "\"" + lineEnd);

        dos.writeBytes("Content-Disposition: form-data; name=\"" + stringFieldName + "\""+ lineEnd);
        dos.writeBytes(lineEnd);

        bytesAvailable = fileInputStream.available(); // create a buffer of maximum size
        bufferSize = (int) sourceFile.length();

        System.out.println("BytesAvail" + bytesAvailable);
        System.out.println("maxBufferSize" + maxBufferSize);
        buffer = new byte[bufferSize];

        // read file and write it into form...
        bytesRead = fileInputStream.read(buffer, 0, bufferSize);

        while (bytesRead > 0) {
            dos.write(buffer, 0, bufferSize);
            bytesAvailable = fileInputStream.available();
            bufferSize = Math.min(bytesAvailable, maxBufferSize);
            bytesRead = fileInputStream.read(buffer, 0, bufferSize);
        }

        // send multipart form data necesssary after file data...
        dos.writeBytes(lineEnd);
        dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);

        // Responses from the server (code and message)
        serverResponseCode = conn.getResponseCode();
        String serverResponseMessage = conn.getResponseMessage();

        System.out.println("Upload file to serverHTTP Response is : "
                + serverResponseMessage + ": " + serverResponseCode);
        // close streams
        System.out.println("Upload file to server"+ fileName + " File is written");
        fileInputStream.close();
        dos.flush();
        dos.close();
    } catch (MalformedURLException ex) {
        ex.printStackTrace();
        Log.e("Upload file to server", "error: " + ex.getMessage(), ex);
    } catch (Exception e) {
        e.printStackTrace();
    }
    // this block will give the response of upload link
    try {
        BufferedReader rd = new BufferedReader(new InputStreamReader(
                conn.getInputStream()));
        String line;
        while ((line = rd.readLine()) != null) {
            System.out.println("RESULT Message: " + line);
        }
        rd.close();
    } catch (IOException ioex) {
        Log.e("Huzza", "error: " + ioex.getMessage(), ioex);
    }
    return; // like 200 (Ok)
}

在我的 servlet 中，我无法访问 user_id 参数。 它没有到达那里。

小程序

public void doPost(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
    response.setContentType("text/html");
    PrintWriter out = response.getWriter();
    File filenameImg = null;
    List<FileItem> items = null;
    try {
        items = new ServletFileUpload(new DiskFileItemFactory())
                    .parseRequest(request);
    } catch (FileUploadException e) {
        throw new ServletException("Cannot parse multipart request.", e);
    }

    for (FileItem item : items) {
        if (item.isFormField()) {
            // Process regular form fields here the same way as
            // request.getParameter().
            // You can get parameter name by

            String fieldname = item.getFieldName();
            String fieldvalue = item.getString(); 
            System.out.println("user_id===fieldname====== "+fieldname);
            System.out.println("user_id====fieldvalue===== "+fieldvalue);
            // You can get parameter value by item.getString();
        } else {
            try{
                // Process uploaded fields here.
                String filename = FilenameUtils.getName(item.getName());
                // Get filename.
                String path = GetWebApplicationPathServlet.getContext().getRealPath("/images");

                File file = new File(path,filename);

                // Define destination file.
                item.write(file);
                System.out.println("filename: "+filename);
                System.out.println("file: "+file);
                request.setAttribute("image", file);
                filenameImg = file;
                // Write to destination file.
                //    request.setAttribute("image", filename);
            }
            catch (Exception e) {
                e.printStackTrace();
            }
        }
    }
    // Show result page.
    System.out.println("request"+request.getAttribute("image"));
    //response.setContentType("image/jpeg"); 
    request.setAttribute("servletName", filenameImg);
    getServletConfig().getServletContext().getRequestDispatcher(
        "/result.jsp").forward(request,response);
}

if (item.isFormField()) {}返回false 。

请帮忙。

Answer 1

如果这与 html 表单提交相同。 那你可以试试这个。

在您的if (item.isFormField()) {在您的 Servlet 中执行此操作。

String user_Id = null; //Create an instance of String variable before initializing (Optional)

if (item.isFormField()) {
    if(item.getFieldName().contentEquals("name")){  //Check if the item in the loop is the user_id
         user_id = item.getString();                //If yes store the value
    }
} else { //continue with your current code

我以前遇到过同样的问题，这是对我有用的问题。 我不确定这是否是您想要的，请尝试一下。