java Android：如何将 .mp3 文件和图像上传到 http 服务器？

Android:How to upload .mp3 file and image to http server?

提问by jennifer

My code for uploading image to Server is :

我将图像上传到服务器的代码是：

String userIdParameter = String.valueOf(userId);
    String fileName = "temporary_holder.jpg";
    HttpURLConnection conn = null;
    DataOutputStream dos = null;

    String lineEnd = "\r\n";
    String twoHyphens = "--";
    String boundary = "*****";
    int bytesRead, bytesAvailable, bufferSize;
    byte[] buffer;
    int maxBufferSize = 1 * 1024 * 1024;

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

        // ------------------ CLIENT REQUEST
        URL url = new URL(upLoadServerUri);

        // Open a HTTP connection to the URL
        conn = (HttpURLConnection) url.openConnection();
        conn.setDoInput(true); // Allow Inputs
        conn.setDoOutput(true); // Allow Outputs
        conn.setUseCaches(false); // Don't use a Cached Copy

        // Use a post method.
        conn.setRequestMethod("POST");
        conn.setRequestProperty("Connection", "Keep-Alive");

        conn.setRequestProperty("ENCTYPE", "multipart/form-data");
        conn.setRequestProperty("Content-Type",
            "multipart/form-data;boundary=" + boundary);
        conn.setRequestProperty("file_name", fileName);
        conn.setRequestProperty("file_name_audio", fileName);
        conn.setRequestProperty("X-myapp-param1", userIdParameter);

        // conn.setFixedLengthStreamingMode(1024);
        // conn.setChunkedStreamingMode(1);

        dos = new DataOutputStream(conn.getOutputStream());

        dos.writeBytes(twoHyphens + boundary + lineEnd);

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

        dos.writeBytes(lineEnd);


            // create a buffer of maximum size
        bytesAvailable = fileInputStream.available();

        int streamSize = (int) sourceFile.length();
        bufferSize = streamSize / 10;

        System.out.println("streamSize" + streamSize);

        buffer = new byte[streamSize];

        // read file and write it into form...
        bytesRead = fileInputStream.read(buffer, 0, bufferSize);
        int count = 0;
        while (bytesRead > 0) {
        progress = (int) (count);
        displayNotification();
        Thread.sleep(500);

        dos.write(buffer, 0, bufferSize);
        bytesAvailable = fileInputStream.available();
        // bufferSize = Math.min(bytesAvailable, maxBufferSize);
        bytesRead = fileInputStream.read(buffer, 0, bufferSize);
        count += 10;

        }

        // 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)

Uploading image to server works fine.. I need to upload both mp3 file and image to the server..Please Help

将图像上传到服务器工作正常.. 我需要将 mp3 文件和图像上传到服务器..请帮助

采纳答案by Stephan

So, you want to send multiple files in one HTTP request? I've never done this myself, but according to the RFC, just add another body to the message in which you send the audio, it should look something like this:

那么，您想在一个 HTTP 请求中发送多个文件吗？我自己从未这样做过，但根据RFC，只需在发送音频的消息中添加另一个正文，它应该如下所示：

    dos = new DataOutputStream(conn.getOutputStream());
    dos.writeBytes(twoHyphens + boundary + lineEnd);
    dos.writeBytes("Content-Disposition: form-data; name=\"file_name\";filename=\""
        + fileName + "\"" + lineEnd);
    dos.writeBytes(lineEnd);
    // Code for sending the image....
    dos.writeBytes(lineEnd);


    dos.writeBytes(twoHyphens + boundary + lineEnd);
    dos.writeBytes("Content-Disposition: form-data; name=\"file_name_audio\";filename=\""
        + fileNameAudio + "\"" + lineEnd);
    dos.writeBytes(lineEnd);
    // Code for sending the MP3
    dos.writeBytes(lineEnd);
    dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);

Make sure that the names of both parts are different (depending on the server software).

确保两个部分的名称不同（取决于服务器软件）。