Android的：如何上传.MP3文件和图像到HTTP服务器？ [英] Android:How to upload .mp3 file and image to http server?

问题描述

我的code上传图像到服务器是：

My code for uploading image to Server is :

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)

上传图片到服务器的正常工作。我需要在两个MP3文件和图像上传到server..Please帮助

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

推荐答案

所以，你要在一个HTTP请求发送多个文件？我从来没有这样做我自己，但根据 RFC ，添加另一身体向消息中发送音频，它应该是这个样子：

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:

    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);

确认这两个部位的名称是不同的（取决于服务器软件）。

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

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