如何将.mp3文件和图像上传到http服务器? [英] How to upload .mp3 file and image to http server?

问题描述

我将图片上传到服务器的代码是:

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)

    

将图像上传到服务器工作正常.我需要将mp3文件和图像都上传到服务器.

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

推荐答案

那么，您要在一个HTTP请求中发送多个文件吗?我从来没有自己做过，但是根据 RFC ，只需添加另一个发送音频消息的正文，它应该类似于以下内容:

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);

确保两个部分的名称不同(取决于服务器软件).

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

这篇关于如何将.mp3文件和图像上传到http服务器?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！