Question

这个问题有点长，所以可能会有混淆，无论如何，如果你有，请不要犹豫，让我知道..

我正在尝试制作代码以将视频上传到网络服务器。为此，我将代码的角色分为三个（通常不包括细节）：

上传视频的活动和名为VideoUploader的内部类，其扩展AsyncTask以执行上传过程，
一个名为VideoManager.java的经理类，负责使用POST方法上传视频，
和PHP部分获取POST视频并将其存储在服务器中。

以下是活动中的uploadVideo()方法。

public void uploadVideo(final String stringPath) {

    class VideoUploader extends AsyncTask<Void, Void, String> {

        ProgressDialog pDialog;

        @Override
        protected void onPreExecute() {
            super.onPreExecute();
            pDialog = new ProgressDialog(UploadVideoActivity.this);
            pDialog.setMessage("Uploading");
            pDialog.setIndeterminate(true);
            pDialog.setProgress(0);
            pDialog.show();

            final int totalProgressTime = 100;
            final Thread thread = new Thread() {
                @Override
                public void run() {
                    int jumpTime = 0;
                    while(jumpTime < totalProgressTime) {
                        try {
                            sleep(200);
                            jumpTime += 5;
                            pDialog.setProgress(jumpTime);
                        } catch(InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                }
            };
            thread.start();
        }

        @Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);
            pDialog.dismiss();
        }

        @Override
        protected String doInBackground(Void... params) {
            VideoManager videoManager = new VideoManager(getApplicationContext());
            final String msg = videoManager.getUploadVideoResponse(stringPath);

            Handler handler = new Handler(Looper.getMainLooper());
            handler.post(new Runnable() {
                @Override
                public void run() {
                    Toast.makeText(getApplicationContext(), msg, Toast.LENGTH_SHORT).show();
                }
            });
            return msg;
        }
    }
    VideoUploader uv = new VideoUploader();
    uv.execute();
}

以下代码是VideoManager.java类。

public class VideoManager {
    private Context context;
    private int serverResponseCode;
    private String urlUploadVideo = ServerURL.UPLOAD_VIDEO;

    public VideoManager(Context context) {
        this.context = context;
    }

    public String getUploadVideoResponse(String filePath) {

        HttpURLConnection conn = null;
        DataOutputStream dos = null;
        String lineEnd = "\r\n";
        String twoHyphens = "--";
        String boundary = "*****";
        int bytesRead, bytesAvailable, bufferSize;
        byte[] buffer;

        /*
          * a buffer is a temporary memory area in which data is stored while it is being processed
          * or transferred, especially one used while streaming video or downloading audio.
          * here the maximum buffer size is 1024 bytes * 1024 bytes = 1048576 bytes ≈ 1.05 megabytes.
          */
        int maxBufferSize = 1 * 1024 * 1024;

        File sourceFile = new File(filePath);

        // check if the source file is a proper file.
        if (!sourceFile.isFile()) {
            Toast.makeText(context, "File does not exist.", Toast.LENGTH_SHORT).show();
            return null;
        }

        try {

            // an input stream that reads bytes from a file
            FileInputStream fileInputStream = new FileInputStream(sourceFile);
            URL url = new URL(urlUploadVideo);
            conn = (HttpURLConnection) url.openConnection();

            // allow the url connection input
            conn.setDoInput(true);
            // allow the url connection output
            conn.setDoOutput(true);
            // not allow the url connection to use cache
            conn.setUseCaches(false);
            conn.setRequestMethod("POST");

            // set the value of the requested header field
            conn.setRequestProperty("Connection", "Keep-Alive");
            conn.setRequestProperty("ENCTYPE", "multipart/form-data");
            conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
            conn.setRequestProperty("myFile", filePath);

            // a data output stream lets an application write primitive Java data types to an output stream in a portable way.
            // an application can then use a data input stream to read the data back in.
            dos = new DataOutputStream(conn.getOutputStream());

            dos.writeBytes(twoHyphens + boundary + lineEnd);
            dos.writeBytes("Content-Disposition: form-data; name=\"myFile\";filename=\"" + filePath + "\"" + lineEnd);
            dos.writeBytes(lineEnd);

            bytesAvailable = fileInputStream.available();
            Log.i("Huzza", "Initial .available : " + bytesAvailable);

            bufferSize = Math.min(bytesAvailable, maxBufferSize);
            buffer = new byte[bufferSize];

            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);
            }

            dos.writeBytes(lineEnd);
            dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);

            serverResponseCode = conn.getResponseCode();

            fileInputStream.close();

            // flush the data output stream.
            dos.flush();
            dos.close();
        } catch (MalformedURLException ex) {
            ex.printStackTrace();
        } catch (Exception e) {
            Toast.makeText(context, e.getMessage(), Toast.LENGTH_SHORT).show();
        }

        // if the server responds OK
        String response;
        if (serverResponseCode == 200) {
            StringBuilder sb = new StringBuilder();
            try {
                BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(conn.getInputStream()));
                String line;
                while ((line = bufferedReader.readLine()) != null) {
                    sb.append(line);
                }
                bufferedReader.close();
            } catch (IOException e) {
                Toast.makeText(context, e.getMessage(), Toast.LENGTH_SHORT).show();
            }
            response = sb.toString();
            return response;
        } else {
            response = "An error occurred while uploading the video.";
            return response;
        }
    }
}

最后，这是接收视频并将其存储在服务器中的PHP端。

<?php

$base_location = dirname(__FILE__);
$directory = '/uploaded_videos/';
$location = $base_location . $directory;

if($_SERVER['REQUEST_METHOD'] == 'POST') {
    $file_name = $_FILES['video']['name'];
    $file_size = $_FILES['video']['size'];
    $file_type = $_FILES['video']['type'];
    $temp_name = $_FILES['video']['tmp_name'];

    move_uploaded_file($temp_name, $location.$file_name);

} else {
    echo $location;
}
?>

如果我运行程序并执行uploadVideo()方法，我设法从public String getUploadVideoResponse(String filePath)方法获取字符串值，如下所示。

根据我的诊断，现在我收到字符串响应，服务器响应代码为200，这意味着请求成功。但我仍然无法在服务器中拥有视频文件。任何超人都可以在这里找到真正的问题吗？

Android：将视频上传到服务器

0 个答案: