将图片网址和视频网址从Android应用程序发送到php服务器

问题描述：

In my android app UI, user can select a video from gallery, after that i am fetching the selected video uri and creating a thumbnail of the video and storing it in users sdcard and getting the uri of the image.

Now i am using the below code to send the video to the server --

//Store video to server
                FileInputStream fileInputStream = new FileInputStream(selectedFile);
                URL url = new URL(SERVER_URL+"?emailid="+uEmailID);
                connection = (HttpURLConnection) url.openConnection();
                connection.setDoInput(true);//Allow Inputs
                connection.setDoOutput(true);//Allow Outputs
                connection.setUseCaches(false);//Don't use a cached Copy
                connection.setRequestMethod("POST");
                connection.setRequestProperty("Connection", "Keep-Alive");
                connection.setRequestProperty("ENCTYPE", "multipart/form-data");
                connection.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
                connection.setRequestProperty("uploaded_file",selectedFilePath);

                //creating new dataoutputstream
                dataOutputStream = new DataOutputStream(connection.getOutputStream());

                //writing bytes to data outputstream
                dataOutputStream.writeBytes(twoHyphens + boundary + lineEnd);
                dataOutputStream.writeBytes("Content-Disposition: form-data; name=\"uploaded_file\";filename=\""
                        + selectedFilePath + "\"" + lineEnd);

                dataOutputStream.writeBytes(lineEnd);

                //returns no. of bytes present in fileInputStream
                bytesAvailable = fileInputStream.available();
                //selecting the buffer size as minimum of available bytes or 1 MB
                bufferSize = Math.min(bytesAvailable,maxBufferSize);
                //setting the buffer as byte array of size of bufferSize
                buffer = new byte[bufferSize];

                //reads bytes from FileInputStream(from 0th index of buffer to buffersize)
                bytesRead = fileInputStream.read(buffer,0,bufferSize);

                //loop repeats till bytesRead = -1, i.e., no bytes are left to read
                while (bytesRead > 0){
                    //write the bytes read from inputstream
                    dataOutputStream.write(buffer,0,bufferSize);
                    bytesAvailable = fileInputStream.available();
                    bufferSize = Math.min(bytesAvailable,maxBufferSize);
                    bytesRead = fileInputStream.read(buffer,0,bufferSize);
                }

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

                serverResponseCode = connection.getResponseCode();
                String serverResponseMessage = connection.getResponseMessage();

                Log.i(TAG, "Server Response is: " + serverResponseMessage + ": " + serverResponseCode);

                //response code of 200 indicates the server status OK
                if(serverResponseCode == 200){
                    runOnUiThread(new Runnable() {
                        @Override
                        public void run() {
                            textViewResponse.setText("File Upload completed.

");
                        }
                    });
                }

The video is successfully uploaded to server, its working. Now i want to send both the video and image url together, is that possible using my above code, any hint or idea is welcome.

答

It should be something like this:

FileInputStream imageStream = new FileInputStream(imageFile);
//Part below should go after the previous file boundary
 dataOutputStream.writeBytes("Content-Disposition: form-data; name=\"uploaded_image\";filename=\""+ imageFilePath + "\"" + lineEnd);
//Change this to whatever the content type is
 dataOutputStream.writeBytes("Content-Type: image/png"+lineEnd);

 dataOutputStream.writeBytes(lineEnd);

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

 bytesRead = imageStream.read(buffer,0,bufferSize);

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

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

In php the file should be available in the $_FILES['uploaded_image'] superglobal.

The multipart/form-data raw request body will look like this:

-----------------------boundary
Content-Disposition: form-data; name="uploaded_file"; filename="video.mp4"
Content-Type: video/mp4

[the bytes]
-----------------------boundary
Content-Disposition: form-data; name="uploaded_image"; filename="filename.png"
Content-Type: image/png

[the bytes]
-----------------------boundary--

Both can have name="uploaded_file" and in PHP your $_FILES['uploaded_file'] will contain details of both files similar to uploading multiple files. Forgot to add, only the last boundary should have the two extra hyphens suffixed.

答

add the into an object model and parse it to json using Gson object then send the POST request as a json:

Data data = new Data();
data.setImageUrl(imageUrl);
data.setVideoUrl(videoUrl);
String jsonObject = new Gson().toJson(data);
// send the jsonObject to server here

将图片网址和视频网址从Android应用程序发送到php服务器

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