如何使用HttpUrl連接在服務器上上傳圖片和視頻？

Question

我想僅使用url連接將視頻連同該視頻的縮略圖一起上傳到服務器。我可以上傳視頻，但我不知道如何同時上傳圖片。

這裏是我的代碼，以供參考：

  HttpURLConnection conn = null; 
      DataOutputStream dos = null; 
      DataInputStream inStream = null; 
      String existingFileName = selectedPath; 
      String lineEnd = "\r\n"; 
      String twoHyphens = "--"; 
      String boundary = "*****"; 
      int bytesRead, bytesAvailable, bufferSize; 
      byte[] buffer; 
      int maxBufferSize = 1*1024*1024; 
      String responseFromServer = ""; 
      String urlString = "http://xxx/yyy/zzz/hh.php; //path of the php file on server 


      try 
      { 
       //------------------ CLIENT REQUEST 
       FileInputStream fileInputStream = new FileInputStream(new File(existingFileName)); 

       // open a URL connection to the Servlet 
       URL url = new URL(urlString); 

       // Open a HTTP connection to the URL 
       conn = (HttpURLConnection) url.openConnection(); 

       // Allow Inputs 
       conn.setDoInput(true); 

       // Allow Outputs 
       conn.setDoOutput(true); 

       // Don't use a cached copy. 
       conn.setUseCaches(false); 

       // Use a post method. 
       conn.setRequestMethod("POST"); 
       conn.setRequestProperty("Connection", "Keep-Alive"); 
       conn.setRequestProperty("Content-Type", "multipart/form-data;boundary="+boundary); 


       bmThumbnail = ThumbnailUtils.createVideoThumbnail(existingFileName, 
          Thumbnails.MICRO_KIND); 


       dos = new DataOutputStream(conn.getOutputStream()); 
       dos.writeBytes(twoHyphens + boundary + lineEnd); 
       dos.writeBytes("Content-Disposition: form-data;name=\"uploadedfile\";filename=\"" + existingFileName +"\"" + lineEnd); 
       // dos.writeBytes(bmThumbnail); 
       dos.writeBytes(lineEnd); 

       // create a buffer of maximum size 
       bytesAvailable = fileInputStream.available(); 
       bufferSize = Math.min(bytesAvailable, maxBufferSize); 
       buffer = new byte[bufferSize]; 

       // read file and write it into form... 
       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); 
       } 



       // send multipart form data necesssary after file data... 
       dos.writeBytes(lineEnd); 
       dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd); 

       //check server response 
       int serverResponseCode = conn.getResponseCode(); 
       String serverResponseMessage = conn.getResponseMessage(); 
       Log.i("file sent to:" +existingFileName,"Code:" + serverResponseCode +"Message: "+ serverResponseMessage); 

       // close streams 
       Log.e("Debug","File is written"); 
       fileInputStream.close(); 
       dos.flush(); 
       dos.close(); 

有什麼建議？提前致謝。

Answer 1

使用multipart http post。您可以使用庫來完成它，或者，您可以手動編寫http請求。