Как написать изображение или файл для httpsurlconnection?

Question

У меня есть изображение из намерений галереи. То есть я получил uri для изображения в android. Как написать изображение или файл в httpsurlconnection для загрузки на сервер. Help.

Answer 1

InputStream iStream = context.getContentResolver().openInputStream(uri); 
//open httpurlconnection and configure. 
OutputStream oos = (OutputStream) httpConnection.getOutputStream(); 
    try 
    { 
     int c; 
     byte bufr[] = new byte[4096]; 
     while((c = iStream.read(bufr, 0, 4096)) != -1) 
     { 
      oos.write(bufr, 0, c); 
     } 
     oos.flush(); 
    } 
    finally 
    { 
     if(oos != null) 
     { 
      oos.close(); 
     } 
     iStream.close(); 
    } 

Answer 2

В коде Java:

      File sourceFile = new File("YOUR IMAGE URL") 
         FileInputStream fileInputStream = new FileInputStream(
           sourceFile); 
         URL url = new URL("your upload url"); 
         // 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 
         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("uploaded_file", filePath); 

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

         // create a buffer of maximum size 
         bytesAvailable = fileInputStream.available(); 
         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); 
         } 
         // send multipart form data necessary after file 
         // data... 
         dos.writeBytes(lineEnd); 
         dos.writeBytes(twoHyphens + boundary + twoHyphens 
           + lineEnd); 
         System.out.println("connection"); 
         // Responses from the server (code and message) 
         serverResponseCode = conn.getResponseCode(); 

         if (serverResponseCode == 200) { 
          String msg = "File Upload Completed.\n\n See uploaded file here : \n\n" 
            + " http://www.androidexample.com/media/uploads/"; 
          System.out.println("message ----" + msg); 
         } 

         fileInputStream.close(); 
         dos.flush(); 
         dos.close(); 

        } catch (FileNotFoundException e) { 
         e.printStackTrace(); 
        } catch (MalformedURLException e) { 
         e.printStackTrace(); 
        } catch (ProtocolException e) { 
         e.printStackTrace(); 
        } catch (IOException e) { 
         e.printStackTrace(); 
        } 
       } 

И если вы используете PHP API

<?php 

$file_path = "PostImage/"; 

$file_path = $file_path . basename($_FILES['uploaded_file']['name']); 
if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $file_path)) { 
    echo "success"; 
} else{ 
    echo "fail"; 
} 

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