從Android設備上傳文件到遠程服務器

Question

我已經檢查了很多問題和答案，將數據上傳到遠程服務器，但大多數都使用已棄用的HttpClient。 我想上傳一個簡單的sqlite數據庫文件，以後可以下載並重新安裝應用程序時恢復。 我知道我應該使用Volley庫，但我只知道如何使用JSONObject而不是文件。請幫忙。

注意：該文件已被複制到我的外部存儲器。

Answer 1

試試這個代碼..

public int uploadFile(String sourceFileUri) { 


     String fileName = sourceFileUri; 

     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; 
     File sourceFile = new File(sourceFileUri); 

     if (!sourceFile.isFile()) { 

      dialog.dismiss(); 

      Log.e("uploadFile", "Source File not exist :" 
           +uploadFilePath + "" + uploadFileName); 

      runOnUiThread(new Runnable() { 
       public void run() { 
        messageText.setText("Source File not exist :" 
          +uploadFilePath + "" + uploadFileName); 
       } 
      }); 

      return 0; 

     } 
     else 
     { 
      try { 

       // open a URL connection to the Servlet 
       FileInputStream fileInputStream = new FileInputStream(sourceFile); 
       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 
       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", fileName); 

       dos = new DataOutputStream(conn.getOutputStream()); 

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

       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); 

       // Responses from the server (code and message) 
       serverResponseCode = conn.getResponseCode(); 
       String serverResponseMessage = conn.getResponseMessage(); 

       Log.i("uploadFile", "HTTP Response is : " 
         + serverResponseMessage + ": " + serverResponseCode); 

       if(serverResponseCode == 200){ 

        runOnUiThread(new Runnable() { 
         public void run() { 

          String msg = "File Upload Completed.\n\n See uploaded file here : \n\n" 
              +" http://www.androidexample.com/media/uploads/" 
              +uploadFileName; 

          messageText.setText(msg); 
          Toast.makeText(UploadToServer.this, "File Upload Complete.", 
             Toast.LENGTH_SHORT).show(); 
         } 
        });     
       }  

       //close the streams // 
       fileInputStream.close(); 
       dos.flush(); 
       dos.close(); 

      } catch (MalformedURLException ex) { 

       dialog.dismiss(); 
       ex.printStackTrace(); 

       runOnUiThread(new Runnable() { 
        public void run() { 
         messageText.setText("MalformedURLException Exception : check script url."); 
         Toast.makeText(UploadToServer.this, "MalformedURLException", 
                  Toast.LENGTH_SHORT).show(); 
        } 
       }); 

       Log.e("Upload file to server", "error: " + ex.getMessage(), ex); 
      } catch (Exception e) { 

       dialog.dismiss(); 
       e.printStackTrace(); 

       runOnUiThread(new Runnable() { 
        public void run() { 
         messageText.setText("Got Exception : see logcat "); 
         Toast.makeText(UploadToServer.this, "Got Exception : see logcat ", 
           Toast.LENGTH_SHORT).show(); 
        } 
       }); 
       Log.e("Upload file to server Exception", "Exception : " 
               + e.getMessage(), e); 
      } 
      dialog.dismiss();  
      return serverResponseCode; 

     } // End else block 
    } 
} 

這是PHP代碼：

<?php 

     $file_path = "uploads/"; 

     $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"; 
    } 
    ?> 

瞭解更多詳情嘗試this例子。

Answer 2

您必須創建一個VolleyMultipartRequest。您可以在Working POST Multipart Request with Volley and without HttpEntity

中找到示例實現您在創建DataPart時需要輸入文件類型，而重寫getByteData（）方法。