如何使用 asynctask 将图像从 android 上传到 php

Posted 2023-05-07

【中文标题】如何使用 asynctask 将图像从 android 上传到 php

【英文标题】：How to upload an image from android to php using asynctask

【发布时间】：2017-03-08 16:47:27

【问题描述】：

每个人。我被困在我正在从事的这个项目上。我希望能够从 android 库上传图像，将该图像编码为 base64 字符串并作为 get 变量发送到 php Web 服务，然后从另一端解码图像并按照我的意愿进行处理。

到目前为止，我能够从图库中选择图像，甚至可以编码为 base64 字符串并存储在 android 首选项中。

问题是，我认为并非所有字符串都被发送到 PHP 服务（有些被截断）。

我为什么这么认为？我的 Log.d 在转储到不同位置时向我显示了不同的字符串。

获取图像并编码的代码是：-

private void galleryIntent()

    Intent intent = new Intent();
    intent.setType("image/*");
    intent.setAction(Intent.ACTION_GET_CONTENT);
    startActivityForResult(Intent.createChooser(intent, "Please select a file"),1);


private String onSelectFromGalleryResult (Intent data) 

    if (data != null) 
        try 
            bitmap = MediaStore.Images.Media.getBitmap(getContext().getContentResolver() , data.getData()) ;
         catch (IOException e) 
            e.printStackTrace();
        

        ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream() ;

        bitmap.compress(Bitmap.CompressFormat.JPEG, 100, byteArrayOutputStream) ;

        byte[] imageBytes = byteArrayOutputStream.toByteArray() ;

        Log.d ("Selected Image Gallery" , Base64.encodeToString(imageBytes, Base64.DEFAULT)) ;

        return Base64.encodeToString (imageBytes, Base64.DEFAULT) ;
     else 
        return null ;
    


@Override
public void onActivityResult(int requestCode, int resultCode, Intent data) 
    super.onActivityResult(requestCode, resultCode, data);

    SharedPreferences sharedPreferences =  getContext().getSharedPreferences("MyOnActivityResultPref" , Context.MODE_PRIVATE);
    SharedPreferences.Editor editor = sharedPreferences.edit() ;

    if (resultCode == Activity.RESULT_OK) 

        if (requestCode == 1) 
            /*Here we handle the image gotten from the gallery*/
            String encodedGalleryImage = onSelectFromGalleryResult(data);

           editor.putString("userEncodedGalleryImage" , encodedGalleryImage);

         else if (requestCode == 0) 
            /*Here we handle the image that was take using the camera*/

        

        editor.apply();
    

这里我们称之为 asynctask 类

private void callAsynctask () 
  SharedPreferences sp = getContext().getSharedPreferences("MyOnActivityResultPref" , Context.MODE_PRIVATE);

    String userQuestionAttachement = sp.getString("userEncodedGalleryImage" , "") ;

Log.d("callingEncodedImage" , userQuestionAttachement) ;

我遇到的问题是来自 Log.d ("Selected Image Gallery" , Base64.encodeToString(imageBytes, Base64.DEFAULT)) 的日志；与 Log.d("callingEncodedImage" , userQuestionAttachement) 不同；

两者的开头相同，但结尾不同。我希望看到相同的字符。

有人可以帮我解决吗？

【问题讨论】：

base64 字符串有时可能太大而无法一次性发送。更好的选择是使用多部分上传图像文件。

你好@VivekMishra 你会怎么做？我想这就是我需要的！...

【参考方案1】：

在 Android 中，

 new UploadFileAsync().execute("");



  private class UploadFileAsync extends AsyncTask<String, Void, String> 

    @Override
    protected String doInBackground(String... params) 

        try 
  String sourceFileUri = "/mnt/sdcard/abc.png";

            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()) 

                try 
                    String upLoadServerUri = "http://website.com/abc.php?";

                    // 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("bill", sourceFileUri);

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

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

                    if (serverResponseCode == 200) 

                        // messageText.setText(msg);
                        //Toast.makeText(ctx, "File Upload Complete.",
                        //      Toast.LENGTH_SHORT).show();

                        // recursiveDelete(mDirectory1);

                    

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

                 catch (Exception e) 

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

                
                // dialog.dismiss();

             // End else block


         catch (Exception ex) 
            // dialog.dismiss();

            ex.printStackTrace();
        
        return "Executed";
    

    @Override
    protected void onPostExecute(String result) 

    

    @Override
    protected void onPreExecute() 
    

    @Override
    protected void onProgressUpdate(Void... values) 
    

在 PHP 中，

    <?php


     if (is_uploaded_file($_FILES['bill']['tmp_name'])) 
    $uploads_dir = './';
                            $tmp_name = $_FILES['bill']['tmp_name'];
                            $pic_name = $_FILES['bill']['name'];
                            move_uploaded_file($tmp_name, $uploads_dir.$pic_name);
                            
               else
                   echo "File not uploaded successfully.";
           

   ?>

【讨论】：

你好@Magash。我已经有一个 Asynctask 代码和一个 PHP 服务，我的问题介于两者之间，编码的字符串被截断。我不知道为什么...

感谢您提供 PHP 代码！这对我有用。我确实向扩展的 AsyncTask 类添加了一个构造函数，它将输入文件路径和 IP 地址作为参数。通过这个微小的更改，代码按预期工作。

【参考方案2】：

要使用 Multipart 上传图片，请按以下步骤操作：

下载 httpmime.jar 文件并将其添加到您的 libs 文件夹中。

下载 http client.jar 文件并将其添加到您的 libs 文件夹中。

从后台线程或 AsyncTask 调用以下方法。

public void executeMultipartPost() throws Exception 
    try 
        ByteArrayOutputStream bos = new ByteArrayOutputStream();
        bm.compress(CompressFormat.JPEG, 75, bos);
        byte[] data = bos.toByteArray();
        HttpClient httpClient = new DefaultHttpClient();
        HttpPost postRequest = new HttpPost(
                "YOUR SERVER URL");
        ByteArrayBody bab = new ByteArrayBody(data, "YOUR IMAGE.JPG");

        MultipartEntity reqEntity = new MultipartEntity(
                HttpMultipartMode.BROWSER_COMPATIBLE);
        reqEntity.addPart("IMAGE", bab);

        postRequest.setEntity(reqEntity);
        HttpResponse response = httpClient.execute(postRequest);
        BufferedReader reader = new BufferedReader(new InputStreamReader(
                response.getEntity().getContent(), "UTF-8"));
        String sResponse;
        StringBuilder s = new StringBuilder();

        while ((sResponse = reader.readLine()) != null) 
            s = s.append(sResponse);
        
        System.out.println("Response: " + s);
     catch (Exception e) 
        // handle exception here
        Log.e(e.getClass().getName(), e.getMessage());