应用程序连接超时，服务器无法处理数据

Question

我构建的应用程序连接到我在 flask 中构建的本地服务器，但是 python 程序需要时间来执行并且客户端应用程序关闭连接而不使用 return 语句然后转到 catch 块并说 Failed!!!。那么我应该怎么做才能保持连接。我想显示一个处理栏或

**

服务器应在处理完成后通知应用程序

** 我想我什至可以构建另一个按钮来获取处理后的数据（所以我需要通知处理完成）所以我该怎么做。有人可以详细指导我吗？谢谢。

enter code here''' `

 import flask
 import werkzeug
 import time
 from flask import flash
 app = flask.Flask(__name__)


 @app.route('/', methods=['GET', 'POST'])
  def handle_request():
    files_ids = list(flask.request.files)
    print("\nNumber of Received Images : ", len(files_ids))
    image_num = 1
    for file_id in files_ids:
        print("\nSaving Image ", str(image_num), "/", len(files_ids))
        imagefile = flask.request.files[file_id]
        filename = werkzeug.utils.secure_filename(imagefile.filename)
        print("Image Filename : " + imagefile.filename)
        timestr = time.strftime("%Y%m%d-%H%M%S")
        imagefile.save(timestr + '_' + filename)
        image_num = image_num + 1
    print("\n")
   // flash("Connected")
   return "Image(s) Uploaded Successfully. Come Back Soon."


 app.run(host="0.0.0.0", port=5000, debug=True)

''' 这是接受传入连接的烧瓶代码。

 public void connectServer(View v) {
    TextView responseText = findViewById(R.id.responseText);
    if (imagesSelected == false) { // This means no image is selected and thus nothing to upload.
        responseText.setText("No Image Selected to Upload. Select Image(s) and Try Again.");
        return;
    }
    responseText.setText("Sending the Files. Please Wait ...");

    EditText ipv4AddressView = findViewById(R.id.IPAddress);
    String ipv4Address = ipv4AddressView.getText().toString();
    EditText portNumberView = findViewById(R.id.portNumber);
    String portNumber = portNumberView.getText().toString();

    Matcher matcher = IP_ADDRESS.matcher(ipv4Address);
    if (!matcher.matches()) {
        responseText.setText("Invalid IPv4 Address. Please Check Your Inputs.");
        return;
    }

    String postUrl = "http://" + ipv4Address + ":" + portNumber + "/";

    MultipartBody.Builder multipartBodyBuilder = new MultipartBody.Builder().setType(MultipartBody.FORM);

    for (int i = 0; i < selectedImagesPaths.size(); i++) {
        BitmapFactory.Options options = new BitmapFactory.Options();
        options.inPreferredConfig = Bitmap.Config.RGB_565;

        ByteArrayOutputStream stream = new ByteArrayOutputStream();
        try {
            // Read BitMap by file path.
            Bitmap bitmap = BitmapFactory.decodeFile(selectedImagesPaths.get(i), options);
            bitmap.compress(Bitmap.CompressFormat.JPEG, 100, stream);
        }catch(Exception e){
            responseText.setText("Please Make Sure the Selected File is an Image.");
            return;
        }
        byte[] byteArray = stream.toByteArray();

        multipartBodyBuilder.addFormDataPart("image" + i, "Android_Flask_" + i + ".jpg", RequestBody.create(MediaType.parse("image/*jpg"), byteArray));
    }

    RequestBody postBodyImage = multipartBodyBuilder.build();

    postRequest(postUrl, postBodyImage);
}

void postRequest(String postUrl, RequestBody postBody) {

    OkHttpClient client = new OkHttpClient();

    Request request = new Request.Builder()
            .url(postUrl)
            .post(postBody)
            .build();

    client.newCall(request).enqueue(new Callback() {
        @Override
        public void onFailure(Call call, IOException e) {
            // Cancel the post on failure.
            call.cancel();
            Log.d("FAIL", e.getMessage());

            // In order to access the TextView inside the UI thread, the code is executed inside runOnUiThread()
            runOnUiThread(new Runnable() {
                @Override
                public void run() {
                    TextView responseText = findViewById(R.id.responseText);
                    responseText.setText("Failed to Connect to Server. Please Try Again.");
                }
            });
        }

        @Override
        public void onResponse(Call call, final Response response) throws IOException {
            // In order to access the TextView inside the UI thread, the code is executed inside runOnUiThread()
            runOnUiThread(new Runnable() {
                @Override
                public void run() {
                    TextView responseText = findViewById(R.id.responseText);
                    try {
                        responseText.setText("Server's Response\n" + response.body().string());
                    } catch (IOException e) {
                        e.printStackTrace();
                    }
                }
            });
        }
    });
}

''' 这是我的 android 工作室连接服务器代码

Answer 1

要从服务器获取图像到您的应用程序，请使用

 Picasso.get()
   .load("*The link of your server*")
   .into(imageView);

应用程序连接超时，服务器无法处理数据

The App connection is timing out before the server could process the data

connection

session

android

push-notification

flask