如何重复发布和接收数据，但使应用程序可响应编辑

Question

我需要反复将数据发布到 php 并取回它们，但使应用程序仍然可以响应。我的代码可以工作，但一段时间后(大约 1 分钟(我的应用程序停止响应。

起初我尝试将 POST(( 函数添加到 onResponse(( 函数(onRespose 调用 POST 函数，它一次又一次地发出新请求(，但这冻结了我的应用程序，所以我添加了每 1 毫秒调用函数的计时器

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编辑后，我的应用程序与以前相同(仅在连接到PC时才有效(，如果我打开应用程序并且我的手机未连接到PC，应用程序在1分钟后下降，所以我仍然遇到与开始时相同的问题。如果有人发现错误，请告诉我在哪里。 .

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我编辑的代码：

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主要活动.java

public class MainActivity extends AppCompatActivity implements Sync.CallBack {
final String URL = "***";
final String KEY = "***";
Data data;
RequestQueue requestQueue;
TextView textView;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
data = new Data();
data.key = KEY;
data.pressed = "0";
textView = findViewById(R.id.status);
requestQueue = Volley.newRequestQueue(this);
StartSync();
}
public void btnClick(View view) {
data.pressed = "1";
}

@Override
public void onCallBack(String data) {
textView.setText(data);
}
public void StartSync(){
Sync thread = new Sync(this,data,this);
thread.start();
}
}

同步.java

public class Sync extends Thread{
private Context context;
boolean wait = false;
final String URL = "***";
public CallBack listener;
RequestQueue requestQueue;
public Data data;
public interface CallBack{
void onCallBack(String data);
}
public Sync(CallBack listener,Data data,Context context){
this.listener = listener;
this.data = data;
this.context = context;
requestQueue = Volley.newRequestQueue(context);
};
@Override
public void run() {
while(true){
wait = true;
POST();
while(wait){}
}
}
private void POST(){
StringRequest stringRequest = new StringRequest(Request.Method.POST, URL,
new Response.Listener<String>() {
@Override
public void onResponse(String response) {
wait = false;
listener.onCallBack(response);
}
},new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
wait = false;
Toast.makeText(context,error.toString(),Toast.LENGTH_LONG).show();
}
}
){
@Override
protected Map<String,String> getParams(){
Map<String,String> params = new HashMap<String, String>();
params.put("key",data.key);
params.put("pressed",data.pressed);
if(data.pressed == "1"){
data.pressed = "0";
}
return params;
}
};
requestQueue.add(stringRequest);
}
}

数据.java

public class Data {
String pressed = "0";
String key = "";
}

Answer 1

好吧，我可以给你一个例子来实现这一点，但你真的需要研究我将使用的东西。

创建线程以发送和接收数据

public class myReceiveThread extends Thread{
public gotNewData listener;
//Create an interface to pass data to Activity
public interface gotNewData(){
void gotNewDataToDisplay(String data); //or int or data or what ever.
}
//CTor
public myThread(gotNewData listener){
this.listener = listener;
}
@Override
public void run(){
while(myAppisRunnung == true){

//Receive Data
listener.gotNewDataToDisplay("New Data");
}
}
}

在主要活动中：

//In Main activity
public class MainActivity extends AppCompactActiity implements gotNewData{
//what ever
public void startThread(){
myReceiveThread thread = new myReceiveThread(this);
thread.start();
}
@Override
public void gotNewDataToDisplay(String data){
someTextView.setText(data);
}
}

创建发送线程

public class mySendingThread extends Thread{
private BlockingQueue<String> toSendMessages= new BlockingQueue<>();
Public mySendingThread (BlockingQueue<String> toSendMessages){
this.toSendMessages = toSendMessages;
}
@Override
public void run(){
while(myAppisRunnung == true){
String message= toSendMessages.take();
//Send message
}
}
}

在主活动中

public void startSendThread(){
mySendingThread threadSend = new mySendingThread(MessageQueue);
thread.start();
}

在主活动中，您需要一个BlockingQueue<String> MessageQueue并且可以添加您想要的每条消息。