应用程序在一次迭代后崩溃。没有可运行的情况，它根本没有运行。现在它第一次迭代后崩溃

Question

package com.example.chirag.example; 

import android.support.v7.app.AppCompatActivity; 
import android.os.Bundle; 
import android.view.View; 
import android.widget.Button; 
import android.widget.TextView; 

import java.util.Random; 


public class MainActivity extends AppCompatActivity { 
TextView TV; 
Button upbutton; 
Button downbutton; 
TextView score; 
String n1; 
int bool=0; 
int a; 
int b; 
int sc=0; 

@Override 
protected void onCreate(Bundle savedInstanceState) 
{ 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.activity_main); 
    TextView TV = (TextView)findViewById(R.id.TV); 
    TextView score = (TextView)findViewById(R.id.score); 
    Button upbutton = (Button) findViewById(R.id.upbutton); 
    Button downbutton = (Button) findViewById(R.id.downbutton); 
    Button start = (Button) findViewById(R.id.start); 
    n1 = generate(); 
    TV.setText(n1); 
    start.setOnClickListener(new View.OnClickListener() { 
     @Override 
     public void onClick(View view) { 
      start(); 
     } 
    }); 
} 
String generate() 
{ 
    TextView TV = (TextView)findViewById(R.id.TV); 
    int n = new Random().nextInt(20); 
    String s = String.valueOf(n); 
    return s; 
} 
void start() { 
    Runnable myRunnable = new Runnable() { 
     @Override 
     public void run() 
     {while (bool == 0) { 
      final TextView TV = (TextView) findViewById(R.id.TV); 
      final TextView score = (TextView) findViewById(R.id.score); 
      Button upbutton = (Button) findViewById(R.id.upbutton); 
      Button downbutton = (Button) findViewById(R.id.downbutton); 
      Button start = (Button) findViewById(R.id.start); 
      String n2 = generate(); 
      TV.setText(n2); 

      a = Integer.valueOf(n1.toString()); 
      b = Integer.valueOf(n2.toString()); 
      if (a > b) { 
       downbutton.setOnClickListener(new View.OnClickListener() { 
        @Override 
        public void onClick(View view) { 
         String n3 = generate(); 
         TV.setText(n3); 
         a = b; 
         b = Integer.valueOf(n3.toString()); 
         score.setText("Score" + (sc + 1)); 
        } 
       }); 
       upbutton.setOnClickListener(new View.OnClickListener() { 
        @Override 
        public void onClick(View view) { 
         TV.setText("Game Over"); 
         bool = 1; 
        } 
       }); 
      } else { 
       upbutton.setOnClickListener(new View.OnClickListener() { 
        @Override 
        public void onClick(View view) { 
         String n3 = generate(); 
         TV.setText(n3); 
         a = b; 
         b = Integer.valueOf(n3.toString()); 
         score.setText("Score" + (sc + 1)); 
        } 
       }); 
       downbutton.setOnClickListener(new View.OnClickListener() { 
        @Override 
        public void onClick(View view) { 
         TV.setText("Game Over"); 
         bool = 1; 
        } 
       }); 
      } 
     } 

     } 
    }; 


    Thread myThread = new Thread(myRunnable); 
    myThread.start(); 

} 

} 

我已经把代码放在一个可运行的。逻辑似乎是正确的，但它不起作用。生成方法会创建一个新号码并将其显示在'TV'中。当点击开始时，会生成一个新号码，并将第二个号码与第一个号码进行比较。如果按下了正确的按钮（向上或向下），分数会增加并且过程继续。如果单击了错误的按钮，则“游戏结束”

Answer 1

您无法触摸另一个线程的UI。 Android中的所有UI任务应在主线程（也称为UI线程）中执行。 这条线（例如）试图从非主线程的run()命令内影响到用户界面：

TV.setText(n2); 

有关如何与UI线程是指this tutorial