Android 在 Activity 之间切换

Question

我想要做的是一个停留 5 秒并转到 Activity 1 的主屏幕。当我在 Activity 1 中单击一个按钮时，我会转到 Activity 2。我尝试了很多次单击该按钮但没有切换发生. 主屏幕(5 秒)=Main_Activity Activity1=selectpets.java Activity2=fishtank.java

onclick 监听器似乎是我不知道它有什么问题的问题

     Main Activity Code
package com.set.petshome;
import android.os.Bundle;
import android.os.Handler;
import android.app.Activity;
import android.content.Intent;
import android.view.Menu;
import android.view.View;
import android.widget.Button;

public class MainActivity extends Activity {
    Button fButton;
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        //Delay Code after 5 seconds
        new Handler().postDelayed(new Runnable() {
            @Override
            public void run() {
                setContentView(R.layout.selectscreen); //where <next> is you target      activity :)
                }
            }, 5000);   
    }
//Delay End
    @Override
    public boolean onCreateOptionsMenu(Menu menu) {
        // Inflate the menu; this adds items to the action bar if it is present.
        getMenuInflater().inflate(R.menu.main, menu);
        return true;
    }
}

现在是 Selectpets 代码

package com.set.petshome;

import android.app.Activity;
import android.content.*;
import android.os.Bundle;
import android.view.*;
import android.widget.Button;
public class SelectPetsScreen extends Activity  {
    Button fButton;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.selectscreen);

      //Button Fishtank Listener Start

        fButton = (Button) findViewById(R.id.button1);

          //Listening to button event
           fButton.setOnClickListener(new View.OnClickListener() {

                public void onClick(View arg0) {
                    //Starting a new Intent
                    Intent nextScreen = new Intent(getApplicationContext(),  fishtank.class);
                    startActivity(nextScreen);

                }
            });     
        //Button Fishtank Listener End

    }
   }

鱼缸类代码

package com.set.petshome;

import android.app.Activity;
import android.content.Intent;
import android.os.Bundle;
import android.view.Menu;
import android.view.View;
import android.widget.Button;

    public class fishtank extends Activity {
        @Override
        public void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);
            setContentView(R.layout.ftank);



        }

    }

顺便说一句，应用程序没有错误，只是点击后没有切换
非常感谢

Answer 1

在这里你永远不会切换到下一个Activity，只是改变当前Activity的layout

 new Handler().postDelayed(new Runnable() {
            @Override
            public void run() {
                setContentView(R.layout.selectscreen); //where <next> is you target      activity :)
                }
            }, 5000);  

您需要使用 Intent 而不是 setContentView()

Intent i = new Intent(MainActivity.this, SelectPetsScreen.this);
startActivity(i);

因为您实际上并没有进入下一个Activity(java 文件)，所以您的onClick() 没有设置。

编辑

这就是你做的

public class MainActivity extends Activity {
Button fButton;
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    //Delay Code after 5 seconds
    new Handler().postDelayed(new Runnable() {
        @Override
        public void run() {
            setContentView(R.layout.selectscreen); //where <next> is you target      activity :)
            }
        }, 5000);   
}

这是你应该做的。注意 run() 函数的区别

public class MainActivity extends Activity {
Button fButton;
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    //Delay Code after 5 seconds
    new Handler().postDelayed(new Runnable() {
        @Override
        public void run() {
            Intent i = new Intent(MainActivity.this, SelectPetsScreen.this);
            startActivity(i);
            }
        }, 5000);   
}