java - Android Studio : App doesn't start with implemented Anonymous Java Class

Question

我发现了错误消息

"5-14 12:39:13.104 2518-2518/com.example.fdai3744.neueleereapp E/AndroidRuntime: FATAL EXCEPTION: main Process: com.example.fdai3744.neueleereapp, PID: 2518 java.lang.RuntimeException: Unable to instantiate activity ..."

这是我的 Java 代码

package com.example.fdai3744.neueleereapp;

import android.net.wifi.p2p.WifiP2pManager;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.TextView;


    public class MainActivity extends AppCompatActivity {

        public Button button_1 = (Button) findViewById(R.id.button1); //Button
        public TextView text1 = (TextView)findViewById(R.id.text1); // Textview

        @Override
        protected void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);
            setContentView(R.layout.activity_main);

            button_1.setOnClickListener(new View.OnClickListener() { // Here I add the ActionListener for my button

                @Override
                public void onClick(View v) {
                    text1.setText("Button 1 wurde geklickt!");
                }
            });
        }

    }

如果我启动我的应用程序，模拟器会抛出错误消息“应用程序已停止”。我应该如何防止这个错误？

Answer 1

嗯，您的 View 层次结构需要处于 Activity 状态，然后才能从中检索各个 View，并且 setContentView() 方法将其激活(或实例化)。

如何？

setContentView(View) is a method exclusively available for Activity. Internally it calls the setContentView(View) of Window. This method sets the activity content to an explicit view. This view is placed directly into the activity's view hierarchy. Calling this function "locks in" various characteristics of the window that can not, from this point forward, be changed. Hence it is called only once.

因此，不要将 View 初始化为实例变量，而是在 setContentView() 之后在 onCreate() 中实例化它们。

另请阅读:Android: setContentView and LayoutInflater