Java android studio 连接 3 游戏

Question

我正在尝试创建井字游戏，当我尝试通过按下按钮再次玩游戏时，应用程序崩溃并且不刷新表格。除此之外，工作正常。但是我找不到这次崩溃的错误所在。这是我的按钮的代码。

public void playAgain(View view) {

    gameIsActive = true;

    LinearLayout layout = (LinearLayout) findViewById(R.id.playAgainLayout);

    layout.setVisibility(View.INVISIBLE);

    activePlayer = 0;


    for (int i = 0; i < gameState.length; i++) {

        gameState[i] = 2;
    }

    GridLayout gridLayout = (GridLayout) findViewById(R.id.gridLayout);

    for (int i = 0; i < gridLayout.getChildCount(); i++) {

        ((ImageView) gridLayout.getChildAt(i)).setImageResource(0);
    }

}

我的按钮属性

 <Button
            android:id="@+id/playAgain"
            android:layout_width="wrap_content"
            android:layout_height="wrap_content"
            android:layout_centerHorizontal="true"
            android:onClick="playAgain"
            android:text="PlayAgain" />

2019-11-03 15:43:41.534 19463-19463/com.example.connect3game E/AndroidRuntime: FATAL EXCEPTION: main
    Process: com.example.connect3game, PID: 19463
    java.lang.IllegalStateException: Could not find method playAgain(View) in a parent or ancestor Context for android:onClick attribute defined on view class androidx.appcompat.widget.AppCompatButton with id 'playAgainButton'
        at androidx.appcompat.app.AppCompatViewInflater$DeclaredOnClickListener.resolveMethod(AppCompatViewInflater.java:424)
        at androidx.appcompat.app.AppCompatViewInflater$DeclaredOnClickListener.onClick(AppCompatViewInflater.java:381)
        at android.view.View.performClick(View.java:7125)
        at android.view.View.performClickInternal(View.java:7102)
        at android.view.View.access$3500(View.java:801)
        at android.view.View$PerformClick.run(View.java:27336)
        at android.os.Handler.handleCallback(Handler.java:883)
        at android.os.Handler.dispatchMessage(Handler.java:100)
        at android.os.Looper.loop(Looper.java:214)
        at android.app.ActivityThread.main(ActivityThread.java:7356)
        at java.lang.reflect.Method.invoke(Native Method)
        at com.android.internal.os.RuntimeInit$MethodAndArgsCaller.run(RuntimeInit.java:492)
        at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:930)
2019-11-03 15:43:41.556 19463-19463/com.example.connect3game I/Process: Sending signal. PID: 19463 SIG: 9

Answer 1

尝试获取对Button而不是LinearLayout的引用：

所以，而不是：

LinearLayout layout = (LinearLayout) findViewById(R.id.playAgainLayout);
layout.setVisibility(View.INVISIBLE);

试试：

Button playAgainButton = (Button) findViewById(R.id.playAgainButton);
playAgainButton.setVisibility(View.INVISIBLE);

Answer 2

public void playAgain(View view){
    Button playAgainButton = (Button) findViewById(R.id.playAgainButton);
    TextView winnerTextView = (TextView) findViewById(R.id.winnerTextView);
    //... Make the Button and Text View Visible .....
    playAgainButton.setVisibility(View.INVISIBLE);
    winnerTextView.setVisibility(View.INVISIBLE);


    GridLayout gridLayout = (GridLayout) findViewById(R.id.gridLayout);

    for (int i = 0; i < gridLayout.getChildCount(); i++) {
        ImageView imageView = (ImageView) gridLayout.getChildAt(i);
        imageView.setImageDrawable(null);
    }
    for (int i = 0; i < gameState.length; i++) {

        gameState[i] = 2;
    }
    activePayer = 0 ; //... 0 :  for yellow , 1 :  for Red...
    //... create a variable to stop the game ........
    gameActive = true;


}