Question

所以我有一个12 x 9网格的“按钮”，称为“瓷砖”。 Tile.java扩展了小部件“Button”。

我现在遇到了一个问题，我正试图让按钮显示一个回忆按下按钮ID的吐司。

网格是动态添加的，我希望它保持这种状态。

以下是GameBoardActivity.java的代码：

这使得12x9网格成为“tile”并为每个网格添加一个监听器。

    public void gridRowButtons(int iterations){
    final Tile[] gridSpaces = new Tile[12];
    LinearLayout.LayoutParams buttonParams = new LinearLayout.LayoutParams(LayoutParams.MATCH_PARENT, LayoutParams.MATCH_PARENT, 1.0f);
    for (int i = 0; i < 12; i++) {
        String idString = String.valueOf(aToI[iterations]) + String.valueOf(i + 1);
        final int id = getResources().getIdentifier(idString, "id", getPackageName());
        gridSpaces[i] = new Tile(getApplicationContext());
        gridSpaces[i].setText("");
        gridSpaces[i].setHeight(40);
        gridSpaces[i].setWidth(40);
        gridSpaces[i].setLayoutParams(buttonParams);
        gridSpaces[i].setId(id);
        OnClickListener showID = new OnClickListener(){
            public void onClick(View view){                 
                TextView text = (TextView) findViewById(R.id.tileIDText);
                String tileID = getApplicationContext().getResources().getResourceEntryName(id);
                text.setText(tileID);
                Tile clickedTile = (Tile) findViewById(id);
                clickedTile.tileClick(0, tileID);
            }
        };
        gridSpaces[i].setOnClickListener(showID);
    }
    LinearLayout buttonRow = new LinearLayout(getApplicationContext());
    buttonRow.setOrientation(LinearLayout.HORIZONTAL);
    buttonRow.setLayoutParams(buttonParams);
    LinearLayout boardSpace = (LinearLayout) this.findViewById(R.id.boardLayout);
    for (int i = 0; i < gridSpaces.length; i++) {
        buttonRow.addView(gridSpaces[i]);
    }
    boardSpace.addView(buttonRow);
}

这是上面提到的tileClick方法：

    public void tileClick(int action, String tileID) {
    switch(action) {
        case 1 :
            //action 1
        case 2 :
            //action 2
        default :
            Context context = getContext();
            Toast toast = new Toast(context);
            Toast.makeText(context, tileID, Toast.LENGTH_SHORT);
            toast.show();
    }
}

LogCat显示以下内容：

02-22 20:45:14.623: E/AndroidRuntime(7868): FATAL EXCEPTION: main
02-22 20:45:14.623: E/AndroidRuntime(7868): java.lang.RuntimeException: setView must have been called
02-22 20:45:14.623: E/AndroidRuntime(7868):     at android.widget.Toast.show(Toast.java:103)
02-22 20:45:14.623: E/AndroidRuntime(7868):     at com.jneal.ecquire.Tile.tileClick(Tile.java:51)
02-22 20:45:14.623: E/AndroidRuntime(7868):     at com.jneal.ecquire.GameBoardActivity$1.onClick(GameBoardActivity.java:55)
02-22 20:45:14.623: E/AndroidRuntime(7868):     at android.view.View.performClick(View.java:3627)
02-22 20:45:14.623: E/AndroidRuntime(7868):     at android.view.View$PerformClick.run(View.java:14329)
02-22 20:45:14.623: E/AndroidRuntime(7868):     at android.os.Handler.handleCallback(Handler.java:605)
02-22 20:45:14.623: E/AndroidRuntime(7868):     at android.os.Handler.dispatchMessage(Handler.java:92)
02-22 20:45:14.623: E/AndroidRuntime(7868):     at android.os.Looper.loop(Looper.java:137)
02-22 20:45:14.623: E/AndroidRuntime(7868):     at android.app.ActivityThread.main(ActivityThread.java:4511)
02-22 20:45:14.623: E/AndroidRuntime(7868):     at java.lang.reflect.Method.invokeNative(Native Method)
02-22 20:45:14.623: E/AndroidRuntime(7868):     at java.lang.reflect.Method.invoke(Method.java:511)
02-22 20:45:14.623: E/AndroidRuntime(7868):     at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:976)
02-22 20:45:14.623: E/AndroidRuntime(7868):     at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:743)
02-22 20:45:14.623: E/AndroidRuntime(7868):     at dalvik.system.NativeStart.main(Native Method)

发生了什么事？我有几乎完全相同的代码，用于在我的MainActivity.java中显示toast并且执行没有问题。是因为我扩展了Button而且它不知道视图已经是什么？另外，Eclipse不会因为某些原因让我添加setView（）。我确定我在这里遇到了封装问题，但我对它们是什么感到困惑。感谢您的帮助。 JRad

Answer 1

将下面的代码行更改为以下代码：

        Toast toast = new Toast(context);
        Toast.makeText(context, tileID, Toast.LENGTH_SHORT);
        toast.show();

更改为：

    Toast toast = Toast.makeText(context, tileID, Toast.LENGTH_SHORT);
    toast.show();

从源代码中可以看出，只有当mNextView为null时才会抛出该异常。函数“makeText”假设设置它，但确实如此，但是您的原始代码不会捕获对它构建的Toast的引用。相反，您的原始代码会创建两个Toasts，并尝试“显示”尚未设置其视图的那个。

public void show() {
    if (mNextView == null) {
        throw new RuntimeException("setView must have been called");
    }

....

public static Toast makeText(Context context, CharSequence text, int duration) {
    Toast result = new Toast(context);

    LayoutInflater inflate = (LayoutInflater)
            context.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
    View v = inflate.inflate(com.android.internal.R.layout.transient_notification, null);
    TextView tv = (TextView)v.findViewById(com.android.internal.R.id.message);
    tv.setText(text);

    result.mNextView = v;
    result.mDuration = duration;

    return result;
}

“必须已经调用了setView” - 自定义类扩展按钮

1 个答案: