我正在尝试写一个具有3个按钮的AlertDialog.如果不满足某个条件,我想要中间的中性按钮被禁用.
这是我的代码：
int playerint = settings.getPlayerInt();
int monsterint = settings.getMonsterInt();
AlertDialog.Builder alertBox = new AlertDialog.Builder(this);
alertBox.setMessage("You have Encountered a Monster");
alertBox.setPositiveButton("Fight!",new DialogInterface.OnClickListener() {
// do something when the button is clicked
public void onClick(DialogInterface arg0,int arg1) {
createMonster();
fight();
}
});
alertBox.setNeutralButton("Try to Outwit",int arg1) {
// This should not be static
// createTrivia();
trivia();
}
});
// Return to Last Saved CheckPoint
alertBox.setNegativeButton("Run Away!",int arg1) {
runAway();
}
});
// show the alert Box
alertBox.show();
// Intellect Check
Button button = ((AlertDialog) alertBox).getButton(AlertDialog.BUTTON_NEUTRAL);
if(monsterint > playerint) {
button.setEnabled(false);
}
}
行：Button button =((AlertDialog)alertBox).getButton(AlertDialog.BUTTON_NEUTRAL);
给出错误：无法从AlertDialog.Builder转换为
AlertDialog
我该如何解决？？我究竟做错了什么？
解决方法
您不能在AlertDialog.Builder上调用getButton().创建后必须在结果AlertDialog上调用它.换一种说法
AlertDialog.Builder alertBox = new AlertDialog.Builder(this); //...All your code to set up the buttons initially AlertDialog dialog = alertBox.create(); Button button = dialog.getButton(AlertDialog.BUTTON_NEUTRAL); if(monsterint > playerint) { button.setEnabled(false); }
构建器只是使构建对话框更容易的类,它不是实际的对话框本身.
HTH
