我正在开发一个无障碍应用.当用户想离开应用程序时,我会显示一个对话框,他必须在其中确认他想离开,如果他在 5 秒后未确认,则对话框应自动关闭(因为用户可能不小心打开了它).这类似于您更改屏幕分辨率时在 Windows 上发生的情况(出现警告,如果您不确认,它会恢复到以前的配置).
I'm working on an accesibility app. When the user wants to leave the app I show a dialog where he has to confirm he wants to leave, if he doesn't confirm after 5 seconds the dialog should close automatically (since the user probably opened it accidentally). This is similar to what happens on Windows when you change the screen resolution (an alert appears and if you don't confirm it, it reverts to the previous configuration).
这就是我显示对话框的方式:
This is how I show the dialog:
AlertDialog.Builder dialog = new AlertDialog.Builder(this).setTitle("Leaving launcher").setMessage("Are you sure you want to leave the launcher?");
dialog.setPositiveButton("Confirm", new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int whichButton) {
exitLauncher();
}
});
dialog.create().show();
如何在显示 5 秒后关闭对话框?
How can I close the dialog 5 seconds after showing it?
final AlertDialog.Builder dialog = new AlertDialog.Builder(this).setTitle("Leaving launcher").setMessage("Are you sure you want to leave the launcher?");
dialog.setPositiveButton("Confirm", new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int whichButton) {
exitLauncher();
}
});
final AlertDialog alert = dialog.create();
alert.show();
// Hide after some seconds
final Handler handler = new Handler();
final Runnable runnable = new Runnable() {
@Override
public void run() {
if (alert.isShowing()) {
alert.dismiss();
}
}
};
alert.setOnDismissListener(new DialogInterface.OnDismissListener() {
@Override
public void onDismiss(DialogInterface dialog) {
handler.removeCallbacks(runnable);
}
});
handler.postDelayed(runnable, 10000);
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