Flutter 防止在按下后关闭对话框

Question

我正在使用showDialog函数来构建一个对话框，但是我需要避免当用户按下后退按钮时，对话框没有关闭，这是我的代码:

showDialog(
      barrierDismissible: false,
      context: context,

      builder: (BuildContext context) {
        // return object of type Dialog
        return AlertDialog(
          title: new Text("Hello"),
          content: new SingleChildScrollView(
            child: Container(),
          actions: <Widget>[
            // usually buttons at the bottom of the dialog

            new FlatButton(
              child: new Text("Close"),
              onPressed: () {
              },
            ),
          ],
        );
      },
    );

我怎样才能让它不关闭？

Answer 1

您需要将 AlertDialon 附在 WillPopScope 上像这样:

showDialog(
      barrierDismissible: false,
      context: context,

      builder: (BuildContext context) {
        // return object of type Dialog
        return WillPopScope(
            onWillPop: (){},
            child:AlertDialog(
            title: new Text("Hello"),
            content: new SingleChildScrollView(
              child: Container(),),
            actions: <Widget>[
              // usually buttons at the bottom of the dialog
              new FlatButton(
                child: new Text("Close"),
                onPressed: () {
                },
              ),
            ],
          )
        )
      },
    );

WillPopScope为您提供 onWillPop参数，您可以在其中传递 child 弹出时所需的函数。在这种情况下，参数接收一个空函数，因此它不会弹出。