Flutter Pop 最佳实践

Flutter pop best practice

我有以下流程屏幕 1 -> 屏幕 2 -> 对话框(在单独的小部件中)。

屏幕 2 显示一个对话框(关闭?是或否)。如果有人按是,我想 return 到屏幕 1,如果他们按否,只需关闭对话框并 return 到屏幕 2。我现在做的是当点击是时,我做Navigator.pop(上下文) 两次。这是一个好习惯吗?有没有办法将屏幕 2 的上下文传递到我的对话框小部件,以便我可以直接弹出那个?

就我个人而言,我认为将对话框的响应传递回页面,让页面处理其余部分会更好。

你可以这样做:

//I'm using a raised button just to call the alert as an example... 
RaisedButton(
    child: Text('Press me'),
    //This part here is the important part
    onPressed: () async {

      //You can return anything when you use Navigator.pop
      //In this case I'm returning a bool indicating if the page should close or not.
      //You have to await this because it depends on user input.
      bool shouldPopResult = await showDialog<bool>(
        context: context,
        builder: (context) => AlertDialog(
          //The content of your dialog

          actions: <Widget>[
            // The value you pass here in Navigator.of(context).pop
            // is the value that will be stored in shouldPopResult,
            // so if "Yes" is pressed, true will return...
            // and if "No", false is returned.
            FlatButton(
              child: Text('Yes'),
              onPressed: () => Navigator.of(context).pop(true),
            ),
            FlatButton(
              child: Text('No'),
              onPressed: () => Navigator.of(context).pop(false),
            )
          ],
        ),
      );

      // This is for if the user dismisses the dialog without pressing a button
      // In that case shouldPopResult would be null, so I'm setting it to false.
      // You can prevent the user from dismissing the dialog
      // setting barrierDismissible to false in the showDialog method.
      if (shouldPopResult == null) shouldPopResult = false;

      // And finally with the dialog already dismissed, you can decide 
      // to go back or not.
      if (shouldPopResult) Navigator.of(context).pop();
    });

像往常一样,您可以将对话框提取为 Widget,或者完全提取处理对话框响应的功能或其他任何功能。

在flutter文档中可以看到页面返回数据的例子here