Question

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”，我想返回到屏幕1，如果他们按“否”，只需关闭对话框并返回到屏幕2。我当前要做的是，当点击“是”时，我将进行两次Navigator.pop（context）。这是一个好习惯吗？有没有办法将屏幕2的上下文传递给我的对话框小部件，这样我就可以直接弹出该小窗口了？

Answer 1

我个人认为，最好将对话框中的响应传递回页面，然后让页面处理其余部分。

您可以这样做：

//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();
    });

通常，您可以将对话框提取为小部件，也可以提取用于完全处理对话框响应的函数或其他任何函数。

您可以在Flutter文档here中看到从页面返回数据的示例。

Flutter Pop最佳做法

1 个答案: