Flutter 中 showDialog 关闭后如何更新父控件？

Question

我有一个带有图标的小部件，当我单击它时，会显示一个对话框小部件，这是对对话框的调用：

// this icon is in widget parent
           IconButton(
              icon: Icon(
                Icons.info_outline,
                size: mobileWidth * 0.07,
              ),
              tooltip: 'information',
              color: Colors.blueGrey,
              onPressed: () {
                showAlertInfo(context, code);
                setState(() {});
              },
            ),

这是我的对话框：

showAlertInfo(BuildContext context, String code) {
  showDialog(
    context: context,
    builder: (context) {
      return StatefulBuilder(
        builder: (context, setState) {
          return AlertDialog(
            title: Text(
              "Information sur le client $code", ),
            content: SingleChildScrollView(
               child: Container( ..... 
               /* ...
               this dialog has some operations that changes info 
               values of the widget that called this in first place
               it is a big code to put here*/
             // here I have a close button
                    actions: [
            FlatButton(
            child: Text(
              "Fermer",
              style: TextStyle(
                color: Colors.red,
                fontSize: mobileWidth * 0.035,
                fontWeight: FontWeight.bold,
              ),
            ),
            onPressed: () {
              Navigator.of(context).pop(); // dismiss dialog
            },
          ),
        ],
      );  

我想要实现的是，当对话框关闭时，我希望更新父小部件，那么setState当对话框小部件关闭时，如何调用父小部件？

Answer 1

对于异步函数

添加await之前showDialog()并setState((){})在下一行调用。

await showDialog(
    context: context,
    builder: (context) {
       return yourWidget;
    }
);
setState((){});

对于同步功能

使用.then()回调并调用setState((){})它。

await showDialog(
    context: context,
    builder: (context) {
       return yourWidget;
    }
);
setState((){});