弹出第二页后触发FirstPage的setState方法 [英] Trigger FirstPage's setState method after popping Second page

问题描述

myapp中有两个页面，分别是FirstPage和SecondPage.

I have two pages in myapp namely FirstPage and SecondPage.

基本上，FirstPage小部件显示带有项目列表的ListView，而SecondPage小部件是我可以在列表中添加/删除项目的位置.

Basically, FirstPage widget displays a ListView with a list of items while SecondPage widget is where I can add/delete items to the list.

我可以通过以下方式导航到SecondPage:

I can navigate to the SecondPage by:

Navigator.of(context).pushNamed("/SecondPage");

我也可以使用以下命令返回到第一页:

And also I can go back to the FirstPage by using:

Navigator.of(context).pop();

我的问题是，在弹出SecondPage以便更新FirstPage的ListView之后，我不知道如何触发FirstPage小部件的setState方法.

My problem is I can't figure out how am I going to trigger the setState method of FirstPage widget after popping the SecondPage so that the FirstPage's ListView is updated.

任何提示，我将不胜感激.

I would appreciate any hint.

推荐答案

每当我们推动时，我们都会获得未来，我们可以利用这一未来来触发setState

Whenever we push, we will get a future, we can use that future to trigger the setState

Future pushNamed = Navigator.of(context).pushNamed("/SecondPage");
pushNamed.then((_)=> setState(() {}));

在此处进行引用，以将数据从secondScreen发送到firstScreen.

Refer here to send data from secondScreen to firstScreen.

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