弹出屏幕后如何调用函数更新值? [英] How to call function to update value after poping screen in flutter?

问题描述

屏幕1:显示带有添加按钮的项目列表屏幕2:将新项目添加到列表的表单.

Screen 1: shows list of item with add button Screen 2: form to add new item to list.

屏幕2》屏幕1在屏幕2中调用navigator.pop()时，如何在屏幕1中调用方法/设置状态(以更新列表)?谁能帮我吗?

Screen 2 》 Screen 1 While calling navigator.pop() in screen 2, how to call method/setstate (to update list) in screen 1? Can anyone help me out?

我不想重新启动屏幕.我只需要运行方法以在弹出上一个屏幕后更新列表?

I didn't want to relaunch screen again.I just need to run method to update my list after popping previous screen?

推荐答案

从屏幕1导航到屏幕2时，使用 push 方法. push 方法返回一个 Future 对象.在将屏幕2从导航堆栈中弹出后，可以使用 Future 的 then 方法执行代码.

When you navigate from Screen 1 to Screen 2, you use the push method. The push method returns a Future object. You can use the then method of Future to execute your code after Screen 2 was popped from the navigation stack.

Navigator.of(context)
  .push(MaterialPageRoute(
   builder: (context) => Screen2(),
  ))
  .then((value) {
    // you can do what you need here
    // setState etc.
  });

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