不使用构造函数将StatefulWidget数据传递给State类 [英] Pass StatefulWidget data to the State class without using constructor

查看:408
本文介绍了不使用构造函数将StatefulWidget数据传递给State类的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我设法通过如下构造函数将有状态类变量的值传递给State类:

I managed to pass Stateful class variables' values to the State class through constructor like below:

import 'package:flutter/material.dart';

void main() => runApp(MyApp());

class MyApp extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    return MaterialApp(
        title: 'Demo',
        home: MyHomePage('John', 'Morison'),
    );
  }
}

class MyHomePage extends StatefulWidget {
  MyHomePage(this.fname, this.lname);

  final String fname;
  final String lname;

  @override
  _MyHomePageState createState() => _MyHomePageState(fname, lname);
}

class _MyHomePageState extends State<MyHomePage> {
  _MyHomePageState(this.fname, this.lname);

  final String fname;
  final String lname;

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      body: Center(
        child: Text('Hello $fname $lname'),
      )
    );
  }
}

这很奇怪,我不得不做很多工作有两个以上的变量。有更好的方法吗?

That's weird and I had to do lot of work as there is more than two variables. Is there a better way?

推荐答案

是的,有 widget


来自文档:

From Doc:

  /// The current configuration.
  ///
  /// A [State] object's configuration is the corresponding [StatefulWidget]
  /// instance. This property is initialized by the framework before calling
  /// [initState]. If the parent updates this location in the tree to a new
  /// widget with the same [runtimeType] and [Widget.key] as the current
  /// configuration, the framework will update this property to refer to the new
  /// widget and then call [didUpdateWidget], passing the old configuration as
  /// an argument.
  T get widget => _widget;
  T _widget;


代码应如下所示:

import 'package:flutter/material.dart';

void main() => runApp(MyApp());

class MyApp extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    return MaterialApp(
        title: 'Demo',
        home: MyHomePage('John', 'Morison'),
    );
  }
}

class MyHomePage extends StatefulWidget {
  MyHomePage(this.fname, this.lname);

  final String fname;
  final String lname;

  @override
  _MyHomePageState createState() => _MyHomePageState();
}

class _MyHomePageState extends State<MyHomePage> {

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      body: Center(
        child: Text('Hello ${widget.fname} ${widget.lname}'),
      )
    );
  }
}

这篇关于不使用构造函数将StatefulWidget数据传递给State类的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆