Flutter 超实用简单菜单弹出框 PopupMenuButton功能

相信在实际开发过程当中,肯定少不了这样的功能:

Flutter 超实用简单菜单弹出框 PopupMenuButton功能_第1张图片

点击 AppBar 右上角的按钮,弹出一个菜单供用户选择。

幸运的是,Flutter 提供给我们了一个 Widget,直接就能实现如上的效果。

PopupMenuButton

还是老规矩,先看官方的说明:

Displays a menu when pressed and calls onSelected [1] when the menu is dismissed because an item was selected. The value passed to  onSelected [2] is the value of the selected menu item.

One of child [3] or  icon [4] may be provided, but not both. If  icon [5] is provided, then  PopupMenuButton [6] behaves like an  IconButton [7] .

If both are null, then a standard overflow icon is created (depending on the platform).

大致意思为:

当按下的时候显示一个菜单,选择了一个项目的时候会回调 onSelected ,传递的值是所选菜单的值。

可以提供 child or  icon ,但是不能同时提供。

如果为空,则提供一个默认的图标,取决于平台。

构造函数

看完了官方说明,再来看构造函数:

const PopupMenuButton({
 Key key,
 @required this.itemBuilder,
 this.initialValue,
 this.onSelected,
 this.onCanceled,
 this.tooltip,
 this.elevation = 8.0,
 this.padding = const EdgeInsets.all(8.0),
 this.child,
 this.icon,
 this.offset = Offset.zero,
 this.enabled = true,
}) : assert(itemBuilder != null),
assert(offset != null),
assert(enabled != null),
assert(!(child != null && icon != null)), // fails if passed both parameters
super(key: key);

这里面每一个参数应该都很好理解,就不做过多的解释了,

唯一必传的参数就是 itemBuilder ,也可以看到后面的断言:

assert(!(child != null && icon != null)) 判断了 child 、icon 是否同时不为空,如果是的话就报错了。

简单 Demo

构造函数理解了,官方也提供了一个 Demo,我们来看一下运行效果:

Flutter 超实用简单菜单弹出框 PopupMenuButton功能_第2张图片

再来看一下代码:

/// 首先定义了一个枚举
enum WhyFarther {
 harder,
 smarter,
 selfStarter,
 tradingCharter,
}
/// ------------------------------------
/// build 方法
Widget build(BuildContext context) {
 return Scaffold(
 appBar: AppBar(
 title: Text('PopupMenuButtonPage'),
 actions: [
 PopupMenuButton(
 onSelected: (WhyFarther result) {
 setState(() {
 _selection = result;
 });
 },
 icon: Icon(Icons.more_vert),
 itemBuilder: (BuildContext context) => >[
 const PopupMenuItem(
 value: WhyFarther.harder,
 child: Text('Working a lot harder'),
 ),
 const PopupMenuItem(
 value: WhyFarther.smarter,
 child: Text('Being a lot smarter'),
 ),
 const PopupMenuItem(
 value: WhyFarther.selfStarter,
 child: Text('Being a self-starter'),
 ),
 const PopupMenuItem(
 value: WhyFarther.tradingCharter,
 child: Text('Placed in charge of trading charter'),
 ),
 ],
 ),
 ],
 ),
 body: Container(),
 );
}

解释一下逻辑:

1. 首先定义了一个枚举

2. 然后在  AppBar  的「actions」里定义了  PopupMenuButton

3. 设置 icon 为  Icon(Icons.more_vert)

4. itemBuilder  需返回一个  List>

5. 这里传入的值就是  PopupMenuItem

6. 然后定义  onSelected  参数接收点击回调

这样整体的逻辑就是定义好了,运行一下:

Flutter 超实用简单菜单弹出框 PopupMenuButton功能_第3张图片

总结

这样就完成了一个超级简单并且实用的菜单弹出框,

其实它的实现逻辑和 DropdownButton 差不多,都是使用了  PopupRoute ,

有对这方面感兴趣的同学,可以查看我以前写的文章: Flutter 源码系列:DropdownButton 源码浅析

完整代码已经传至GitHub:https://github.com/wanglu1209/WFlutterDemo

References

[1] onSelected: https://api.flutter.dev/flutter/material/PopupMenuButton/onSelected.html
[2] onSelected: https://api.flutter.dev/flutter/material/PopupMenuButton/onSelected.html
[3] child: https://api.flutter.dev/flutter/material/PopupMenuButton/child.html
[4] icon: https://api.flutter.dev/flutter/material/PopupMenuButton/icon.html
[5] icon: https://api.flutter.dev/flutter/material/PopupMenuButton/icon.html
[6] PopupMenuButton: https://api.flutter.dev/flutter/material/PopupMenuButton-class.html
[7] IconButton: https://api.flutter.dev/flutter/material/IconButton-class.html

总结

以上所述是小编给大家介绍的Flutter 超实用简单菜单弹出框 PopupMenuButton功能,希望对大家有所帮助,如果大家有任何疑问请给我留言,小编会及时回复大家的。在此也非常感谢大家对脚本之家网站的支持!
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