我正在尝试创建一种功能,当我在FlatButton
小部件内单击Expanded
时,其flex
会更改为 2 兄弟姐妹FlatButton
的弹性更改为 1 。
import 'package:flutter/material.dart';
void main() {
runApp(MyApp());
}
class MyApp extends StatelessWidget {
@override
Widget build(BuildContext context) {
return Directionality(
textDirection: TextDirection.ltr,
child: Column(
crossAxisAlignment: CrossAxisAlignment.stretch,
children: <Widget>[
new ButtonWidget(text: "Hello",selectedColor: Colors.yellow),
new ButtonWidget(text: "This is", selectedColor: Colors.red),
new ButtonWidget(text: "Button", selectedColor: Colors.blue),
],
),
);
}
}
class ButtonWidget extends StatefulWidget {
String text;
MaterialColor selectedColor;
ButtonWidget({this.text, this.selectedColor});
createState() =>
ButtonState(text: this.text, selectedColor: this.selectedColor);
}
class ButtonState extends State<ButtonWidget> {
String text;
MaterialColor selectedColor;
int _flexValue = 1;
ButtonState({this.text, this.selectedColor});
@override
Widget build(BuildContext context) {
return Expanded(
flex: _flexValue,
child: FlatButton(
color: selectedColor,
child: Text(
text,
),
onPressed: () {
setState(() {
_flexValue = 2;
});
},
),
);
}
}
我试图找到一种方法,也许以阵列或某种方式跟踪所有它们。我搜索发现InheritedWidget
方法适用于继承的小部件,而不是兄弟姐妹。
我敢肯定有一种干净的方法可以做到这一点。但是我无法动手。
答案 0 :(得分:1)
这里有另一种方法,将主窗口小部件更改为Stateful
,将按钮更改为Stateless
。
class MyApp2 extends StatefulWidget {
@override
_MyApp2State createState() => _MyApp2State();
}
class _MyApp2State extends State<MyApp2> {
int selectedIndex = -1;
onIndexChanged(int index){
setState(() {
selectedIndex = index;
});
}
@override
Widget build(BuildContext context) {
return Directionality(
textDirection: TextDirection.ltr,
child: Column(
crossAxisAlignment: CrossAxisAlignment.stretch,
children: <Widget>[
new ButtonWidget(
index: 1,
text: "Hello",
selectedColor: Colors.yellow,
selectedIndex: selectedIndex,
onChanged: onIndexChanged
),
new ButtonWidget(
index: 2,
text: "This is",
selectedColor: Colors.red,
selectedIndex: selectedIndex,
onChanged: onIndexChanged),
new ButtonWidget(
index: 3,
text: "Button",
selectedColor: Colors.blue,
selectedIndex: selectedIndex,
onChanged: onIndexChanged),
],
),
);
}
}
class ButtonWidget extends StatelessWidget {
final String text;
final MaterialColor selectedColor;
final int index;
final ValueChanged<int> onChanged;
final int selectedIndex;
ButtonWidget(
{this.text,
this.selectedColor,
this.index,
this.onChanged,
this.selectedIndex});
@override
Widget build(BuildContext context) {
return Expanded(
flex: selectedIndex != null && selectedIndex == index ? 2 : 1,
child: FlatButton(
color: selectedColor,
child: Text(
text,
),
onPressed: () => onChanged(index),
),
);
}
}
答案 1 :(得分:1)
不要尝试保持按钮本身是否处于选中状态。如您所见,很难使3个按钮的状态保持同步。在小部件树的上方找到一个位置,您可以在其中维护一次该状态。在这种情况下,它在您的应用程序中。将应用设为有状态,以便它可以记住状态,然后您的按钮无需记住它。可以告诉他们在构造函数中是选择(大)还是未选择(小)。
那么,该按钮如何告诉父项它现在已成为选定项?有多种策略,但都涉及:
尝试一下:
import 'package:flutter/material.dart';
void main() {
runApp(MyApp());
}
class MyApp extends StatefulWidget {
@override
State createState() => MyAppState();
}
class MyAppState extends State<MyApp> {
int selected = 0;
@override
Widget build(BuildContext context) {
return Directionality(
textDirection: TextDirection.ltr,
child: Column(
crossAxisAlignment: CrossAxisAlignment.stretch,
children: <Widget>[
ButtonWidget(0, flexValue(0), 'Hello', Colors.yellow, onClick),
ButtonWidget(1, flexValue(1), 'This is', Colors.red, onClick),
ButtonWidget(2, flexValue(2), 'Button', Colors.blue, onClick),
],
),
);
}
void onClick(int i) {
setState(() {
selected = i;
});
}
int flexValue(int index) => (index == selected) ? 2 : 1;
}
class ButtonWidget extends StatelessWidget {
ButtonWidget(this.index, this._flexValue, this.text, this.selectedColor,
this.notifyClick);
final Function notifyClick;
final int index;
final int _flexValue;
final String text;
final MaterialColor selectedColor;
@override
Widget build(BuildContext context) {
return Expanded(
flex: _flexValue,
child: FlatButton(
color: selectedColor,
child: Text(
text,
),
onPressed: () {
notifyClick(index);
},
),
);
}
}