刷新ListView.builder按钮点击Flutter

Question

我正在使用下面的代码，其中我有一个带有开关的ListView。我想实现一些类似于当我点击RaisedButton时的东西——它将重新加载ListView，并且switch.value的所有值都应该更改为true或false。用户可以从ListView中的项目或单击按钮更改开关的值。我不知道应该如何更改ListView中的值或所有开关。

return Column(
children: <Widget>[
Container(
width: MediaQuery.of(context).size.width / 2,
height: 100,
padding: EdgeInsets.all(20),
child: RaisedButton(
onPressed: () {
},
child: Text(
BTN_START_TRIP,
style: new TextStyle(
fontSize: 20.0,
),
),
textColor: buttonFontColor,
color: buttonColor,
shape: new RoundedRectangleBorder(
borderRadius: new BorderRadius.circular(15.0))
),
),
Expanded(
child: ListView.builder(
padding: EdgeInsets.all(3.0),
// Let the ListView know how many items it needs to build.
itemCount: snapshot.data.results.length,
// Provide a builder function. This is where the magic happens.
// Convert each item into a widget based on the type of item it is.
itemBuilder: (context, index){
return Container(
height: 120,
child: Card(
elevation: 10,
child: InkWell(
splashColor: Colors.blue.withAlpha(30),
onTap: () {
print(snapshot.data.results[index].original_title);
},
child: Container(
height: 120,
child: Row(
children: <Widget>[
Column(
mainAxisAlignment: MainAxisAlignment.center,
crossAxisAlignment: CrossAxisAlignment.start,
children: <Widget>[
Container(
padding: EdgeInsets.fromLTRB(20, 0, 5, 0),
width: MediaQuery.of(context).size.width -90,
child: Align(
alignment: Alignment.topLeft,
child: Text(snapshot.data.results[index].original_title,
textAlign: TextAlign.left,
style: TextStyle(fontSize: defaultTitleFontsize, color: defaultFontColor),
maxLines: 5),
),
),
Container(
padding: EdgeInsets.fromLTRB(20, 0, 5, 0),
child:  Align(
alignment: Alignment.topLeft,
child: Text(snapshot.data.results[index].original_language,textAlign: TextAlign.left,style: TextStyle(fontSize: defaultsubTitleFontsize, color: defaultFontColor)),
),
),
],
),
Row(
mainAxisAlignment: MainAxisAlignment.end,
crossAxisAlignment: CrossAxisAlignment.end,
children: <Widget>[
Container(
child: Switch(
value: false,
onChanged: (value){
setState(() {
print(value);
});
}
),
),
],
),
]
),
),
),
),
);
},
),
)
],
);

Answer 1

您需要一个变量来决定开关是打开还是关闭；通过调用setState来重新触发构建(重新绘制UI(。您需要有一个有状态小部件的上述逻辑部分来实现这一点。