本文整理自https://stackoverflow.com/questions/62499593/
我试图在我的应用程序中实现listView
单一选择,以便一旦点击列表中的某个项目,从而使按下的项目颜色状态与其他项目不同。我已经尽力了,但是效果不好。问题在于,即使我的实现在按下时更新了每个项目状态,也不会将其他状态重置为初始状态。
class BoxSelection{
bool isSelected;
String title;
String options;
BoxSelection({this.title, this.isSelected, this.options});
}
class _AddProjectState extends State<AddProject> {
List<BoxSelection> projectType = new List();
@override
void didChangeDependencies() {
super.didChangeDependencies();
projectType
.add(BoxSelection(title: "Building", isSelected: false, options: "A"));
projectType
.add(BoxSelection(title: "Gym House", isSelected: false, options: "B"));
projectType
.add(BoxSelection(title: "School", isSelected: false, options: "C"));
}
child: ListView.builder(
itemCount: projectType.length,
itemBuilder: (BuildContext context, int index) {
return GestureDetector(
onTap: () {
setState(() {
//here am trying to implement single selection for the options in the list but it don't work well
for(int i = 0; i < projectType.length; i ) {
if (i == index) {
setState(() {
projectType[index].isSelected = true;
});
} else {
setState(() {
projectType[index].isSelected = false;
});
}
}
});
},
child: BoxSelectionButton(
isSelected: projectType[index].isSelected,
option: projectType[index].options,
title: projectType[index].title,
),
);
},
),
最佳答案
您的问题是您正在使用index
来访问projectType
元素,但是您应该使用i
if (i == index) {
setState(() {
projectType[i].isSelected = true;
});
} else {
setState(() {
projectType[i].isSelected = false;
});
}
无论如何,我认为您的代码可以改进,因为它没有效率那么高。您要遍历整个列表,并在每次迭代中两次调用setState
,一次完成一次就不必要地重新创建了小部件树很多次。
- 将当前选择保存在类级别变量中BoxSelection _selectedBox
- 简化代码,使其直接作用于迭代整个列表的当前选择 onTap: () => setState(() { if (_selectedBox != null) { _selectedBox.isSelected = false; } projectType[index].isSelected = !projectType[index].isSelected; _selectedBox = projectType[index]; });
关于flutter - 单选ListView Flutter,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/62499593/