我想以特定模式显示小部件。我尝试使用GridView,但似乎GridView只能在crossAxisCount上设置一个值。我希望它是3、2模式。我在this处引用了@chunhunghan答案,但该模式垂直为00、111、22、333、44、555模式。我希望它垂直为012、34、567、89模式。我已经附上了预期的输出。谢谢。
GridView.builder(
gridDelegate: SliverGridDelegateWithFixedCrossAxisCount(
crossAxisCount: 3, childAspectRatio: 1),
itemCount: 10,
itemBuilder: (context, index) {
return Column(children: <Widget>[
Icon(Icons.access_alarm,
color: Colors.redAccent, size: 100.0),
Text(index.toString())
],);
})
也许您可以只用2个项目用空白Container()
“填充”行,如果该解决方法对您有用?
GridView.builder(
gridDelegate: SliverGridDelegateWithFixedCrossAxisCount(
crossAxisCount: 3, childAspectRatio: 1),
itemCount: 10,
itemBuilder: (context, index) {
return Column(children: <Widget>[
Icon(Icons.access_alarm,
color: Colors.redAccent, size: 100.0),
Text(index.toString()),
Container()
],);
})
或检查索引是否为奇数或偶数,然后具有两个构建器函数,一个返回3个项目,另一个返回仅2个
保持简单,只需将Row
和Column
与MainAxisAlignment.center
一起使用。这是工作中的DartPad和代码:
itemBuilder(index) {
return Column(children: [
Icon(Icons.access_alarm, color: Colors.redAccent, size: 100.0),
Text(index.toString())
]);
}
rowBuilder(start, end) {
var length = end - start + 1;
var itemArray = new List<Widget>(length);
for (var index = 0; index < length; index++) {
itemArray[index] = itemBuilder(start + index);
}
return Row(
mainAxisAlignment: MainAxisAlignment.center, children: itemArray);
}
@override
Widget build(BuildContext context) {
return Column(children: [
rowBuilder(0, 2),
rowBuilder(3, 4),
rowBuilder(5, 7),
rowBuilder(8, 9),
]);
}
根据您的要求修改以下代码!更新
int itemCount = 100, lastCount = 3, lastNum = 0;
@override
void initState(){
super.initState();
itemCount++;
}
@override
Widget build(BuildContext context) {
return ListView.builder(
itemCount: ((itemCount / 3 / 2) + (itemCount / 2 / 2)).toInt(),
itemBuilder: (con, ind) {
if (lastCount == 3) {
lastCount = 2;
//Add below condition to show exact pattern by avoiding extra numbers
return (itemCount - lastNum) >= lastCount + 1
? Row(
mainAxisAlignment: MainAxisAlignment.center,
children: List.generate(lastCount + 1, (cInd) {
lastNum++;
//Below condition to show upto itemCount by showing remaining numbers in next line ! It won't work if you use above formula. i.e Showing Exact Pattern
return lastNum < itemCount + 2
? Column(mainAxisSize: MainAxisSize.min, children: [
Icon(Icons.check_circle),
Text('${lastNum - 1}')
])
: SizedBox();
}))
: SizedBox();
} else {
lastCount = 3;
//Add below condition to show exact pattern by avoiding extra numbers
return (itemCount - lastNum) >= lastCount - 1
? Row(
mainAxisAlignment: MainAxisAlignment.center,
children: List.generate(lastCount - 1, (cInd) {
lastNum++;
//Below condition to show upto itemCount by showing remaining numbers in next line ! It won't work if you use above formula. i.e Showing Exact Pattern
return lastNum < itemCount + 2
? Column(mainAxisSize: MainAxisSize.min, children: [
Icon(Icons.check_circle),
Text('${lastNum - 1}')
])
: SizedBox();
}))
: SizedBox();
}
});
}