我有2个Text
里面的小部件GestureDetector
. The onTap
回调仅在我触摸时通知Text
但不是我里面的空白空间Container
。如何让这个通知就像我触摸按钮一样?
+------------------------------------------------+
| Very very very long long ◎ng long text view |
| Short ◎xt ⦿ |
+------------------------------------------------+
我的源代码:
ListView.separated(
itemCount: _listModel.length,
padding: EdgeInsets.only(left: NOTIFICATION_LEFT_SPACE),
separatorBuilder: (context, index) => Divider(),
itemBuilder: (BuildContext context, int index) => GestureDetector(
child: Container(
child: Hero(
tag: _listModel[index].title,
child: Column(
crossAxisAlignment: CrossAxisAlignment.start,
children: <Widget>[
Text('Very very very long long long long text view'),
SizedBox(height: 10),
Text('Short text')
],
),
),
),
onTap: () {
print('Tapped');
},
),
)
您可以使用behavior: HitTestBehavior.opaque
的财产GestureDetector
小部件,有助于点击内部的占位符Container
即使Container
小部件没有任何孩子。
GestureDetector(
behavior: HitTestBehavior.opaque,
child: Container(
child: Hero(
tag: 'test',
child: Column(
crossAxisAlignment: CrossAxisAlignment.start,
children: <Widget>[
Text('Very very very long long long long text view'),
SizedBox(height: 10),
Text('Short text')
],
),
),
),
onTap: () {
print('Tapped');
},
),
本文内容由网友自发贡献,版权归原作者所有,本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容,请联系:hwhale#tublm.com(使用前将#替换为@)