我已经制作了一个警报对话框,用户可以在其中更新其个人资料。其中有图片容器和图标按钮小部件。 我想要的是,当用户单击图标按钮时,会显示弹出菜单,并提供添加/删除图像选项。以下是我的警报对话框代码:
showDialog<void>(
builder: (BuildContext context) {
return AlertDialog(
title: Text('Update details'),
shape: RoundedRectangleBorder(borderRadius: BorderRadius.all(Radius.circular(8.0))),
content: StatefulBuilder(
builder: (context, setState) { return Container(
width: 400,
child: Form(
key: _formKey,
child: Column(
mainAxisAlignment: MainAxisAlignment.spaceAround,
children: <Widget>[
Stack(
alignment: Alignment.center,
children: [
Container(
width: 100.0,
height: 100.0,
decoration: new BoxDecoration(
shape: BoxShape.circle,
image: new DecorationImage(
fit: BoxFit.cover,
colorFilter: new ColorFilter.mode(Colors.black.withOpacity(0.2), BlendMode.darken),
image: data != null ? MemoryImage(data) : AssetImage("web/icons/contactsDefaultImage.png")
)
)
),
IconButton(icon: Icon(Icons.edit), onPressed: () async {
//display option here
_showPopupMenu();
})
]),
Container(
child: TextFormField(
decoration: InputDecoration(
labelText: 'name'
),
),
),
TextFormField(
decoration: InputDecoration(
labelText: 'email'
),
),
],
),
),
);},
),
actions: <Widget>[
FlatButton(
child: Text('Cancel'),
onPressed: () {
Navigator.of(context).pop();
},
),
FlatButton(child: Text('Save'),
onPressed: () {
// save
},
)
],
);
},
);
我试过使用showMenu来实现,但由于位置必须硬编码,所以我不想使用它。 我尝试过:
void _showPopupMenu() async {
await showMenu(
context: context,
position: RelativeRect.fromLTRB(100, 100, 100, 100),
items: [
PopupMenuItem(
child: Text("add"),
),
PopupMenuItem(
child: Text("remove"),
),
],
elevation: 8.0,
);
现在,我想知道如何在图标按钮被点击时显示它(而不是硬编码值)。是否有另一种方法来实现它,即不使用showMenu。