Display pop up menu when icon button widget is clicked Flutter - flutter

Question

I have made an alert dialog where user can update their profile details. In that with image container there is icon button widget. What I want is that when user clicks icon button, pop up menu will display with add/remove image option. Here is my code for alert dialog:
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
},
)
],
);
},
);
I tried to user showMenu for that. But as the position has to be hard-coded I don't was to use it.
what I tried:
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,
);
}
Now, what i want to know is how can i display it where the icon-button is tapped (without hard-coding the value). And is there another way to do it .i.e without using showMenu.

Answer 1

You can write a method like this and call it on your icon button's onPressed
showPopupMenu(){
showMenu<String>(
context: context,
position: RelativeRect.fromLTRB(25.0, 25.0, 0.0, 0.0), //position where you want to show the menu on screen
items: [
PopupMenuItem<String>(
child: const Text('menu option 1'), value: '1'),
PopupMenuItem<String>(
child: const Text('menu option 2'), value: '2'),
PopupMenuItem<String>(
child: const Text('menu option 3'), value: '3'),
],
elevation: 8.0,
)
.then<void>((String itemSelected) {
if (itemSelected == null) return;
if(itemSelected == "1"){
//code here
}else if(itemSelected == "2"){
//code here
}else{
//code here
}
});
}
Edit: (to show menu at the position where user tapped)
We can have a method like so -
void showPopUpMenuAtTap(BuildContext context, TapDownDetails details) {
showMenu(
context: context,
position: RelativeRect.fromLTRB(
details.globalPosition.dx,
details.globalPosition.dy,
details.globalPosition.dx,
details.globalPosition.dy,
),
// other code as above
);
}
and use it with GestureDetector like so -
GestureDetector(
child: const Icon(Icons.menu),
onTapDown: (details) => showPopUpMenuAtPosition(context, details),
);

Answer 2

Solution if you wish to re-use your button and not a Gesture detector:
Create a key and assign your button the key. Then:
TextButton(
key: _accKey,
text: "Account",
onPressed: () {
final RenderBox renderBox =
_accKey.currentContext?.findRenderObject() as RenderBox;
final Size size = renderBox.size;
final Offset offset = renderBox.localToGlobal(Offset.zero);
showMenu(
context: context,
position: RelativeRect.fromLTRB(
offset.dx,
offset.dy + size.height,
offset.dx + size.width,
offset.dy + size.height),
items: [
PopupMenuItem<String>(
child: const Text('menu option 1'), value: '1'),
PopupMenuItem<String>(
child: const Text('menu option 2'), value: '2'),
PopupMenuItem<String>(
child: const Text('menu option 3'), value: '3'),
]);
}),

Answer 3

what you are looking to is showdialog and alertdialog.
Void<String> testdialog(BuildContext context) {
return showDialog(
barrierDismissible: false,
context: context,
builder: (context) {
return StatefulBuilder(builder: (context, setState) {
return AlertDialog(
title: ....

Answer 4

There are a lot of options you can choose:
You can use:
Banner
Card
Dialog
PopupMenuButton
Or even BottomSheet
I hope it will help