Flutter UI Create menu with function after click icon more

Question

Halo,

First, I wanna show you my current UI :

And here's the code for red circle function :

InkWell(
  onTap: null,
  child: Icon(Icons.more_horiz,
    size: 18,
    color: Color.fromRGBO(0, 0, 0, 0.25),
  ),
),

What action I needed is to show pop-up menu like this after click the icon, for example :

I need that to create edit and delete with function to navigate to another page, and need to know how to control menu position.

I already check on pub.dev, but it didn't resolve my problem. Can anyone give me an advice?

Answer 1

Try this code : the package is here : https://pub.dev/packages/pull_down_button

PullDownButton(
  itemBuilder: (context) => [
    PullDownMenuItem(
      title: 'Menu item',
      onTap: () => action(),
    ),
    const PullDownMenuDivider(),
    PullDownMenuItem(
      title: 'Menu item 2',
      onTap: () => action2(),
    ),
  ],
  position: PullDownMenuPosition.under,
  buttonBuilder: (context, showMenu) => CupertinoButton(
    onPressed: showMenu,
    padding: EdgeInsets.zero,
    child: const Icon(CupertinoIcons.ellipsis_circle),
  ),
);

result :

Answer 2

You can try using PopupMenuButton you can find here an example

Answer 3

Try below code hope its help to you. and used PopupMenuButton

Declare Key Variable:

  final GlobalKey menuKey = GlobalKey();

Widget:

InkWell(
  onTap: () {
    dynamic state = menuKey.currentState;
    state.showButtonMenu();
  },
  child: PopupMenuButton(
    key: menuKey,
    itemBuilder: (_) => const <PopupMenuItem<String>>[
      PopupMenuItem<String>(
        child: Text('Show Test'),
        value: 'Test',
      ),
      PopupMenuItem<String>(
        child: Text('Edit Test Solution'),
        value: 'Edit',
      ),
    ],
    onSelected: (_) {},
  ),
),

Result->