Flutter dropdown button selected value position

Issue

How can I make the selected dropdown item appear first in the list when its opened?
I’m using GetX for state,

is this the proper way of doing it anyway?

This is the code:

var dropdownvalue = 'Default'.obs;
  var items = [
    'Default',
    'Difficulty',
    'Time',
    'Distance',
  ];
  RxInt selectedIndex = 0.obs;

Obx(() => DropdownButton(
                    value: dropdownvalue.value,
                    items: items.map((String items) {
                      return DropdownMenuItem(
                        value: items,
                        child: Text(items),
                      );
                    }).toList(),
                    onChanged: (value) {
                      dropdownvalue.value = value.toString();
                    },
                  )

This is how it looks now:

enter image description here

The list should stay in one place when its opened, and values should change places.

Edit: now it shows the same value

enter image description here

enter image description here

Solution

you should rearrange your list after selecting an item

onChanged: (value) {
  dropdownvalue.value = value.toString();
  items.remove("Time");
  items.insert(0, "Time");
  setState((){});
},

Answered By – Ahmed Mohamed

Answer Checked By – Robin (FlutterFixes Admin)

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