Flutter dropdown button selected value position

Question

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:

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

Edit: now it shows the same value

score 2 · Accepted Answer · answered Sep 19 '22 at 11:05

2

you should rearrange your list after selecting an item

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

answered Sep 19 '22 at 11:05

Ahmed Elhenidy

219
1
8

this does the trick, it could get confusing cause the order is different every time you pick a different option – Gryva Sep 19 '22 at 11:39
1

yeah, but you can make a temp list with the original items and every time you select an item from the drop down you put that item in the first and assign this temp list to the dropdown list – Ahmed Elhenidy Sep 20 '22 at 08:18

Tasnuva Tavasum oshin · Answer 2 · 2022-09-19T09:40:17.380

0

var selected value = "";

initState(){
value = dropdownvalue.value;
}

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


Change there your problem will be solved

edited Sep 19 '22 at 09:40

answered Sep 19 '22 at 09:29

Tasnuva Tavasum oshin

1
1
17
28

It didn't do the trick unfortunately, I added images so you can see the problem. – Gryva Sep 19 '22 at 09:36
i have changed the answer can you try with initialise e value like this – Tasnuva Tavasum oshin Sep 19 '22 at 09:40
did you change anything? I didn't notice changes, tried again but the same problem emerges – Gryva Sep 19 '22 at 09:53

Flutter dropdown button selected value position

2 Answers2