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

Ahmed Elhenidy · Accepted Answer · 2022-09-19 11:05:23Z

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, 2022 at 11:05

Ahmed Elhenidy

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2 Comments

Gryva Over a year ago

this does the trick, it could get confusing cause the order is different every time you pick a different option

Ahmed Elhenidy Over a year ago

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

Tasnuva Tavasum oshin · Accepted Answer · 2022-09-19 09:40:17Z

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, 2022 at 9:40

answered Sep 19, 2022 at 9:29

Tasnuva Tavasum oshin

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3 Comments

Gryva Over a year ago

It didn't do the trick unfortunately, I added images so you can see the problem.

Tasnuva Tavasum oshin Over a year ago

i have changed the answer can you try with initialise e value like this

Gryva Over a year ago

did you change anything? I didn't notice changes, tried again but the same problem emerges

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