Update value of a single entry in a List> in dart

Question

I have a list of type Map<String,dynamic>, I fill this list with List.filled.

[{gstatus: 1, vermerk: }, {gstatus: 1, vermerk: }, {gstatus: 1, vermerk: }, {gstatus: 1, vermerk: }]

But if I want to change e.g. the value "vermerk" of the second entry of the list with ".update()", all values of the key "vermerk" change like:

[{gstatus: 1, vermerk: test}, {gstatus: 1, vermerk: test}, {gstatus: 1, vermerk: test}, {gstatus: 1, vermerk: test}]

I would like to have the following result

[{gstatus: 1, vermerk: }, {gstatus: 1, vermerk: test}, {gstatus: 1, vermerk: }, {gstatus: 1, vermerk: }]

Here the Code

void main() {
  
  late List<Map<String, dynamic>> deviceForm;
  
  deviceForm = List.filled(4,
          {"gstatus": 1, "vermerk": ""});
  
  print(deviceForm);
  
  deviceForm[1].update('vermerk', (value) => 'test');
  
  print(deviceForm);
  
}

Kaushik Chandru · Answer 1 · 2022-07-24T13:13:25.557

1

You can try to use generate and update


void main() {
  
  List<Map<String, dynamic>> deviceForm = List.generate(4,
         (index)=> {"gstatus": 1, "vermerk": ""});
  
  print(deviceForm);
  
  deviceForm[1].update('vermerk', (value) => 'test');
  
  print(deviceForm);
  //Output : [{gstatus: 1, vermerk: }, {gstatus: 1, vermerk: test}, {gstatus: 1, vermerk: }, {gstatus: 1, vermerk: }]
}

edited Jul 24 '22 at 13:13

answered Jul 24 '22 at 12:15

Kaushik Chandru

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unfortunately it does not work like this! when i do it like this i also get the following output ´´´[{gstatus: 1, vermerk: test}, {gstatus: 1, vermerk: test}, {gstatus: 1, vermerk: test}, {gstatus: 1, vermerk: test}]´´´ – mascy Jul 24 '22 at 12:21
Thank you very much. I would vote up your answer but I don't have enough points yet. – mascy Jul 24 '22 at 14:16

score 1 · Accepted Answer · answered Jul 24 '22 at 12:27

This is very interesting issue, seems like List.filled will produce identical data, because of this somehow every data will be changed if you try to change one of them... and once i check like this, it return TRUE

print(deviceForm[1] == deviceForm[2]); //TRUE

but i try another shot with List.generate, the result will be different

print(deviceForm[1] == deviceForm[2]); //FALSE

If you have specific reason why you use List.filled, maybe this answer will be useless, or in case you have no problem to switch using List.generate, it will solve your problem

deviceForm = List.generate(4, (index) => {"gstatus": 1, "vermerk": ""});

Thank you very much. It seems to point exactly at the object and then overwrite the data. — mascy, Jul 24 '22 at 14:16

Md. Yeasin Sheikh · Answer 3 · 2022-07-24T16:00:03.417

List.filled: All elements of the created list share the same [fill] value.

The issue lies on List.filled, you can use List.generate instead

  deviceForm1 = List.generate(
    3,
    (_) => {"gstatus": 1, "vermerk": ""},
  );

Or create new map and assign on index 1.

void main() {
  late List<Map<String, dynamic>> deviceForm;

  deviceForm = List.filled(2, {"gstatus": 0, "vermerk": ""});

  print(deviceForm[1] == Map.from(deviceForm[0])); //false
  print(deviceForm[1] == deviceForm[0]); //true

  deviceForm[1] = Map.from(deviceForm[1])
    ..update("vermerk",
        (value) => "Test"); // we are creating new Map and assingng it

  print(deviceForm);
  //[{gstatus: 0, vermerk: }, {gstatus: 0, vermerk: newValue}]
}

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Update value of a single entry in a List> in dart

3 Answers3