Flutter; replace item in list

Question

I have a list in my app. It containe a couple of items.

I want to replace every item that is equal to a user-input(A String) , with another user-input(Another String). (If the solution is to remove the item, and add a new one, it needs to be in the same location in the list)

How do I do that?

Thanks :)

Answer 1

You can also use replaceRange for this:

void main() {
  var list = ['Me', 'You', 'Them'];
  print(list); // [Me, You, Them]
  var selected = 'You'; // user input A (find)
  var newValue = 'Yours'; // user input B (replace)
  
  // find the item you want to replace, in this case, it's the value of `selected`.
  var index = list.indexOf(selected);
  
  // if replacing only one item, the end index should always be `startIndex` +1.
  // `replaceRange` only accepts iterable(list) so `newValue` is inside the array.
  list.replaceRange(index, index + 1, [newValue]);
  
  print(list); // [Me, Yours, Them]
}

Answer 2

for (int i = 0; i < LIST.length; i++){
    if (LIST[i] == USERINPUT1){
        LIST[i] = USERINPUT2;
    }
}

Basically iterate through the list in the app to check if the input is equal to the user's input.

OR

index = LIST.indexOf(USERINPUT1);

if (index != -1){
    LIST[index] = USERINPUT2;
}

This only works for the first occurrence though.

Answer 3

Here is how I would do it:

void main() {
  String inputString = 'myInput'; // User input to replace inside the list
  List<String> list = ['1', inputString, '3', '4', '5', inputString, '7']; // List of items
  List<String> filteredList = list.where((e) => e == inputString).toList();
  for (final e in filteredList) {
    final index = list.indexOf(e);
    list.removeAt(index);
    list.insert(index, 'newInput'); // Replacing inputString by 'newInput'
  }
}

Basically what I'm doing is creating a sublist filteredList containing only the occurrence of the user input. Then I iterate over my filteredList, before removing the item from list I'm keeping its index to insert my new element at the correct position.

Try the code on DartPad

Answer 4

First occurrence simplest solution

final l = [1,2,3,4];
l[l.indexOf(2)] = 5;
print(l); // [1,5,3,4]