Flutter for loop to generate list of widgets

Question

I have code like this but I want it to iterate over an integer array to display a dynamic amount of children:

return Container(
  child: Column(
    children: <Widget>[
      Center(
        child: Text(text[0].toString(),
            textAlign: TextAlign.center),
      ),
      Center(
        child: Text(text[1].toString(),
            textAlign: TextAlign.center),
      ),
    ],
  ),
)

Where the text variable is a list of integers converted to string here. I tried adding a function to iterate through the array and display the 'children' but was getting a type error. Not sure how to do it since I'm new to Dart and Flutter.

Abbas.M · Accepted Answer · 2019-07-09 07:46:46Z

235

You can try this :

@override
  Widget build(BuildContext context) {
    List<int> text = [1,2,3,4];
    return Scaffold(
      appBar: AppBar(
        title: Text(widget.title),
      ),
      body: Container(
        child: Column(
          children: [
            for ( var i in text ) Text(i.toString())
          ],
        ),
      ),
    );

Note that this was added with the updated of dart to version 2.3. You can read about some of the best changes in this article

Another method that was provided before dart 2.3 is this:

@override
  Widget build(BuildContext context) {
    List<int> text = [1,2,3,4];
    return Scaffold(
      appBar: AppBar(
        title: Text(widget.title),
      ),
      body: Container(
        child: Column(
          children: List.generate(text.length,(index){
            return Text(text[index].toString());
          }),
        ),
      ),
    );

edited Jul 9, 2019 at 7:46

answered Jul 9, 2019 at 7:16

Abbas.M

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4 Comments

Hasen Over a year ago

Weird, I tried the first method and I got an error RangeError (index): Invalid value: Not in range 0..3, inclusive: 4 with List<int> text = [1, 2, 3, 4];

Abbas.M Over a year ago

ah crap my bad, i edited the code to fix it, i used a for each loop. Let me know if you still needed anything!

Hasen Over a year ago

It should be Text(i.toString() since i here is actually the value and not the key.

RnDrx Over a year ago

This requires 'control-flow-collections' to be enabled

handsomepotato · Accepted Answer · 2022-04-11 08:24:43Z

32

Assuming you want to loop some widgets (e.g Text()) in the Column widget, you can add a loop inside the children property. See a sample below:

Column(
   children: <Widget>[
      for (int i=0; i<3; i++)
         Text("Hello" + i)
   ],
)

edited Apr 11, 2022 at 8:24

handsomepotato

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answered Feb 13, 2021 at 16:24

Eaweb

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2 Comments

adamu_fura Over a year ago

The i inside the text widget has to be converted to string: Text("Hello" + i.toString())

Jani Over a year ago

Alternatively, you can iterate over a list, like so: for(final text in texts) Text(text)

Sahan Amarsha · Accepted Answer · 2022-02-20 16:25:03Z

19

You can try .map Method here,

class Example extends StatelessWidget {

  List <int> exampleList =  [1,2,3,4];

  @override
  Widget build(BuildContext context) {
    return
      Container(
        child: Column(
            children: exampleList.map((i) => new Text(i.toString())).toList()
        ),
      );
  }
}

This method will come in handy if you have objects inside your list. Also with the .map() method .toList() must be used at the end.

edited Feb 20, 2022 at 16:25

answered May 8, 2020 at 7:58

Sahan Amarsha

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Comments

Cassio Seffrin · Accepted Answer · 2020-09-29 20:16:51Z

5

You could use the map method of your list of Strings.

 Widget _getListWidgets(List<String> yourList){
    return Row(children: yourList.map((i) => Text(i)).toList());
  }

When your List have a complex Object:

Widget _getListWidgets( List<YourObject> lstItens) {
  return Row(children: lstItens.map((i) => Text(i.name)).toList());
}

answered Sep 29, 2020 at 20:16

Cassio Seffrin

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Comments

RnDrx · Accepted Answer · 2020-05-14 04:40:38Z

2

The best way to do this is to make use of the List.map()

That way you do not have to enable 'control-flow-collections'

 Container(
        child: Column(
            children: myList.map((e) => new Text(e)).toList(),
          ),
   );

answered May 14, 2020 at 4:40

RnDrx

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Comments

Pedro Martín · Accepted Answer · 2023-09-05 15:59:20Z

2

  @override
  Widget build(BuildContext context) {
    List<int> text = [1, 2, 3, 4];
    return Scaffold(
      appBar: AppBar(
        title: Text(widget.title),
      ),
      body: SizedBox(
        // better use SizedBox instead of Container
        child: Column(
          // children: text.map((e) => Text("$e")).toList() // OPTION 1
          children: [...text.map((e) => Text("$e"))],       // OPTION 2 using spread operator
        ),
      ),
    );
  }

answered Sep 5, 2023 at 15:59

Pedro Martín

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