Insert into generated list in Flutter
At the moment I have this list
List.generate(72, (index) {
retrun Container(
child: new Text('$index'),
)
})
as the children of a GridView widget. What I however would like to do is return a different value than the $index for certain index values.
For this I have a List that looks like [{index: 2, value: test}{index: 5, value: hello}] with a lot of index/value pairs. So here is the question:
Is there a way now to display the value from the list in the GridView field if the matching index is in the list and if it is not simply return $index?
Just as an example, the field with the index 1 in the GridView should display its index, so it displays 1. The field with the index 2 however should display the matching value from the list, which is test and so on.
dart
flutter asked Nov 21 '18 at 21:52
stefanmukestefanmuke
172113
add a comment |
At the moment I have this list
List.generate(72, (index) {
retrun Container(
child: new Text('$index'),
)
})
as the children of a GridView widget. What I however would like to do is return a different value than the $index for certain index values.
For this I have a List that looks like [{index: 2, value: test}{index: 5, value: hello}] with a lot of index/value pairs. So here is the question:
Is there a way now to display the value from the list in the GridView field if the matching index is in the list and if it is not simply return $index?
Just as an example, the field with the index 1 in the GridView should display its index, so it displays 1. The field with the index 2 however should display the matching value from the list, which is test and so on.
dart
flutter asked Nov 21 '18 at 21:52
stefanmukestefanmuke
172113
add a comment |
At the moment I have this list
List.generate(72, (index) {
retrun Container(
child: new Text('$index'),
)
})
as the children of a GridView widget. What I however would like to do is return a different value than the $index for certain index values.
For this I have a List that looks like [{index: 2, value: test}{index: 5, value: hello}] with a lot of index/value pairs. So here is the question:
Is there a way now to display the value from the list in the GridView field if the matching index is in the list and if it is not simply return $index?
Just as an example, the field with the index 1 in the GridView should display its index, so it displays 1. The field with the index 2 however should display the matching value from the list, which is test and so on.
dart
flutter asked Nov 21 '18 at 21:52
stefanmukestefanmuke
172113
At the moment I have this list
List.generate(72, (index) {
retrun Container(
child: new Text('$index'),
)
})
as the children of a GridView widget. What I however would like to do is return a different value than the $index for certain index values.
For this I have a List that looks like [{index: 2, value: test}{index: 5, value: hello}] with a lot of index/value pairs. So here is the question:
Is there a way now to display the value from the list in the GridView field if the matching index is in the list and if it is not simply return $index?
Just as an example, the field with the index 1 in the GridView should display its index, so it displays 1. The field with the index 2 however should display the matching value from the list, which is test and so on.
dart
flutter dart
flutter asked Nov 21 '18 at 21:52
stefanmukestefanmuke
172113
asked Nov 21 '18 at 21:52
stefanmukestefanmuke
172113
asked Nov 21 '18 at 21:52
stefanmukestefanmuke
172113
asked Nov 21 '18 at 21:52
stefanmukestefanmuke
172113
asked Nov 21 '18 at 21:52
stefanmukestefanmuke
172113
172113
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
It looks like you should preprocess the list into a Map. If necessary, iterate the list adding each entry to a map.
Then you can:
Map m = <int, String>{
2: 'test',
5: 'hello',
};
List<Container>.generate(72, (int index) {
String s = m[index];
return Container(
child: Text(s != null ? s : '$index'),
);
});
or, more succinctly with the null aware operator
List<Container>.generate(
72,
(int index) => Container(child: Text(m[index] ?? '$index')),
);
answered Nov 22 '18 at 0:06
Richard HeapRichard Heap
7,67121019
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It looks like you should preprocess the list into a Map. If necessary, iterate the list adding each entry to a map.
Then you can:
Map m = <int, String>{
2: 'test',
5: 'hello',
};
List<Container>.generate(72, (int index) {
String s = m[index];
return Container(
child: Text(s != null ? s : '$index'),
);
});
or, more succinctly with the null aware operator
List<Container>.generate(
72,
(int index) => Container(child: Text(m[index] ?? '$index')),
);
answered Nov 22 '18 at 0:06
Richard HeapRichard Heap
7,67121019
add a comment |
It looks like you should preprocess the list into a Map. If necessary, iterate the list adding each entry to a map.
Then you can:
Map m = <int, String>{
2: 'test',
5: 'hello',
};
List<Container>.generate(72, (int index) {
String s = m[index];
return Container(
child: Text(s != null ? s : '$index'),
);
});
or, more succinctly with the null aware operator
List<Container>.generate(
72,
(int index) => Container(child: Text(m[index] ?? '$index')),
);
answered Nov 22 '18 at 0:06
Richard HeapRichard Heap
7,67121019
add a comment |
It looks like you should preprocess the list into a Map. If necessary, iterate the list adding each entry to a map.
Then you can:
Map m = <int, String>{
2: 'test',
5: 'hello',
};
List<Container>.generate(72, (int index) {
String s = m[index];
return Container(
child: Text(s != null ? s : '$index'),
);
});
or, more succinctly with the null aware operator
List<Container>.generate(
72,
(int index) => Container(child: Text(m[index] ?? '$index')),
);
answered Nov 22 '18 at 0:06
Richard HeapRichard Heap
7,67121019
It looks like you should preprocess the list into a Map. If necessary, iterate the list adding each entry to a map.
Then you can:
Map m = <int, String>{
2: 'test',
5: 'hello',
};
List<Container>.generate(72, (int index) {
String s = m[index];
return Container(
child: Text(s != null ? s : '$index'),
);
});
or, more succinctly with the null aware operator
List<Container>.generate(
72,
(int index) => Container(child: Text(m[index] ?? '$index')),
);
answered Nov 22 '18 at 0:06
Richard HeapRichard Heap
7,67121019
answered Nov 22 '18 at 0:06
Richard HeapRichard Heap
7,67121019
answered Nov 22 '18 at 0:06
Richard HeapRichard Heap
7,67121019
answered Nov 22 '18 at 0:06
Richard HeapRichard Heap
7,67121019
7,67121019
add a comment |
add a comment |
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