sorted participants list alphabetically #87

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MineTec merged 3 commits from develop-sortedParticipants into develop 2025-06-24 10:47:07 +00:00
3 changed files with 15 additions and 8 deletions
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@ -13,16 +13,23 @@ class ParticipantsListView extends StatefulWidget {
class _ParticipantsListViewState extends State<ParticipantsListView> { class _ParticipantsListViewState extends State<ParticipantsListView> {
@override @override
Widget build(BuildContext context) => Scaffold( Widget build(BuildContext context) {
final participants = widget.participantsResponse.data.map((participant) => ListTile(
leading: UserAvatar(id: participant.actorId),
title: Text(participant.displayName),
subtitle: participant.statusMessage != null ? Text(participant.statusMessage!) : null,
)).toList();
lastname(participant) => participant.title.toString().split(' ').last;
participants.sort((a, b) => lastname(a).compareTo(lastname(b)));
Pupsi marked this conversation as resolved Outdated

ggf. kann das sort aus der Build methode rausgezogen werden und das Widget zu einem Stateless werden

ggf. kann das sort aus der Build methode rausgezogen werden und das Widget zu einem Stateless werden
return Scaffold(
appBar: AppBar( appBar: AppBar(
title: const Text('Teilnehmende'), title: const Text('Teilnehmende'),
), ),
body: ListView( body: ListView(
children: widget.participantsResponse.data.map((participant) => ListTile( children: participants,
leading: UserAvatar(id: participant.actorId),
title: Text(participant.displayName),
subtitle: participant.statusMessage != null ? Text(participant.statusMessage!) : null,
)).toList(),
), ),
); );
}
} }