list - Flutter未处理的异常: type 'Future<Group>' is not a subtype of type 'Group'

Question

在使用 Firebase Firestore 时，我遇到以下错误:

E/flutter ( 4477): [ERROR:flutter/shell/common/shell.cc(93)] Dart Unhandled Exception: type 'Future<Group>' is not a subtype of type 'Group', stack trace: #0      new List.from (dart:core-patch/array_patch.dart:41:5)
E/flutter ( 4477): #1      GroupService.getGroups (package:money_manager/core/services/group_service.dart:56:21)
E/flutter ( 4477): <asynchronous suspension>
E/flutter ( 4477): #2      GroupViewModel._getAllGroups (package:money_manager/core/view_models/group_viewmodel.dart:44:26)
E/flutter ( 4477): <asynchronous suspension>
E/flutter ( 4477):

虽然我认为我已经在必要的地方正确使用了 async-await，但我无法找出这个错误。相关代码:

在group-service.dart中:

Future<List<Group>> getGroups() async {
    final userData = await getUserData();
    List<Group> groups = [];

    if (userData['groups'] != null) {
      List<String> groupIds = List.from(userData['groups'].map((e) => e['id']));
      groups = List.from(groupIds.map((e) async {
        return await getGroupFromId(e);
      }));
    }

    return groups;
  }

  Future<Group> getGroupFromId(String groupId) async {
    final groupData = await getGroupData(groupId);
    return Group.fromJson(groupData);
  }

  Future<Map<String, dynamic>> getGroupData(String groupId) async {
    DocumentReference groupDoc =
        FirebaseFirestore.instance.collection('groups').doc(groupId);
    final snapshot = await groupDoc.get();
    Map<String, dynamic> groupData = snapshot.data() as Map<String, dynamic>;
    return groupData;
  }

在group_viewmodel.dart中:

List<Group> _userGroups = [];
 void _getAllGroups() async {
    List<Group> groups = await _groupService.getGroups();
    _userGroups = groups;
  }

Answer 1

问题出在以下行:

groups = List.from(groupIds.map((e) async {
   return await getGroupFromId(e);
}));

即使您使用 await之前return map()的函数，它仍然会返回 Future 。 map()是一个同步函数，它不会异步运行内部函数。因此，它将返回 Iterable<Future<Group>> ，无法转换为 List<Group>在 List.from()功能。

有一个方便的函数，它接受可迭代的 future 并等待其中的每一个，Future.wait() 。使用它后您的代码将如下所示:

groups = List.from(await Future.wait(groupIds.map((e) async {
   return await getGroupFromId(e);
})));

撕下效果更好:

groups = List.from(await Future.wait(groupIds.map(getGroupFromId)));