如何在Flutter中结合来自两个Firestore集合的数据？

Question

在Flutter中，我有一个使用Firestore的聊天应用程序，并且有两个主要集合：

• chats（在自动ID上键入），并具有message，timestamp和uid字段。
• users，在uid上键入并具有一个name字段

在我的应用中，我使用以下小部件显示消息列表（来自messages集合）

class ChatList extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    var messagesSnapshot = Firestore.instance.collection("chat").orderBy("timestamp", descending: true).snapshots();
    var streamBuilder = StreamBuilder<QuerySnapshot>(
          stream: messagesSnapshot,
          builder: (BuildContext context, AsyncSnapshot<QuerySnapshot> querySnapshot) {
            if (querySnapshot.hasError)
              return new Text('Error: ${querySnapshot.error}');
            switch (querySnapshot.connectionState) {
              case ConnectionState.waiting: return new Text("Loading...");
              default:
                return new ListView(
                  children: querySnapshot.data.documents.map((DocumentSnapshot doc) {
                    return new ListTile(
                      title: new Text(doc['message']),
                      subtitle: new Text(DateTime.fromMillisecondsSinceEpoch(doc['timestamp']).toString()),
                    );
                  }).toList()
                );
            }
          }
        );
        return streamBuilder;
  }
}

但是现在我想显示每封邮件的用户名（来自users集合）。

我通常称其为客户端联接，尽管我不确定Flutter是否为其指定名称。

我已经找到了一种执行此操作的方法（我在下面发布了此方法），但是想知道在Flutter中是否有另一种/更好/更惯用的方法来进行这种操作。

那么：在Flutter中，惯用的方式是如何在上述结构中查找每条消息的用户名？

Answer 1

我可以使用另一个版本，该版本似乎比my answer with the two nested builders更好。

这里，我使用专用的Message类来保存来自消息Document和可选的关联用户Document的信息，从而以自定义方法隔离了数据加载。

class Message {
  final message;
  final timestamp;
  final uid;
  final user;
  const Message(this.message, this.timestamp, this.uid, this.user);
}
class ChatList extends StatelessWidget {
  Stream<List<Message>> getData() async* {
    var messagesStream = Firestore.instance.collection("chat").orderBy("timestamp", descending: true).snapshots();
    var messages = List<Message>();
    await for (var messagesSnapshot in messagesStream) {
      for (var messageDoc in messagesSnapshot.documents) {
        var message;
        if (messageDoc["uid"] != null) {
          var userSnapshot = await Firestore.instance.collection("users").document(messageDoc["uid"]).get();
          message = Message(messageDoc["message"], messageDoc["timestamp"], messageDoc["uid"], userSnapshot["name"]);
        }
        else {
          message = Message(messageDoc["message"], messageDoc["timestamp"], "", "");
        }
        messages.add(message);
      }
      yield messages;
    }
  }
  @override
  Widget build(BuildContext context) {
    var streamBuilder = StreamBuilder<List<Message>>(
          stream: getData(),
          builder: (BuildContext context, AsyncSnapshot<List<Message>> messagesSnapshot) {
            if (messagesSnapshot.hasError)
              return new Text('Error: ${messagesSnapshot.error}');
            switch (messagesSnapshot.connectionState) {
              case ConnectionState.waiting: return new Text("Loading...");
              default:
                return new ListView(
                  children: messagesSnapshot.data.map((Message msg) {
                    return new ListTile(
                      title: new Text(msg.message),
                      subtitle: new Text(DateTime.fromMillisecondsSinceEpoch(msg.timestamp).toString()
                                         +"\n"+(msg.user ?? msg.uid)),
                    );
                  }).toList()
                );
            }
          }
        );
        return streamBuilder;
  }
}

与solution with nested builders相比，此代码更具可读性，主要是因为数据处理和UI构建器更好地分开了。它还仅为发布消息的用户加载用户文档。不幸的是，如果用户发布了多条消息，它将为每条消息加载文档。我可以添加一个缓存，但是认为这段代码已经完成了一些工作。

Answer 2

您可以使用RxDart这样操作。.https://pub.dev/packages/rxdart

import 'package:rxdart/rxdart.dart';

class Messages {
  final String messages;
  final DateTime timestamp;
  final String uid;
  final DocumentReference reference;

  Messages.fromMap(Map<String, dynamic> map, {this.reference})
      : messages = map['messages'],
        timestamp = (map['timestamp'] as Timestamp)?.toDate(),
        uid = map['uid'];

  Messages.fromSnapshot(DocumentSnapshot snapshot)
      : this.fromMap(snapshot.data, reference: snapshot.reference);

  @override
  String toString() {
    return 'Messages{messages: $messages, timestamp: $timestamp, uid: $uid, reference: $reference}';
  }
}

class Users {
  final String name;
  final DocumentReference reference;

  Users.fromMap(Map<String, dynamic> map, {this.reference})
      : name = map['name'];

  Users.fromSnapshot(DocumentSnapshot snapshot)
      : this.fromMap(snapshot.data, reference: snapshot.reference);

  @override
  String toString() {
    return 'Users{name: $name, reference: $reference}';
  }
}

class CombineStream {
  final Messages messages;
  final Users users;

  CombineStream(this.messages, this.users);
}

Stream<List<CombineStream>> _combineStream;

@override
  void initState() {
    super.initState();
    _combineStream = Observable(Firestore.instance
        .collection('chat')
        .orderBy("timestamp", descending: true)
        .snapshots())
        .map((convert) {
      return convert.documents.map((f) {

        Stream<Messages> messages = Observable.just(f)
            .map<Messages>((document) => Messages.fromSnapshot(document));

        Stream<Users> user = Firestore.instance
            .collection("users")
            .document(f.data['uid'])
            .snapshots()
            .map<Users>((document) => Users.fromSnapshot(document));

        return Observable.combineLatest2(
            messages, user, (messages, user) => CombineStream(messages, user));
      });
    }).switchMap((observables) {
      return observables.length > 0
          ? Observable.combineLatestList(observables)
          : Observable.just([]);
    })
}

Answer 3

如果我正确地阅读了此内容，则问题将抽象为：如何转换需要进行异步调用以修改流中数据的数据流？

在问题的上下文中，数据流是消息的列表，异步调用是获取用户数据并使用流中的此数据更新消息。

可以使用asyncMap()函数在Dart流对象中直接执行此操作。这是一些演示如何实现的纯Dart代码：

import 'dart:async';
import 'dart:math' show Random;

final random = Random();

const messageList = [
  {
    'message': 'Message 1',
    'timestamp': 1,
    'uid': 1,
  },
  {
    'message': 'Message 2',
    'timestamp': 2,
    'uid': 2,
  },
  {
    'message': 'Message 3',
    'timestamp': 3,
    'uid': 2,
  },
];

const userList = {
  1: 'User 1',
  2: 'User 2',
  3: 'User 3',
};

class Message {
  final String message;
  final int timestamp;
  final int uid;
  final String user;
  const Message(this.message, this.timestamp, this.uid, this.user);

  @override
  String toString() => '$user => $message';
}

// Mimic a stream of a list of messages
Stream<List<Map<String, dynamic>>> getServerMessagesMock() async* {
  yield messageList;
  while (true) {
    await Future.delayed(Duration(seconds: random.nextInt(3) + 1));
    yield messageList;
  }
}

// Mimic asynchronously fetching a user
Future<String> userMock(int uid) => userList.containsKey(uid)
    ? Future.delayed(
        Duration(milliseconds: 100 + random.nextInt(100)),
        () => userList[uid],
      )
    : Future.value(null);

// Transform the contents of a stream asynchronously
Stream<List<Message>> getMessagesStream() => getServerMessagesMock()
    .asyncMap<List<Message>>((messageList) => Future.wait(
          messageList.map<Future<Message>>(
            (m) async => Message(
              m['message'],
              m['timestamp'],
              m['uid'],
              await userMock(m['uid']),
            ),
          ),
        ));

void main() async {
  print('Streams with async transforms test');
  await for (var messages in getMessagesStream()) {
    messages.forEach(print);
  }
}

大多数代码都将来自Firebase的数据模拟为消息映射流，并具有异步功能来获取用户数据。这里的重要功能是getMessagesStream()。

由于它是流中出现的消息列表，因此代码有些复杂。为了防止调用获取用户数据的同步发生，该代码使用Future.wait()来收集List<Future<Message>>并在所有期货都完成后创建一个List<Message>。

在Flutter中，您可以在getMessagesStream()中使用来自FutureBuilder的流来显示消息对象。

Answer 4

理想情况下，您希望排除任何业务逻辑，例如将数据加载到单独的服务中或遵循BloC模式，例如：

class ChatBloc {
  final Firestore firestore = Firestore.instance;
  final Map<String, String> userMap = HashMap<String, String>();

  Stream<List<Message>> get messages async* {
    final messagesStream = Firestore.instance.collection('chat').orderBy('timestamp', descending: true).snapshots();
    var messages = List<Message>();
    await for (var messagesSnapshot in messagesStream) {
      for (var messageDoc in messagesSnapshot.documents) {
        final userUid = messageDoc['uid'];
        var message;

        if (userUid != null) {
          // get user data if not in map
          if (userMap.containsKey(userUid)) {
            message = Message(messageDoc['message'], messageDoc['timestamp'], userUid, userMap[userUid]);
          } else {
            final userSnapshot = await Firestore.instance.collection('users').document(userUid).get();
            message = Message(messageDoc['message'], messageDoc['timestamp'], userUid, userSnapshot['name']);
            // add entry to map
            userMap[userUid] = userSnapshot['name'];
          }
        } else {
          message =
              Message(messageDoc['message'], messageDoc['timestamp'], '', '');
        }
        messages.add(message);
      }
      yield messages;
    }
  }
}

然后，您可以仅在组件中使用Bloc并收听chatBloc.messages流。

class ChatList extends StatelessWidget {
  final ChatBloc chatBloc = ChatBloc();

  @override
  Widget build(BuildContext context) {
    return StreamBuilder<List<Message>>(
        stream: chatBloc.messages,
        builder: (BuildContext context, AsyncSnapshot<List<Message>> messagesSnapshot) {
          if (messagesSnapshot.hasError)
            return new Text('Error: ${messagesSnapshot.error}');
          switch (messagesSnapshot.connectionState) {
            case ConnectionState.waiting:
              return new Text('Loading...');
            default:
              return new ListView(children: messagesSnapshot.data.map((Message msg) {
                return new ListTile(
                  title: new Text(msg.message),
                  subtitle: new Text('${msg.timestamp}\n${(msg.user ?? msg.uid)}'),
                );
              }).toList());
          }
        });
  }
}

Answer 5

允许我提出我的RxDart解决方案版本。我使用combineLatest2和ListView.builder来构建每个消息小部件。在构造每个消息小部件的过程中，我用相应的uid查找用户的名称。

在此代码段中，我对用户名使用了线性查找，但是可以通过创建uid -> user name映射来改进

import 'package:cloud_firestore/cloud_firestore.dart';
import 'package:flutter/widgets.dart';
import 'package:rxdart/rxdart.dart';

class MessageWidget extends StatelessWidget {
  // final chatStream = Firestore.instance.collection('chat').snapshots();
  // final userStream = Firestore.instance.collection('users').snapshots();
  Stream<QuerySnapshot> chatStream;
  Stream<QuerySnapshot> userStream;

  MessageWidget(this.chatStream, this.userStream);

  @override
  Widget build(BuildContext context) {
    Observable<List<QuerySnapshot>> combinedStream = Observable.combineLatest2(
        chatStream, userStream, (messages, users) => [messages, users]);

    return StreamBuilder(
        stream: combinedStream,
        builder: (_, AsyncSnapshot<List<QuerySnapshot>> snapshots) {
          if (snapshots.hasData) {
            List<DocumentSnapshot> chats = snapshots.data[0].documents;

            // It would be more efficient to convert this list of user documents
            // to a map keyed on the uid which will allow quicker user lookup.
            List<DocumentSnapshot> users = snapshots.data[1].documents;

            return ListView.builder(itemBuilder: (_, index) {
              return Center(
                child: Column(
                  children: <Widget>[
                    Text(chats[index]['message']),
                    Text(getUserName(users, chats[index]['uid'])),
                  ],
                ),
              );
            });
          } else {
            return Text('loading...');
          }
        });
  }

  // This does a linear search through the list of users. However a map
  // could be used to make the finding of the user's name more efficient.
  String getUserName(List<DocumentSnapshot> users, String uid) {
    for (final user in users) {
      if (user['uid'] == uid) {
        return user['name'];
      }
    }
    return 'unknown';
  }
}

Answer 6

我工作的第一个解决方案是嵌套两个StreamBuilder实例，每个集合/查询一个实例。

class ChatList extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    var messagesSnapshot = Firestore.instance.collection("chat").orderBy("timestamp", descending: true).snapshots();
    var usersSnapshot = Firestore.instance.collection("users").snapshots();
    var streamBuilder = StreamBuilder<QuerySnapshot>(
      stream: messagesSnapshot,
      builder: (BuildContext context, AsyncSnapshot<QuerySnapshot> messagesSnapshot) {
        return StreamBuilder(
          stream: usersSnapshot,
          builder: (context, usersSnapshot) {
            if (messagesSnapshot.hasError || usersSnapshot.hasError || !usersSnapshot.hasData)
              return new Text('Error: ${messagesSnapshot.error}, ${usersSnapshot.error}');
            switch (messagesSnapshot.connectionState) {
              case ConnectionState.waiting: return new Text("Loading...");
              default:
                return new ListView(
                  children: messagesSnapshot.data.documents.map((DocumentSnapshot doc) {
                    var user = "";
                    if (doc['uid'] != null && usersSnapshot.data != null) {
                      user = doc['uid'];
                      print('Looking for user $user');
                      user = usersSnapshot.data.documents.firstWhere((userDoc) => userDoc.documentID == user).data["name"];
                    }
                    return new ListTile(
                      title: new Text(doc['message']),
                      subtitle: new Text(DateTime.fromMillisecondsSinceEpoch(doc['timestamp']).toString()
                                          +"\n"+user),
                    );
                  }).toList()
                );
            }
        });
      }
    );
    return streamBuilder;
  }
}

正如我的问题中所述，我知道这种解决方案虽然不好，但至少可以奏效。

我看到一些问题：

• 它将加载所有用户，而不只是加载消息的用户。在小型数据集中这不是问题，但是随着我收到更多消息/用户（并使用查询显示其中的一部分），我将加载越来越多的未发布任何消息的用户。 / li>
• 在两个构建器的嵌套中，代码的可读性不是很高。我怀疑这是惯用的Flutter。

如果您知道更好的解决方案，请发布答案。