Question

我通过以下方式设置了Firestore：

频道[集合] ----> channelID->消息[集合]-> messageID

如何将snapshotListener添加到子集合“消息”中？

  Firestore.firestore().collection("Channels").document().collection("Messages").addSnapshotListener { (querySnapshot, error) in
        guard let snapshot = querySnapshot else {
            print("Error listening for channel updates: \(error?.localizedDescription ?? "No error")")
            return
        }

        snapshot.documentChanges.forEach { change in 
           print(change)
        }
    }

这对我不起作用

Answer 1

您不可能只有一个侦听器从未知数量的子集合中接收更新。集合中没有用于侦听器的“通配符”运算符。您必须选择一个特定的集合或查询，并在其上附加一个侦听器。

Answer 2

正如道格在正确答案中指出的那样，您无法让单个侦听器接收到来自未知子集（数量不明或未指定）的更新。

但是，如果可以确定那些子集合名称，那么答案就非常简单了。

想法是读取Channels的子节点，它们将是channel_0，channel_1等，并使用这些文档ID来建立对您有兴趣收听的节点的引用。

因此，给出此结构（与问题中的结构匹配）：

Channels
  channel_0
    Messages
      message_0
        msg: "chan 0 msg 0"
      message_1
        msg: "chan 0 msg 1"
      message_2
        msg: "chan 0 msg 2"
  channel_1
    Messages
      message_0
        msg: "chan 1 msg 0"
      message_1
        msg: "chan 1 msg 1"

这是将侦听器添加到每个频道的代码，并响应该频道中的事件，并在控制台中通知消息ID，msg文本和事件发生所在的频道。

func addChannelCollectionObserver() {
    let channelsRef = self.db.collection("Channels")
    channelsRef.getDocuments(completion: { snapshot, error in

        guard let documents = snapshot?.documents else {
            print("Collection was empty")
            return
        }

        for doc in documents {
            let docID = doc.documentID
            let eachChannelRef = channelsRef.document(docID)
            let messagesRef = eachChannelRef.collection("Messages")

            messagesRef.addSnapshotListener { querySnapshot, error in

                querySnapshot?.documentChanges.forEach { diff in
                    if diff.type == .added {
                        let doc = diff.document
                        let msgId = doc.documentID
                        let channelId = messagesRef.parent!.documentID
                        let msg = doc.get("msg") as? String ?? "no message"
                        print(" added msgId: \(msgId) with msg: \(msg) in channel: \(channelId)")
                    }

                    if diff.type == .modified {
                        let doc = diff.document
                        let msgId = doc.documentID
                        let msg = doc.get("msg") as? String ?? "no message"
                        print(" modified msgId: \(msgId) with msg: \(msg)")
                    }

                    if diff.type == .removed {
                        let doc = diff.document
                        let msgId = doc.documentID
                        let msg = doc.get("msg") as? String ?? "no message"
                        print(" removed msgId: \(msgId) with msg: \(msg)")
                    }
                }
            }
        }
    })
}

第一次运行时，输出将按预期显示每个子节点。从那时起，它将输出任何添加，修改或删除。

 added msgId: message_0 with msg: chan 0 msg 0 in channel: channel_0
 added msgId: message_1 with msg: chan 0 msg 1 in channel: channel_0
 added msgId: message_2 with msg: chan 0 msg 2 in channel: channel_0
 added msgId: message_0 with msg: chan 1 msg 0 in channel: channel_1
 added msgId: message_1 with msg: chan 1 msg 1 in channel: channel_1

对于某些可选选项，代码还需要进行其他错误检查，但它应该提供解决方案。

Answer 3

为此，请使用collection group query来收听所有具有相同名称的收藏集

    db.collectionGroup("Messages"). docRef.addSnapshotListener(new EventListener<DocumentSnapshot>() {
  @Override
  public void onEvent(@Nullable DocumentSnapshot snapshot,
                      @Nullable FirestoreException e) {
    if (snapshot != null && snapshot.exists()) {
      System.out.println("Current data: " + snapshot.getData());
    } else {
      System.out.print("Current data: null");
    }
  }
});

Answer 4

我认为是因为

Firestore.firestore().collection("Channels").document().collection("Messages")

您未定义正确的CollectionReference，因为您未标识“渠道”集合的文档。

您应该这样做：

Firestore.firestore().collection("Channels").document(channelID).collection("Messages")

子集合的Firestore侦听器

4 个答案: