如何在 firebase firestore 中获取内部集合

Question

我正在尝试获取 firestore 中特定用户的设备令牌，该设备令牌存储在“客户”或“律师”集合中的令牌集合中。当我从链中删除第二个 .collection("tokens") 时，我得到了用户对象，但是随着链中的令牌集合，我似乎无法获得任何用户（客户或律师）回来，即使用户和它的令牌存在。我做错了什么

  exports.onReceiveChatMessage = functions.database
  .ref("/messages/{uid}")
  .onCreate(async (snapshot, context) => {
    const newMessage = snapshot.val();
    console.log("NEW_MESSAGE", newMessage);
    const senderName = newMessage.sender_name;
    const messageContent = newMessage.content;
    console.log("SENDER'S_NAME", senderName);
    console.log("MESSAGE_BODY", messageContent);
    const uid = context.params.uid;
    console.log("RECEIVERS_ID", uid);

    if (newMessage.sender_id == uid) {
      //if sender is receiver, don't send notification
      console.log("sender is receiver, dont send notification...");
      return;
    } else if (newMessage.type === "text") {
      console.log(
        "LETS LOOK FOR THIS USER, STARTING WITH CLIENTS COLLECTION..."
      );
      let userDeviceToken;
      await firestore
        .collection("clients")
        .doc(uid)
        .collection("tokens")
        .get()
        .then(async (snapshot) => {
          if (!snapshot.exists) {
            console.log(
              "USER NOT FOUND IN CLIENTS COLLECTION, LETS CHECK LAWYERS..."
            );
            await firestore
              .collection("lawyers")
              .doc(uid)
              .collection("tokens")
              .get()
              .then((snapshot) => {
                if (!snapshot.exists) {
                  console.log(
                    "SORRY!!!, USER NOT FOUND IN LAWYERS COLLECTION EITHER"
                  );
                  return;
                } else {
                  snapshot.forEach((doc) => {
                    console.log("LAWYER_USER_TOKEN=>", doc.data());
                    userDeviceToken = doc.data().token;
                  });
                }
              });
          } else {
            snapshot.forEach((doc) => {
              console.log("CLIENT_USER_TOKEN=>", doc.data());
              userDeviceToken = doc.data().token;
            });
          }
        });
      // console.log("CLIENT_DEVICE_TOKEN", userDeviceToken);
    } else if (newMessage.type === "video_session") {
    }
     })

Answer 1

这一行

if (!snapshot.exists) {

应该是：

if (snapshot.empty) {

因为您是在 CollectionReference（returns 和 QuerySnapshot）上调用 get()，而不是在 DocumentReference（returns一个DocumentSnapshot).

如果在示例中从链中删除 .collection('tokens')，它确实有效，因为 DocumentSnapshot 确实有成员 exists，但 CollectionReference 没有t.

在这里看看他们的成员：

https://googleapis.dev/nodejs/firestore/latest/CollectionReference.html#get

然后：

https://googleapis.dev/nodejs/firestore/latest/QuerySnapshot.html

作为建议，我曾经混淆快照并因为使用 Javascript 而不是 Typescript 而遇到这个问题。所以我习惯了调用文档时调用结果 snap，调用集合时调用结果 snaps。这让我想起了我正在研究什么样的反应。像这样：

// single document, returns a DocumentSnapshot
const snap = await db.collection('xyz').doc('123').get();
if (snap.exists) {
  snap.data()...
}

// multiple documents, returns a QuerySnapshot
const snaps = await db.collection('xyz').get();
if (!snaps.empty) { // 'if' actually not needed if iterating over docs
  snaps.forEach(...);
  // or, if you need to await, you can't use the .forEach loop, use a plain for:
  for (const snap of snaps.docs) {
    await whatever(snap);
  }
}

如何在 firebase firestore 中获取内部集合

How to get inner collection in firebase firestore

node.js

firebase

google-cloud-functions

google-cloud-firestore