Firestore安全规则被拒绝

时间:2020-08-22 21:08:28

标签: javascript react-native google-cloud-firestore firebase-security

我希望得到一些编写安全规则的帮助。这很简单,但是我为消息规则编写的每个变体似乎都被拒绝了。我想说的是“仅当您是邮件的发送者或接收者时才授予读/写访问权限。”

此安全性要涵盖的基本规则:

  • 用户无法编辑其他人的数据
  • 用户可以与其他人一起创建和阅读消息
  • 如果用户未通过身份验证,则无法访问任何数据

错误消息:

[未处理的承诺被拒绝:FirebaseError:缺少权限或权限不足。]

到目前为止我所拥有的:

service cloud.firestore {
  match /databases/{database}/documents {

    function isLoggedIn() {
        return request.auth != null && request.auth.uid != null;
    }

    function isSender() {
        return resource != null && resource.data.user._id == request.auth.uid;
    }

    function isReceiver() {
      return resource != null && request.auth.uid in resource.data.receiver
    }
      
    match /Users/{userId} {
      //Only authenticated users can access/write data
      allow delete: if request.auth.uid == userId;
      allow read, write: if request.auth != null && request.auth.uid == userId;
    }

    match /messages/{messageId} {
      // allow read, write: if isLoggedIn() && (isSender() || isReceiver());
        allow read, write: if isLoggedIn() && (resource.data.user._id == request.auth.uid || request.auth.uid in resource.data.receiver);
    }    
  }
}

消息数据结构如下:

id: ""
createdAt:""
index: 1
key: ""
receiver: 
    [0: receiverIDgoeshere]
message: ""
user:
    {_id: ""}

查询:

let query = config.db
.collection(messages)
.where("key", "==", uid)
.orderBy("index", "desc");
if (typeof index === "number") {
    query = query.where("index", ">", index);
}
const chats = await query.get();

还有一个传入消息的侦听器:

const listener = config.db
  .collection(messages)
  .where("key", "==", uid)
  .where("receiver", "array-contains", userId)
  .onSnapshot((snapshot) => {
    const msgs = snapshot.docChanges().map(({doc, type}) => {
      if (type === "added") {
        return fixData({
          id: doc.id,
          match,
          fromRealTime: true,
          ...doc.data(),
        });
      } else {
        console.log("type is NOT added...");
      }
      return null;
    });
    setMessages(msgs.filter((msg) => msg !== null));
  });

以及当我们添加聊天时:

const fs = config.db;
const doc = fs.collection(messages).doc();
await doc.set(chat);

2 个答案:

答案 0 :(得分:1)

查询必须更新如下:

let query = config.db
.collection(messages)
.where("key", "==", uid)
.where("user._id", "==", uid)
.orderBy("index", "desc");
//rest is same

我认为侦听器的查询没有任何问题。

答案 1 :(得分:0)

我发现我的安全规则很好,我试图对其进行细化。以上所有安全规则均已涵盖。