Swift Firebase using an unspecified index

Question

So, I have a chat app and my data is structured like so

chats : {
    chat12345 : {
        members : {
            user123 : true,
            user456 : true,
        }
    }
}

users : {
    user123 : {
        chats : {
            chat12345 : true
        }
    }
}

So when I want to grab a list of the user's chats, I simply use:

let ref = dbref.child("users").child("user123").child("chats")
ref.observe(.childAdded, with: { (snapshot) -> Void in

    let chatRef = dbref.child("chats").child(snapshot.key)

        chatRef.observeSingleEvent(of: .value, with: { (chatSnapshot) in

            //I now have the chat object

        })
})

So you see above, that I am able to query each chat ID to get the chat object from Firebase successfully.

But I also need to observe changes to chats that I am a member of, for instance if the chat title, image, or other properties change.

So when I try the following:

let queryRef = chatRef.queryOrdered(byChild: "members/user123").queryEqual(toValue: true)

queryRef.observe(.childChanged, with: { (snapshot) -> Void in

})

This technically works, but the console outputs the following:

Using an unspecified index. Your data will be downloaded and filtered on the client. Consider adding ".indexOn": "members/user123" at /chats to your security rules for better performance.

How do either modify security rules or my query to observe changes to chats of which I am a member?

Here are my security rules right now:

"chats": {

    ".read": "auth != null",
    ".write": "auth != null",

    "$chatId" : {
        "members": {
          ".indexOn": ".value"
        }
    }

 },

wrosb · Accepted Answer · 2018-10-02 19:44:55Z

1

you should not queryOrderedBy(/key/$variable) you need to structure your data something like your variables stay at chatRef. Then you could use static indexes.

answered Oct 2, 2018 at 19:44

wrosb

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