83

The structure of the table is:

  • chats
  • --> randomId
  • -->--> participants
  • -->-->--> 0: 'name1'
  • -->-->--> 1: 'name2'
  • -->--> chatItems

etc

What I am trying to do is query the chats table to find all the chats that hold a participant by a passed in username string.

Here is what I have so far:

 subscribeChats(username: string) {
    return this.af.database.list('chats', {
        query: {
            orderByChild: 'participants',
            equalTo: username, // How to check if participants contain username
        }
    });
 }
0

2 Answers 2

101

Your current data structure is great to look up the participants of a specific chat. It is however not a very good structure for looking up the inverse: the chats that a user participates in.

A few problems here:

  • you're storing a set as an array
  • you can only index on fixed paths

Set vs array

A chat can have multiple participants, so you modelled this as an array. But this actually is not the ideal data structure. Likely each participant can only be in the chat once. But by using an array, I could have:

participants: ["puf", "puf"]

That is clearly not what you have in mind, but the data structure allows it. You can try to secure this in code and security rules, but it would be easier if you start with a data structure that implicitly matches your model better.

My rule of thumb: if you find yourself writing array.contains(), you should be using a set.

A set is a structure where each child can be present at most once, so it naturally protects against duplicates. In Firebase you'd model a set as:

participants: {
  "puf": true
}

The true here is really just a dummy value: the important thing is that we've moved the name to the key. Now if I'd try to join this chat again, it would be a noop:

participants: {
  "puf": true
}

And when you'd join:

participants: {
  "john": true,
  "puf": true
}

This is the most direct representation of your requirement: a collection that can only contain each participant once.

You can only index known properties

With the above structure, you could query for chats that you are in with:

ref.child("chats").orderByChild("participants/john").equalTo(true)

The problem is that this requires you to define an index on `participants/john":

{
  "rules": {
    "chats": {
      "$chatid": {
        "participants": {
          ".indexOn": ["john", "puf"]
        }
      }
    }
  }
}

This will work and perform great. But now each time someone new joins the chat app, you'll need to add another index. That's clearly not a scaleable model. We'll need to change our data structure to allow the query you want.

Invert the index - pull categories up, flattening the tree

Second rule of thumb: model your data to reflect what you show in your app.

Since you are looking to show a list of chat rooms for a user, store the chat rooms for each user:

userChatrooms: {
  john: {
    chatRoom1: true,
    chatRoom2: true
  },
  puf: {
    chatRoom1: true,
    chatRoom3: true
  }
}

Now you can simply determine your list of chat rooms with:

ref.child("userChatrooms").child("john")

And then loop over the keys to get each room.

You'll like have two relevant lists in your app:

  • the list of chat rooms for a specific user
  • the list of participants in a specific chat room

In that case you'll also have both lists in the database.

chatroomUsers
  chatroom1
    user1: true
    user2: true
  chatroom2
    user1: true
    user3: true
userChatrooms
  user1:
    chatroom1: true
    chatroom2: true
  user2:
    chatroom1: true
  user2:
    chatroom2: true

I've pulled both lists to the top-level of the tree, since Firebase recommends against nesting data.

Having both lists is completely normal in NoSQL solutions. In the example above we'd refer to userChatrooms as the inverted index of chatroomsUsers.

Cloud Firestore

This is one of the cases where Cloud Firestore has better support for this type of query. Its array-contains operator allows filter documents that have a certain value in an array, while arrayRemove allows you to treat an array as a set. For more on this, see Better Arrays in Cloud Firestore.

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3 Comments

This is an awesome explanation. I have my data stored like this, which allows me to grab a chatroom from a user's list, or grab a user from a chat room's members. But what it doesn't allow is to observe .childChanged for the chatrooms of which I am a member. If I try to set up an observer for chats where members/myId = true, then I get the unspecified index error. How would you go about this? I have a question posted here: stackoverflow.com/questions/47769044/…
@frank-van-puffelen referencing your last code snippet, let's say I am "user1" and I fetch all keys of chatrooms I'm a member of. Then for each chatroom key I fetch the full chatroom object, and display the list of chatrooms in a list. That's good, however, what if I also needed to access full user objects that are a part of a chatroom. It seems very strange that I would have to do another for-loop (2-level nested for-loop!) fetch for every single user in every single chatroom to get that data. Is there a better way?
Great explanation. This a brilliant example of how you need to model your data differently to take advantage of NoSQL (and the cheap platforms that use it).
0

I believe you can simply use

{
  "rules": {
    "chats": {
      "$chatid": {
        "participants": {
          ".indexOn": ".value"
        }
      }
    }
  }
}

in database rules and then in your code use

ref.child("chats").orderByChild("participants/john").equalTo(true)

assuming both .orderByChild() and .orderByValue() methods can be used with ".indexOn" : ".value". According to the examples from the official Firebase website

https://firebase.google.com/docs/database/security/indexing-data

.orderByChild() is used with ".indexOn" : "childName" and .orderByValue() is used with ".indexOn" : ".value" but who tested the opposite to see if it works or not? since the server in either case is eventually indexing the end value the above solution should be working fine, if someone knows more about this please tell us

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