Firebase querying

Question

Let's say I have a structure like this:

-users
  -user1_uid
    name
    distance
    age

How would I do a query like (Find users with distance <100 and age between 20 and 25)?

I have tried the standard method

        let recentPostsQuery = (ref?.child("users").queryOrderedByChild("age").queryStartingAtValue("20"))!

M problem is, that is does not appear to be possible to query multiple childs (like combining age and distance filtering). Did something change in this regard compared to Firebase a few months ago? Filtering them locally after the first query is not an option, I believe, as there could be potentially thousands of objects.

Answer 1

My first choice would be to query for all the users between 20 and 25 and then filter in code for ones with a distance < 100.

The question states filtering in code is not an option but I wanted to include it for completeness for situations where it was several thousand nodes or less:

    struct User { //starting with a structure to hold user data
        var firebaseKey : String?
        var theAge: Int?
        var theDistance: Int?
    }

    var userArray = [User]() //the array of user structures

    usersRef.queryOrderedByChild("age").queryStartingAtValue(20)
     .queryEndingAtValue(25).observeEventType(.Value, withBlock: { snapshot in

        for child in snapshot.children { //.Value so iterate over nodes

            let age = child.value["age"] as! Int
            let distance = child.value["distance"] as! Int
            let fbKey = child.key!

            let u = User(firebaseKey: fbKey, theAge: age, theDistance: distance)

            userArray.append(u) //add the user struct to the array
        }

        //the array to contain the filtered users
        var filteredArray: [User] = []
        filteredArray = userArray.filter({$0.theDistance < 100}) //Filter it, baby!

        //print out the resulting users as a test.
        for aUser in filteredArray {

            let k = aUser.firebaseKey
            let a = aUser.theAge
            let d = aUser.theDistance

            print("array: \(k!)  \(a!)  \(d!)")

        }

    })
}

Now a potential super simple answer.

    let usersRef = self.myRootRef.childByAppendingPath("users")

    usersRef.queryOrderedByChild("age").queryStartingAtValue(20)
     .queryEndingAtValue(25).observeEventType(.ChildAdded, withBlock: { snapshot in

        let distance = snapshot.value["distance"] as! Int

        if distance < 100 {
            let age = snapshot.value["age"] as! Int
            let fbKey = snapshot.key!

            print("array: \(fbKey)  \(age)  \(distance)")
        }
    })

Notice that we are leveraging .ChildAdded instead of .Value so each node is read in one at a time - if the distance is not what we want, we can ignore it and move on to the next.

Answer 2

Firebase can't combine always the conditions. How to query based on multiple conditions in Firebase?

But, using the new firebase API, this post can give some hints: Query based on multiple where clauses in firebase