Receive limited query with custom start point - ios

Question

I'm building a basic chat app with swift for iOS with firebase realtime database.
The Messages are observed with a limit for the least 10.
Now, I want to implement the functionality of loading earlier send messages. Currently I'm trying to achieve this by using this function:
let query = threadRef.child("messages").queryOrderedByKey().queryStarting(atValue: "2").queryLimited(toLast: 2)
Which returns this query:
(/vYhNJ3nNQlSEEXWaJAtPLhikIZi1/messages {
i = ".key";
l = 2;
sp = 2;
vf = r;
})
And this should give me the data:
query.observeSingleEvent(of: .value, with: { (snap) in
But it just limits the query and not set the start point to a specific position.
Here is the firebase database structure:
messages
-Kgzb3_b26CnkTDglNd8
date:
senderId:
senderName:
text:
-Kgzb4Qip6_jQdKRWFey
-Kgzb4ha0KZkLZeBIaxW
-Kgzb577KlNKOHxsQo9W
-Kgzb5cqIVMhRmU019Jf
Anyone have an idea on how to implement a feature like that?

Answer 1

Okay I finally found a way to do what I wanted.
First of all I misunderstood the way to access data from Firebase.
This is now how I get the query:
let indexValue = messages.first?.fireBaseKey
let query = messageRef.queryOrderedByKey().queryEnding(atValue:indexValue).queryLimited(toLast: 3)
1) get the FireBase key I previously saved to my custom chat messages
2) construct the query:
order it by key
set the ending to oldest message
limit the array to query to desired length
Then to actually get the query I used:
query.observeSingleEvent(of: .value, with: { snapshot in
for child in snapshot.children.dropLast().reversed() {
let fireSnap = (child as! FIRDataSnapshot)
//do stuff with data
}
})
1) get the query as a single event
2) iterate over children and I needed to dropLast() to make sure I don't have any duplicated messages and reverse it to get the correct order.
3) cast the current child as a FIRDataSnapshot to access the data
Since I couldn't find a simple example for this so I thought I leave my solution here incase other people running into the same problem.