I am making a chat app where 10 messages are loaded per page. The messages are sorted from oldest to latest, so the new messages come in bottom and old messages are in top. I tried to use skip() and limit() methods it didnt work.
let count = page*10
MessageModel.find(find_query).sort({createdAt: 'asc'}).skip(count-10).limit(10).exec();for example:
let these be my messages sorted oldest to newest: ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12"]
the above query returns the data like this in
1st page: ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10"]
2nd page : ["11", "12"]
but I want it to return
in 1st page: ["3", "4", "5", "6", "7", "8", "9", "10", "11", "12"]
in 2nd page: ["1", "2"]
EDIT: I tried sorting it into descending and reversing, this seemed to fix the issue.
let count = page*10
const messages = await MessageModel.find(find_query).sort({createdAt: 'desc'}).skip(count-10).limit(10).exec();
const re_messages = messages.reverse()Since you want to display messages 3 to 12 first, you need to sort using 'desc' instead of 'asc', so as to get the newest messages first. This will give you ["12", "11", ... "4", "3"]. Now all you have to do is to invert this array to get ["3", "4", ... "11", "12"]:
let count = page*10
let messages = await MessageModel
.find(find_query)
.sort({createdAt: 'desc'}) // get the latest ones
.skip(count-10)
.limit(10)
.exec()
.lean(); // Get a simple array
messages.reverse(); // ["3", "4", ... "11", "12"]