Mongoose $in [ObjectIds] returns 0 records - mongodb

Question

In our Mongoose model, we have a product referring to an article.
this is a piece of the schema:
const product = new Schema({
article_id: Schema.Types.ObjectId,
title: String,
description: String,
...
In the API we are looking for products that are referring to a list of specific articles, and I wanted to use the $in operator:
const articles = ["5dcd2a95d7e2999332441825",
"5dcd2a95d7e2999332441827",
"5dcd2a96d7e2999332441829"]
filter.article_id = {
$in: articles.map(
article => new mongoose.Types.ObjectId(article)
),
};
return Product.find({ ...filter })
This returns 0 records, whereas I know for sure it should have returned at least 3. Looking at the console log, all that has happened is that the double quotes have been removed from the array during the ObjectId conversion.
Then I tried a different approach by returning {$oid: "id goes here"} for each mapped array item:
const articles = ["5dcd2a95d7e2999332441825",
"5dcd2a95d7e2999332441827",
"5dcd2a96d7e2999332441829"]
filter.article_id = {
$in: articles.map(
article => ({$oid: article})
),
};
return Product.find({ ...filter })
This gives a different array:
console.log(filter);
// {article_id: {$in: [{$oid: "5dcd2a95d7e2999332441825"}, {$oid: "5dcd2a95d7e2999332441827"}, {$oid: "5dcd2a96d7e2999332441829"}]}}
But in this case I get following error:
CastError: Cast to ObjectId failed for value "\"{$oid: \"5dcd2a95d7e2999332441825\"}\"".
Though - if I take that particular console logged filter and pass it in Studio 3T as a filter, I do indeed get the desired results.
Any idea what I doing wrong in this case?

Answer 1

Frick me! I am just a big idiot.. Apparently there was a .skip(10) added after the find() method -.-'.... Now I understand why 0 records where returned... Been spending hours on this..
For future references, Mongoose casts strings to ObjectIds automatically if defined in Schema. Therefor following is working exactly as it should given you don't skip the first 10 records:
const articles = ["5dcd2a95d7e2999332441825",
"5dcd2a95d7e2999332441827",
"5dcd2a96d7e2999332441829"]
filter.article_id = {
$in: articles
};
return Product.find({ ...filter }) // Note that I DON'T put .skip() here..