Variable not defined MongoDB $function

Question

Ok I am confused. I am trying to find documents using the $function operator as described here: S_function.

My model:

const newsSchema = new mongoose.Schema({
news_id: {
    type: mongoose.Schema.Types.ObjectId,
    unique: true,
    index: true
},
created_at: { type: String, default: Date.now().toString() },
updated_at: { type: String, default: Date.now().toString() },
Company_id: [{type: mongoose.Schema.Types.ObjectId}],
Stock_id: [{type: mongoose.Schema.Types.ObjectId}],
newsItem: {
    date: {type: String},
    title: { type: String },
    summary: { type: String },
    url: { type: String, unique: true },
    content: { type: String }
}
};

My code attempt:

newsModel.aggregate([
        {$match: {$function: {
            body: matchArrToArr(matchArr, Company_id),
            args: [ "$Company_id" ],
            lang: "js"
         }}}
    ])

matchArr is a variable I pass in from the surrounding function and Company_id is supposed to be the array 'Company_id' from the news document. However when I execute the function, the error 'Company_id is not defined' is returned. Which makes sense, as I dont define it but how do I get the Company_id field into the function.

I also tried find($expr:{$function:... with the same result and $where:... with this.Company_id. First case same error. Second case 'this' has the state of my calling JS function and not the DB document.

What I am actually trying to do: I want to return all documents, where one of the ids in the Company_id array matches one of the ids I pass in. If there is a nicer aggregate way without $function, I am open to that. But I also want to understand what I am doing wrong with my $function expression.

Thanks!

Function declaration and function invocation are two different this. You need to define a function not invoke it. MongoDB will invoke the function you passed it as an argument to body property. Do not call matchArrToArr function. Replace it with function definition. — Dheemanth Bhat
– Dheemanth Bhat, Commented Feb 27, 2021 at 12:21
Update the question with your attempt. If still not working I will help u. — Dheemanth Bhat
– Dheemanth Bhat, Commented Feb 27, 2021 at 12:24
Omg yes you are completely right, stupid me. Maybe my 3 days no food is showing. Thanks a lot! — user13548229
– user13548229, Commented Feb 27, 2021 at 13:30

Dheemanth Bhat · Accepted Answer · 2021-02-27 14:01:22Z

1

This is what you want:

let someCompany_id = "60278ce8b370ff29b83226e7";

db.news.find(
    {
        $expr: {
            $function: {
                body: function(company_ids, someCompany_id) {
                    let _ids= company_ids.map(company_id => company_id.valueOf());
                    return _ids.includes(someCompany_id);
                },
                args: ["$company_id", someCompany_id],
                lang: "js"
            }
        }
    }
)

But this can be done easily like this:

db.news.find({ 
    company_id: {
        $in: [someCompany_id]
    }
});

So why do you want to pass function?

edited Feb 27, 2021 at 14:01

answered Feb 27, 2021 at 13:03

Dheemanth Bhat

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1 Comment

user13548229 Over a year ago

Yes I was calling instead of passing the function. As usual when I am stuck it is the dumbest of mistakes. The reason I cannot just do a simple find is that both input and match IDs are arrays and I want a match if just one of them matches. So [someId1,someId2] and [someId2, someId3] should return true,

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