Mongodb $lookup on multiple fields

Question

I have two collections, products and properties.

I'm doing a lookup such as:

[{
        $match: {
            category_id: mongoose.Types.ObjectId(category_id)
        }
    },
    {
        $lookup: {
            from: "properties",
            localField: "category_id",
            foreignField: "category_id",
            as: "properties"
        }
    }
]

This is basically getting me all of the products that match the category_id and including the properties that match the same category_id.

I need to add an additional check, for some_id on the properties result. In otherwords, the properties should be grouped by the some_id that is returned from the products collection and matches the same key in properties. Does that make sense? Basically having the ability to have multiple local/foreign field definitions.

any idea how I could this?

Any sample data available? Like input data and then desired output. — dnickless
– dnickless, Commented Sep 8, 2017 at 19:13
Have you tried $filter ? Something like { $addFields: { properties: { $filter: { input: "$properties", as: "result", cond: { $eq: ["$$result.some_id ", "$some_id"] } } } } } — s7vr
– s7vr, Commented Sep 8, 2017 at 20:30

F.H. · Accepted Answer · 2022-09-15 09:09:07Z

6

Since version 3.6 we can use uncorrelated sub-queries

    {
       $lookup:
         {
           from: '<collection to join>',
           let: { <var_1>: '<expression>', …, <var_n>: '<expression>' },
           pipeline: [ '<pipeline to execute on the collection to join>' ],
           as: <output array field>
         }
    }

This allows us to to have more than a single equality match on the lookup. See also: https://docs.mongodb.com/manual/reference/operator/aggregation/lookup/

edited Sep 15, 2022 at 9:09

answered Apr 8, 2019 at 14:53

F.H.

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