mongoDB find documents based on a condition with a value from linked document of another collection

Question

I have collections of the following structure:

objects:

[{"type": "someTypeOne", "menuId": 1},
{"type": "someTypeTwo", "menuId": 1},
{"type": "someTypeOne", "menuId": 2}]

menus:

[{"id":1, "type": "someTypeOne"},
{"id":2, "type": "someTypeOne"}]

I need to find all objects where "type" property doesn't match its menus "type". In this case the desired output would be:

[{"type": "someTypeTwo", "menuId": 1}]

I think that I should use aggregation for this one and I'm fiddling with it at the moment but I was not able to formulate a working query so far. Thanks

mickl · Accepted Answer · 2019-06-18 12:28:19Z

2

You can try below aggregation:

db.objects.aggregate([
    {
        $lookup: {
            from: "menus",
            localField: "menuId",
            foreignField: "id",
            as: "menu"
        }
    },
    {
        $unwind: "$menu"
    },
    {
        $match: {
            $expr: {
                $ne: [ "$menu.type", "$type" ]
            }
        }
    },
    {
        $project: {
            menu: 0
        }
    }
])

$lookup allows you to get data from both collections, then you can run $unwind on menu array to get single menu per document and you can apply you inequality condition using $match and $expr

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answered Jun 18, 2019 at 12:28

mickl

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1 Comment

mickl Over a year ago

@FIIFE cause it's extremely easy to answer when you clearly define what you need (by examples)

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