CODEWITHSUNDEEP
mongodb
Lempkin
Lempkin

Reputation: 1588

Find id that are not present in DB

Following query gives me in return objects that are present in DB and matches one of the ids in the list :

db.intervention.find( { _id: { $in: [ ObjectId("5f7fc2c3e21da01ffe711e48"), ObjectId("5f7fc255555da01ffe711e48") ] } } )

What I need is the contrary : to pass a list of ids, and get in response a list of ids that could not be found in DB.

Example : In DB I have ["A", "B", "C", "D", "E"]

Query :

db.intervention.find( { _id: { $something: ["D", "E", "F", "G"] } })

What I want in response is ["F", "G"] -> id that are in the list (in the query) but not in DB.

I'm new to Mongo and have no idea about how to achieve that.

Many Thx

Upvotes: 1

Views: 1178

Answers (1)

Sunil K Samanta
Sunil K Samanta

Reputation: 171

This will do the job. But you'll have to supply the Array of Ids 2 times in the query. Suppose you have ids. [84, 75, 76, 70, 71, 50, 12]. And Database don't have 50 & 12 in Database. Then..

db.intervention.aggregate([
    {
        $match: {
            _id: {$in: [84, 75, 76, 70, 71, 50, 12]}
        }
    },
    {
        $group: {
            _id: null,
            foundIds: {$push: "$_id"}
        }
    },
    {
        $project: {
            notFoundIds: {$setDifference: [ [84, 75, 76, 70, 71, 50, 12], "$foundIds"]}
        }
    }

])

This will result:

{
    "_id" : null,
    "notFoundIds" : [
        50,
        12
    ]
}

Upvotes: 2

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