Mongodb Aggregation : Project an array without the first element

Question

{
    "_id" : ObjectId("5e3179e83708dc0dcfefaf"),

    "CaseNumber" : "T978045628",

    "ServiceReferences" : [ 

        ObjectId("5e317d48bc13eaf17712a786"),

        ObjectId("5e317f0ec08d57f4a88c3444"),

        ObjectId("5e317f0ec08d57f4a88c3234")

    ]

}

I have a collection with above fields. I want to project the 'ServiceReferences' array except first element as I want to skip first indexed element of the array. Please help me. I have tried a couple of use cases like :

db.getCollection('cases').aggregate([{
             "$project": {

                  "ServRefs": { $pop: { ServiceReferences: -1 } }
              }
       },
       {
        $lookup: {

            from: 'services',

            localField: 'ServRefs',

            foreignField: '_id',

            as: 'serviceDoc'

        }
    }, { $unwind: '$serviceDoc' }])

Can you please give an example of the expected result?

Ashwanth Madhav
– Ashwanth Madhav

2020-02-05 11:36:25 +00:00
Commented Feb 5, 2020 at 11:36 — Ashwanth Madhav
– Ashwanth Madhav, Commented Feb 5, 2020 at 11:36

Ashwanth Madhav · Accepted Answer · 2020-02-05 11:51:19Z

7

Just try with this one

db.getCollection('cases').aggregate([
{ $project: { ServiceReferences: 1, SerRef: { $slice: [ "$ServiceReferences", 1, {$subtract: [ {$size:"$ServiceReferences"}, 1 ]} ] }} }
])

answered Feb 5, 2020 at 11:51

Ashwanth Madhav

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3 Comments

Swapna Over a year ago

Thank you. It is working fine... First I checked the existence of the length then it worked :: { "$match": { "ServicesReferences.1":{$exists:true}} }, Otherwise it will through :: errmsg" : "Third argument to $slice must be positive: 0

Ashwanth Madhav Over a year ago

Good. It's better to check the existence.

Frozen Flame Over a year ago

It might be more convenient omitting the $subtract operation. Just { $slice: [ "$ServiceReferences", 1, {$size:"$ServiceReferences"}]}.

Paplusc · Accepted Answer · 2020-02-05 10:46:42Z

0

You can remove the first element of an array by using $pop

The $pop operator removes the first or last element of an array. Pass $pop a value of -1 to remove the first element of an array and 1 to remove the last element in an array.

You can apply it to your projection.

Here is the documentaiton: $pop documentation mongoDb

answered Feb 5, 2020 at 10:46

Paplusc

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3 Comments

Swapna Over a year ago

I cann't remove first element from array because I need it in the next projection which is in the same aggregate query. I just want to ignore the first element.

Swapna Over a year ago

I want to Project all items in an array except first one, to new field which is exact opposite of this query :: { "$project": { "ServRef" : { "$arrayElemAt" : ["$ServicesReferences", 0] } } }

Paul Feb 7 at 18:24

$pop can't be used in an aggregation.

Niels Abildgaard · Accepted Answer · 2024-12-05 14:24:43Z

0

use slice

https://www.mongodb.com/docs/manual/reference/operator/aggregation/slice/

The following syntax returns elements from the specified position in the array:

db.in_house_beta.aggregate([ 
     { $project: { "experience_combined": { $slice: [ "$experience_combined", 1 ] } } },
     { $unwind: "$experience_combined" },
     { $unwind: "$experience_combined.Company" },
     { $group:{ _id:{ Company:'$experience_combined.Company'}, count: { $sum:1} } 
}])

edited Dec 5, 2024 at 14:24

Niels Abildgaard

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answered May 20, 2022 at 12:21

Pirai Sudie

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1 Comment

Frozen Flame Over a year ago

{ $slice: [ "$experience_combined", 1 ] } means 1 element from position 0. If $experience_combined equals [1,2,3,4], { $slice: [ "$experience_combined", 1 ] } gives [1]

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