$group by one parameter and get all data of it

MongoDB Realm
, ,
#1

Hello Everyone,

I am using the below query to group data by resource but I want all data in that particular resource how can I do that

    `var m = await model.aggregate([{$group : {_id: {resource : '$resource', project: '$project', reporting_to: '$reporting_to', remark: '$remark', occupancy: '$occupancy', status: '$status'}, count: { $sum: 1 }}}]);`

what I want is to group it by the resource : '$resource' and get all the items in that particular resource entry, in case with my above query its is showing all 6 entries but i dont want duplicated entry if there are two resource of same name so it should give me the latest one

Thank you in Advance
Juned Adenwalla,

#2

@Juned_Adenwalla hi there!
I am not sure I understand, if this is not what you need, please include input documents (at least one) and expected output.

model.aggregate({
  $group: {
    _id: {
      key1: "$key1",
      key2: "$key2",
      
    },
    all: {
      $push: "$$ROOT"
    }
  }
})

This groups from your collection documents with the same fields in _id. This is a low match, i.e, documents with those fields + extra fields would be grouped also. Hope that makes sense.

If that resonates, check out carefully this live example in MongoPlayground.

You can also append your accumulator field.

#3

Thank you so much for your help sir, but fortunatly when i got this message I also found solution from google.

I used the code below

doc: { $first: "$$ROOT" }

thank you so much for your reply

#4

I actually missed that, great that you found the right solution.

#5

Note that using $first is only meaningful (and correct to get latest entry) if you guarantee your data is sorted by date descending - otherwise you’ll get whatever happens to be the first entry in some arbitrary order.

Asya