How could use aggregates in mongo to do operations in an array embedded inside a document?

Question

Let us say I have a collection of users. Each user has assets. There are some statistical measures that I want to perform on assets. But in order to be able to do this I need to only have the assets in the projection as an input for the aggregate. How could this be achieved?

I want to be able to do a bunch of aggregate operations on assets. These operations would include filtering, limit and skip so I would be using operators like $skip, $match, $limit, and others. But in order for me to perform these operations on the embedded array, I would need an operator that takes a user document and outputs the assets array embedded into it.

Simply I want an operator that that takes a user document with the following schema and outputs an array of assets that I further perform aggregate operations on it.

user:{
   name: string,
   assets: Asset[],
   age: number
}

I tried this aggregate operations but it is not working as I expected.

db.getCollection('users').aggregate( [ 
{ $match: {_id: ObjectId("5ae837dca9a8e04dd0689824")} } ,
{ $project : { assets: 1  } },
] )

It outputs an array that have one user document in which the assets are embedded inside like this

[{
"_id" : ObjectId("5ae837dca9a8e04dd0689824"),
"assets" : [{_id:ObjectId("1"), assetName: 'tt' }, {_id : ObjectId("2"), assetName: 'rr'}]
}]

I need to have the output like the one below

[
{_id:ObjectId("1"), assetName: 'tt'}, 
{_id:ObjectId("2"), assetName: 'rr'}
]

It's not really enough of an explanation. Better to show a few sample documents and the result you expect to achieve. Ideally we really want you to "show your attempt" as well, since we're happy to help you with your code but less happy to do your job for "free" as it were. See How do I ask a good question? and it's various linked topics including How to create a Minimal, Complete, and Verifiable example — Neil Lunn
– Neil Lunn, Commented May 2, 2018 at 3:21

mickl · Accepted Answer · 2018-05-02 06:09:13Z

1

Two steps to do that (plus $match):

$unwind to transform an array of subdocuments to multiple documents having one subdocument
$replaceRoot to promote subdocument to a new root

Try:

db.users.aggregate([
    { $match: {_id: ObjectId("5ae837dca9a8e04dd0689824")} },
    { $unwind: "$assets" },
    { $replaceRoot: { newRoot: "$assets" } }
])

answered May 2, 2018 at 6:09

mickl

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2 Comments

Yoosif Sherif Over a year ago

Thank you for this answer and thumbs up, but don't you think there is a more performant solution.

mickl Over a year ago

@YoussefSherif nothing comes to my mind since I believe there are no replacements for those two operators. What do you mean by performant ? If you start with $match and perform $unwind on single document then your result set shouldn't be so huge I guesa

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