What is the equivalent of findOne using .aggregate in Mongodb?

Question

Basically with this I manage to return all my objects from a collection. How can I return a single element, for example in the style of findOne({_ id:" 5e82d378527bb420a4001aaf ")?

I know how to use $match, but this returns various results.

let _id="5e82d378527bb420a4001aaf"

Noticia.aggregate([
    {
        $addFields: {
            like: {
                $cond: [{ $in: [_id, "$likes"] }, true, false]
            },
            dislike: {
                $cond: [{ $in: [_id, "$dislikes"] }, true, false]
            }

        }
    }

], (err, noticia) => {

    // console.log(trans);
    if (err) {
        return res.status(400).json({
            ok: false,
            err
        });
    }
    return res.status(200).json({
        ok: true,
        data: noticia
    });

})

.findOne returns null or object, while .aggregate returns list — Valijon
– Valijon, Commented Mar 31, 2020 at 22:58
@Valijon I just need to search for a record with a unique value. i don't know if i can use match in this case or if there is a better findOne () style path — yavg
– yavg, Commented Mar 31, 2020 at 23:02

prasad_ · Accepted Answer · 2020-04-01 03:44:47Z

12

Consider a sample names collection:

{ _id: 1, name: "Jack", favoriteColor: "blue" },
{ _id: 2, name: "James", favoriteColor: "red" },
{ _id: 3, name: "John", favoriteColor: "blue" }

and run the following three queries using findOne:

db.names.findOne( { _id: 1 } )
db.names.findOne()
db.names.findOne( { favoriteColor : "blue" } )

the result is same for the three queries:

{ "_id" : 1, "name" : "Jack", "favoriteColor" : "blue" }

The equivalent queries respectively using aggregation are the following, with the same result:

db.names.aggregate( [
  { $match: { _id: 1 } },
] )

db.names.aggregate( [
  { $limit: 1 }
] )


db.names.aggregate( [
  { $match: { "favoriteColor" : "blue" } },
  { $limit: 1 }
] )

db.collection.findOne definition says -

Returns one document that satisfies the specified query criteria on the collection or view. If multiple documents satisfy the query, this method returns the first document according to the natural order which reflects the order of documents on the disk.

With findOne if no document is found it returns a null. But an aggregation returns a cursor, and you can apply the cursor methods on the result.

answered Apr 1, 2020 at 3:44

prasad_

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5 Comments

Zeeshan Ahmad Khalil Over a year ago

It returns array, how we can get only first document w/o in an array?

prasad_ Over a year ago

@ZeeshanAhmadKhalil If your code is returning an array, use the array element access code; e.g., myarray[0] gets the first element of the array. If your code is returning a cursor then use the cursor methods (there is a link in the above answer).

Zeeshan Ahmad Khalil Over a year ago

It returns the array but I want to get the first doc within the aggregate instead of returning the myarray and doing myarray[0]. Because I want to avoid using asyc await in my service function.

prasad_ Over a year ago

Aggregation query returns a cursor in case of native shell query. In case of some APIs the query can return an array. You can use $limit: 1 aggregation stage to return one document only - but it will be in a cursor or an array (as per the software you are using). You can create a single call, for example, db.names.aggregate( [ ... ] ).toArray()[0] in case of a cursor - note that if the aggregation query returns no data this will throw an error.

prasad_ Over a year ago

@ZeeshanAhmadKhalil (continued from previous comment) I suggest you create a new question explaining with code samples you are working on, software you are using, and other details for a proper answer you are looking for.

whoami - fakeFaceTrueSoul · Accepted Answer · 2020-03-31 22:49:14Z

1

Just use $limit to limit no.of docs to be returned from aggregation pipeline :

Noticia.aggregate([
    {
        $addFields: {
            like: {
                $cond: [{ $in: [_id, "$likes"] }, true, false]
            },
            dislike: {
                $cond: [{ $in: [_id, "$dislikes"] }, true, false]
            }

        }
    } , {$limit :1} // will return first doc from aggregation result of '$addFields' stage

], (err, noticia) => {

    // console.log(trans);
    if (err) {
        return res.status(400).json({
            ok: false,
            err
        });
    }
    return res.status(200).json({
        ok: true,
        data: noticia
    });
 })

Or if you wanted to return random single doc try $sample.

answered Mar 31, 2020 at 22:49

whoami - fakeFaceTrueSoul

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5 Comments

yavg Over a year ago

Thank you, your responses are as good as ever. But I think I have not made myself understood. I just need to search for a record with a unique value. i don't know if i can use match in this case or if there is a better findOne () style way

whoami - fakeFaceTrueSoul Over a year ago

@yavg : So you just wanted to return one doc from aggregation or multiple docs ? If you wanted to return multiple docs then what do you wanted to do with matched doc from $match?

whoami - fakeFaceTrueSoul Over a year ago

@yavg : Then all you can do is $match

yavg Over a year ago

can you help me please with this hard question for me? stackoverflow.com/questions/61000005/…

yavg Over a year ago

You always shake my hand when I need you the most. Can you help me please? stackoverflow.com/questions/61004259/…

Kleber R · Accepted Answer · 2021-12-11 02:28:44Z

1

//Express and mongoose
exports.getData = (req, res) => {

    Noticia.aggregate([
        {
            $addFields: {
                like: {
                    $cond: [{ $in: [_id, "$likes"] }, true, false]
                },
                dislike: {
                    $cond: [{ $in: [_id, "$dislikes"] }, true, false]
                }
    
            }
        }  
    ]
    ).then( data=>{
        res.send(data[0]); //return one document in format { ..... }
    }).catch(err => {
        res.status(500).send({ 
          message: err.message 
        });
    });
}

edited Dec 11, 2021 at 2:28

answered Dec 11, 2021 at 2:22

Kleber R

112 bronze badges

1 Comment

Kleber R Over a year ago

This code uses the express and mongoose framework and returns a single document, without using brackets. Show the document in braces

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