Run a loop through an array of objects in the $project pipeline of MongoDB

Question

I have used the $group pipeline and inside it, I have used $addToSet which has created an array of objects which contains two values - {subGenre (string), flag (bool)}. In the next pipeline of $project, I need to run a loop through this array and I need to select only that element of the array where flag is false.

So my code looks like this:

let data = await Books.aggregate(
   [
    {
        $group: {
          _id:  genre,
          price: { $sum: "$price" },
           data: {
            $addToSet: {
              subGenre: "$subGenre",
              flag: "$flagSelectGenre"
            }
          }
        }
       }
      ]
    );

This would return documents like:

    _id: {
      genre: "suspense",
    },
   price: 10210.6,
   data: [
      {
        subGenre: "Thriller",
        flag: false,
      },
      {
        subGenre: "jumpScare",
        flag: true,
      }
      {
        subGenre: "horror",
        flag: false,
      }
    ]

After this, I need to run a $project pipeline where I have to only project that element of the data array where the flag is true. The flag will be true for only one element.

$project: {
          price: "$price",
          subGenre: {$...... } // some condition on data array??
}

The final output should look like this:

   price: 10210.6,
   subGenre: "jumpScare",

NeNaD · Accepted Answer · 2022-12-29 10:39:42Z

2

You can do it like this:

$filter - to filter items from data array, where flag property is equal to true.
$first - to get first item from above array.
$getField - to get value of subGenre property of the above item.

db.collection.aggregate([
  {
    "$project": {
      "_id": 0,
      "price": 1,
      "data": {
        "$getField": {
          "field": "subGenre",
          "input": {
            "$first": {
              "$filter": {
                "input": "$data",
                "cond": "$$this.flag"
              }
            }
          }
        }
      }
    }
  }
])

Working example

edited Dec 29, 2022 at 10:39

answered Dec 29, 2022 at 10:33

NeNaD

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4 Comments

Roomi Over a year ago

How to select that element where the flag does not exist? I am unable to use $exist with cond here. My syntax is similar to yours with difference in $filter - $filter: { input: "$data", cond: { $exists: [ "$$this.flag", false ] } } its giving error - Unrecognized expression '$exist'

NeNaD Over a year ago

$exists is not an Aggregate operator. Why didn't you use my solution? flag is Boolean anyway, so only the items where flag is true will be filtered.

Roomi Over a year ago

I want to select the element where the flag does not exist. Actually, the real problem is the flag will be set to true if that element of the array is edited. I need to select the element which was not edited. So I need the element where flag does not exist.

NeNaD Over a year ago

But you said the opposite in the question. In that case, you can use the $type operator: "cond": { "$eq": [ { "$type": "$$this.flag"}, "bool" ] }

turivishal · Accepted Answer · 2022-12-29 10:42:20Z

2

You can use $filter array operator to loop and filter elements by required conditions,

$filter to iterate loop of data array, if flag is true then return element
$let to define a variable and store the above filter result
$first to return the first element from the filtered result, you can also use $arrayElemAt if you are using lower version of the MongoDB

  {
    $project: {
      price: 1,
      subGenre: {
        $let: {
          vars: {
            data: {
              $filter: {
                input: "$data",
                cond: "$$this.flag"
              }
            }
          },
          in: { $first: "$$data.subGenre" }
        }
      }
    }
  }

Playground

Another approach using $indexOfArray and $arrayElemAt operators,

$indexOfArray will find the matching element index of the array
$arrayElemAt to get specific element by specifying the index of the element

  {
    $project: {
      price: 1,
      subGenre: {
        $arrayElemAt: [
          "$data.subGenre",
          { $indexOfArray: ["$data.flag", true] }
        ]
      }
    }
  }

Playground

edited Dec 29, 2022 at 10:42

answered Dec 29, 2022 at 10:34

turivishal

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2 Comments

NeNaD Over a year ago

Nice solution with $let!

turivishal Over a year ago

Yeah, there are many ways to do this, yours is also a nice solution with $getField.

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