如何使用 mongodb 在聚合中计算布尔值

Question

Here is my query to get count of only true in accept_one (boolean type) reject_one (boolean type) value in collection of Mongodb.

Collection looks like:

 {accept_one:true,
    reject_one: true,
    desc:"abcd"

    }
    {accept_one:false,
    reject_one: false
    desc:"efghi"
    },
    {accept_one:true,
    reject_one: true,
    desc:"efghi"
    },
    {accept_one: false,
    reject_one: false,
    desc:"abcd"
    },
     {accept_one: true,
    reject_one: false,
    desc:"abcd"
    },
    {accept_one: false,
        reject_one: true,
        desc:"xyz"
        }

Here is my query:

db.aggregate([
            {
              $facet: {
                 "accept_one": [
                   { $group: { _id: '$accept_one', c: { $sum:1 } }
                 ],
                 "reject_one": [
                   { $group: { _id: '$reject_one', c: { $sum: 1 } } }
                 ],
                 "desc": [
                   { $group: { _id: '$desc', c: { $sum: 1 } } }
                 ],
                 }
    }
 ])

By this query every value whether if it's false or true it returns me count of every false and true but I only want to count true in fields, I mentioned above.

Expecting Output:

[
accept_one:{
_id:true,
c: 3
},
reject_one: {
_id: true,
c: 3
}
]

Answer 1

Add $match stage to filter true value for accept_one and reject_one in $facet .

db.collection.aggregate([
  {
    $facet: {
      "accept_one": [
        {
          $match: {
            "accept_one": true
          }
        },
        {
          $group: {
            _id: "$accept_one",
            c: {
              $sum: 1
            }
          }
        }
      ],
      "reject_one": [
        {
          $match: {
            "reject_one": true
          }
        },
        {
          $group: {
            _id: "$reject_one",
            c: {
              $sum: 1
            }
          }
        }
      ],
      "desc": [
        {
          $group: {
            _id: "$desc",
            c: {
              $sum: 1
            }
          }
        }
      ],
      
    }
  }
])

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