Question 115848 in Mongodb Question posted in
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How to compare two objects in MongoDB (ignoring order of keys)?

What is the best way to find all the documents where objA is the same as objB (order of keys is not important)?

Inspired by another question by @Digvijay, I was looking for a way to compare two objects on MongoDB query and could not find a relevant solution on SO.

Sample data:

[
  {
    objA: {a: 1, b: 2},
    objB: {a: 1, b: 2}
  },
  {
    objA: {m: "g", c: 5},
    objB: {c: 5, m: "g"}
  },
  {
    objA: {m: "g", c: 7},
    objB: {c: 5, m: "g"}
  },
  {
    objA: {m: "g", c: 7},
    objB: {b: "g", c: 7}
  }
]

Expected results:

[
  {
    objA: {a: 1, b: 2},
    objB: {a: 1, b: 2}
  },
  {
    objA: {m: "g", c: 5},
    objB: {c: 5, m: "g"}
  },
]
Tags:

2

Answers


  1. 0

    Maybe use the old-school trick of converting them into an array by $objectToArray. Use $sortArray to sort by key. Compare by the sorted array to get the matches.

    db.collection.aggregate([
  {
    $addFields: {
      "sortedA": {
        $sortArray: {
          input: {
            "$objectToArray": "$objA"
          },
          sortBy: {
            k: 1
          }
        }
      },
      "sortedB": {
        $sortArray: {
          input: {
            "$objectToArray": "$objB"
          },
          sortBy: {
            k: 1
          }
        }
      }
    }
  },
  {
    "$match": {
      $expr: {
        $eq: [
          "$sortedA",
          "$sortedB"
        ]
      }
    }
  },
  {
    "$unset": [
      "sortedA",
      "sortedB"
    ]
  }
])

    Mongo Playground

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  2. 0

    You can do it like this:

    • $objectToArray – to transform objA and objB to arrays.
    • $setEquals – to compare if above arrays have the same distinct elements.
    db.collection.aggregate([
  {
    $match: {
      $expr: {
        $setEquals: [
          { $objectToArray: "$objA" },
          { $objectToArray: "$objB" }
        ]
      }
    }
  }
])

    Working example

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