MongoDB sort data by most number of records-CodePudding

I want to sort with the most occurrences of any value in a collection.

Eg.

{
  "id": "ID",
  "fruit": 'Apple'
}, 
{
  "id": "ID",
  "fruit": 'Banana'
}, 
{
  "id": "ID",
  "fruit": 'Apple'
}, 
{
  "id": "ID",
  "fruit": 'Orange'
}, 
{
  "id": "ID",
  "fruit": 'Apple'
}, 
{
  "id": "ID",
  "fruit": 'Banana'
}, 
{
  "id": "ID",
  "fruit": 'Apple'
}

I need fruit with the name Apple should on top because Apple occurs more time than other fruit in above collection

CodePudding user response：

You could use the "$sortByCount" aggregation stage to accomplish this.

db.collection.aggregate([
  {
    "$sortByCount": "$fruit"
  }
])

Example output:

[
  {
    "_id": "Apple",
    "count": 4
  },
  {
    "_id": "Banana",
    "count": 2
  },
  {
    "_id": "Orange",
    "count": 1
  }
]

Try it on mongoplayground.net.

CodePudding user response：

You'll need to use the aggregation framework to do that.

You can use this MongoDB playground to test it, you need to run the following aggregation pipeline:

db.collection.aggregate([
  {
    $group: {
      _id: "$fruit",
      count: {
        $sum: 1
      },
      documents: {
        $push: "$$ROOT"
      }
    }
  },
  {
    $sort: {
      count: -1
    }
  },
  {
    $unwind: {
      path: "$documents"
    }
  },
  {
    $replaceWith: "$documents"
  }
])