'How to find Distinct No of Document on specific field MongoDB and Implement in Java

I have data and need to group by two fields and in response display unique filed along with the count on this field.

Sample data:

[
   {
      "categoryCode":"categoryCode1",
      "categoryName":"categoryName1",
      "cat_id": "A"
   },
   {
      "categoryCode":"categoryCode1",
      "categoryName":"categoryName1",
      "cat_id": "B"
   },
   {
      "categoryCode":"categoryCode2",
      "categoryName":"categoryName2",
      "cat_id": "C"
   }
]

Expected Output:

[
   {
      "categoryCode":"categoryCode1",
      "categoryName":"categoryName1",
      "cat_count": 2
   },
   {
      "categoryCode":"categoryCode2",
      "categoryName":"categoryName2",
      "cat_count": 1
   }
]

and how to implement it in Java with Aggregation Object.

GroupOperation groupOperation = Aggregation.group("categoryCode", "categoryName").sum(1).as("cat_count");

but getting compile error:

1 is not taken inside sum method

mongodbmongodb-queryspring-dataaggregation-framework


Solution 1:[1]

  1. $group - Group by categoryCode and categoryName. And add aggregate field cat_count for $sum when match with group.
  2. $project - Display document for desired output field.
db.collection.aggregate([
  {
    $group: {
      _id: {
        "categoryCode": "$categoryCode",
        "categoryName": "$categoryName"
      },
      "cat_count": {
        $sum: 1
      }
    }
  },
  {
    $project: {
      "_id": 0,
      "categoryCode": "$_id.categoryCode",
      "categoryName": "$_id.categoryName",
      "cat_count": "$cat_count"
    }
  }
])

Sample Mongo Playground

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Yong Shun

Related Questions