MongoDB using duplicate function to print different field-CodePudding

db.collection.aggregate([{"$group" : {"_id": "$fieldA", "count": {"$sum": 1}}},{"$match":{"_id" :{"$ne":null}, "count" : {"$gt":1}}},{"$project": {"fieldA":"$_id", "_id" : 0}}]);

I am currently using this code to find duplicates however I actually want to print fieldB and not fieldA. By using the above I believe I have created a subset? Is there a simple way to print only fieldB?

ie. I want to print {"_id":0, "fieldB":1}

The only method I can think of is naming this subset and then joining with the original and then printing just fieldB but that doesn't seem efficient.

CodePudding user response：

To see the last fieldB you can do something like this:

$last: playground1 ( But here if there is different fieldB per duplicated fieldA you will not find out )

Or to see the all fieldB for the duplicated filedA you can do something like this:

$push: playground2

db.collection.aggregate([
{
"$group": {
  "_id": "$fieldA",
  "count": {
    "$sum": 1
  },
  fieldB: {
    $push: "$fieldB"
  }
 }
},
{
 "$match": {
  "_id": {
    "$ne": null
   },
  "count": {
    "$gt": 1
   }
  }
 },
 {
  "$project": {
  "fieldA": "$_id",
  "fieldB": 1,
  "_id": 0
  }
 }
])