20

I have a collection containing entries in following format:

{ 
    "_id" : ObjectId("5538e75c3cea103b25ff94a3"), 
    "userID" : "USER001", 
    "userName" : "manish", 
    "collegeIDs" : [
        "COL_HARY",
        "COL_MARY",
        "COL_JOHNS",
        "COL_CAS",
        "COL_JAMES",
        "COL_MARY",
        "COL_MARY",
        "COL_JOHNS"
    ]
}

I need to find out the collegeIDs those are repeating. So the result should give "COL_MARY","COL_JOHNS" and if possible the repeating count. Please do give a mongo query to find it.

2
  • 1
    possible duplicate of How to remove duplicate entries from an array? Commented Sep 10, 2015 at 13:09
  • 2
    Please, search for other similar questions before posting your own. I found this through Googling "mongodb find duplicate values array" in under a minute. There are plenty of resources out there to help you with this. Attempt also to show us what you have done. That way we can better guide you. Commented Sep 10, 2015 at 13:11

1 Answer 1

25

Probably there would be many of these documents and thus you want it per ObjectId.

db.myCollection.aggregate([
  {"$project": {"collegeIDs":1}},
  {"$unwind":"$collegeIDs"},
  {"$group": {"_id":{"_id":"$_id", "cid":"$collegeIDs"}, "count":{"$sum":1}}},
  {"$match": {"count":{"$gt":1}}},
  {"$group": {"_id": "$_id._id", "collegeIDs":{"$addToSet":"$_id.cid"}}}
])

This might be what you want to, not clear from your question:

db.myCollection.aggregate([
  {"$match": {"userID":"USER001"}},
  {"$project": {"collegeIDs":1, "_id":0}},
  {"$unwind":"$collegeIDs"},
  {"$group": {"_id":"$collegeIDs", "count":{"$sum":1}}},
  {"$match": {"count":{"$gt":1}}},
])
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5 Comments

yes.., i want to find it for a particular user in that collection. i mean filter by giving userid in the query.
Then there is your solution, just ignore the down voter whoever he is he did without thinking. With your data result is: { "_id" : ObjectId("5538e75c3cea103b25ff94a3"), "collegeIDs" : [ "COL_MARY", "COL_JOHNS" ] }
where should i give the user id in that query ?
Ahh you want it for a particular UserID? Then add a $match at top.
this perfectly works.

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