Find duplicate records in MongoDB

Use aggregation on name and get name with count > 1:

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

To sort the results by most to least duplicates:

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

To use with another column name than "name", change "$name" to "$column_name"

You can find the list of duplicate names using the following aggregate pipeline:

• Group all the records having similar name.
• Match those groups having records greater than 1.
• Then group again to project all the duplicate names as an array.

The Code:

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

o/p:

{ "duplicateNames" : [ "ksqn291", "ksqn29123213Test" ] }

The answer anhic gave can be very inefficient if you have a large database and the attribute name is present only in some of the documents.

To improve efficiency you can add a $match to the aggregation.

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

db.getCollection('orders').aggregate([  
    {$group: { 
            _id: {name: "$name"},
            uniqueIds: {$addToSet: "$_id"},
            count: {$sum: 1}
        } 
    },
    {$match: { 
        count: {"$gt": 1}
        }
    }
])

First Group Query the group according to the fields.

Then we check the unique Id and count it, If count is greater then 1 then the field is duplicate in the entire collection so that thing is to be handle by $match query.