Myujth

I want to find the duplicated document in my mongodb based on name, I have the following code:

def Check_BFA_DB(options):
 issue_list=
 client = MongoClient(options.host, int(options.port))
 db = client[options.db]
 collection = db[options.collection]
 names = ['$project': 'name':'$name']
 name_cursor = collection.aggregate(names, cursor=)
 for name in name_cursor:
 issue_list.append(name)
 print(name)

It will print all names, how can I print only the duplicated ones?

Appritiated for any help!

The following query will show only duplicates:

db['collection_name'].aggregate(['$group': '_id':'$name', 'count': '$sum': 1, '$match': 'count': '$gt': 1])

How it works:

Step 1:
Go over the whole collection, and group the documents by the property called name, and for each name count how many times it is used in the collection.

Step 2:
filter (using the keyword match) only documents in which the count is greater than 1 (the gt operator).

An example (written for mongo shell, but can be easily adapted for python):

db.a.insert(name: "name1")
db.a.insert(name: "name1")
db.a.insert(name: "name2")
db.a.aggregate(["$group": _id:"$name", count: "$sum": 1, $match: count: "$gt": 1])

Result is "_id" : "name1", "count" : 2

So your code should look something like this:

def Check_BFA_DB(options):
 issue_list=
 client = MongoClient(options.host, int(options.port))
 db = client[options.db]
 name_cursor = db[options.collection].aggregate([
 '$group': '_id': '$name', 'count': '$sum': 1,
 '$match': 'count': '$gt': 1
 ])

 for document in name_cursor:
 name = document['_id']
 issue_list.append(name)
 print(name)

BTW (not related to the question), python naming convention for function names is lowercase letters, so you might want to call it check_bfa_db()