MongoDB : Can't insert twice the same document - mongodb

Question

On my pymongo code, inserting twice the same doc raises an error :
document = {"auteur" : "romain",
"text" : "premier post",
"tag" : "test2",
"date" : datetime.datetime.utcnow()}
collection.insert_one(document)
collection.insert_one(document)
raises :
DuplicateKeyError: E11000 duplicate key error collection: test.myCollection index: _id_ dup key: { : ObjectId('5aa282eff1dba231beada9e3') }
inserting two documents with different content works fine.
Seems like according to https://docs.mongodb.com/manual/reference/method/db.collection.createIndex/#options I should do something aobut option of indexes:
unique boolean
Optional. Creates a unique index so that the collection will not accept insertion or update of documents where the index key value matches an existing value in the index.
Specify true to create a unique index. The default value is false.
The option is unavailable for hashed indexes.

Answer 1

Adding to Peba's answer, you can use the .copy() method of python dictionary to avoid the mutation of the document itself.
document = {"auteur" : "romain",
"text" : "premier post",
"tag" : "test2",
"date" : datetime.datetime.utcnow()}
collection.insert_one(document.copy())
collection.insert_one(document.copy())
This way, each insert_one call get's a shallow copy of the document and at the same time keeps your code more pythonic.

Answer 2

Inserting a document implicitly generates an _id.
So after inserting the document it will mutate to
document = {"_id" : ObjectId('random_id_here'),
"auteur" : "romain",
"text" : "premier post",
"tag" : "test2",
"date" : datetime.datetime.utcnow()}
Trying to insert said document again will result in an error due to the duplicated _id.
You can create a new document with the same values and insert it.
document = {"auteur" : "romain",
"text" : "premier post",
"tag" : "test2",
"date" : datetime.datetime.utcnow()}
collection.insert_one(document)
document = {"auteur" : "romain",
"text" : "premier post",
"tag" : "test2",
"date" : datetime.datetime.utcnow()}
collection.insert_one(document)