How to Ignore Duplicate Key Errors Safely Using insert_many

user_190894 0
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I need to ignore duplicate inserts when using insert_many with pymongo, where the duplicates are based on the index. I've seen this question asked on stackoverflow, but I haven't seen a useful answer.

Here's my code snippet:

try:
    results = mongo_connection[db][collection].insert_many(documents, ordered=False, bypass_document_validation=True)
except pymongo.errors.BulkWriteError as e:
    logger.error(e)

I would like the insert_many to ignore duplicates and not throw an exception (which fills up my error logs). Alternatively, is there a separate exception handler I could use, so that I can just ignore the errors. I miss "w=0"...

Thanks

pymongo mongodb python
3 Answers

Adding more to Neil's solution.

Having 'ordered=False, bypass_document_validation=True' params allows new pending insertion to occur even on duplicate exception.

from pymongo import MongoClient, errors

DB_CLIENT = MongoClient()
MY_DB = DB_CLIENT['my_db']
TEST_COLL = MY_DB.dup_test_coll

doc_list = [
    {
        "_id": "82aced0eeab2467c93d04a9f72bf91e1",
        "name": "shakeel"
    },
    {
        "_id": "82aced0eeab2467c93d04a9f72bf91e1",  # duplicate error: 11000
        "name": "shakeel"
    },
    {
        "_id": "fab9816677774ca6ab6d86fc7b40dc62",  # this new doc gets inserted
        "name": "abc"
    }
]

try:
    # inserts new documents even on error
    TEST_COLL.insert_many(doc_list, ordered=False, bypass_document_validation=True)
except errors.BulkWriteError as e:
    print(f"Articles bulk insertion error {e}")

    panic_list = list(filter(lambda x: x['code'] != 11000, e.details['writeErrors']))
    if len(panic_list) > 0:
        print(f"these are not duplicate errors {panic_list}")

And since we are talking about duplicates its worth checking this solution as well.

user_95928010

The correct solution is to use a WriteConcern with w=0 and ordered=False:

import pymongo
from pymongo.write_concern import WriteConcern


mongodb_connection[db][collection].with_options(write_concern=WriteConcern(w=0)).insert_many(messages, ordered=False)
user_1908940
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