How to handle DuplicateKeyError in MongoDB (pyMongo)? - python

Could anybody please tell me how to handle DuplicateKeyError in MongoDB?
I am writing a python script, where I move several docs from two different collections into a third one. There is a small overlap between the two collections due to having a few identical documents (with identical ObjectId). This results in the following:
DuplicateKeyError: E11000 duplicate key error collection: admin.collection_test index: id dup key: { : ObjectId('593a920b529e170d4b8fbf72') }
In order to get rid of the error I use:
try:
do something
except pymongo.errors.DuplicateKeyError:
pass
I expect by using of the "try-except" to move all non-crossing documents to the third collection, but instead the script just peacefully stops running once a first overlap (an already existing document in the collection) appears.
Would appreciate any help a lot!

If you're iterating over the documents, try using continue instead of pass.
for doc in documents:
try:
# insert into new collection
except pymongo.errors.DuplicateKeyError:
# skip document because it already exists in new collection
continue

for doc in documents:
client.update_one({'_id': doc['_id']}, doc, upsert=True)
You can use update_one with upsert=True. This updates doc with new doc if doc exists already otherwise it creates new doc.

CONNECT TO DATABASE
connection = pymongo.MongoClient("192.168.2.202", 27017)
CREATE DATABASE
database = connection['my_database']
CREATE COLLECTION
collection = database['my_collection']
INSERT DOCUMENTS IN COLLECTION
url="http://some/api/url/path/?format=json"
data = {
'_id': url,
'timestamp': datetime.datetime.now(),
'data': {
'XX': 1,
'YY': 2,
'ZZ': 3
}
}
TO AVOID DUPLICATES - THIS WILL CREATE NEW DOCUMENT IF SAME ID NOT EXIST
collection.update_one({'_id': url}, {"$set": data}, upsert=True)

Related

How to prevent python from inserting duplicates into mongodb?

I have json data stored in a variable that gets inserted into a MongoDB, once per day with Python. But the json data in the variable often does not change, but it still get's inserted into the MongoDB ... which creates masses of duplicates of the same entries.
Every entry in the json data variable each has 1 unique key: uuid.
How do you prevent Python from inserting duplicates into MongoDB? I looked into db.collection.update(), but I'm not sure if its suitable and I don't know how to use it with a variable?
As long as you can check its id for uniqueness, you can use the method update_one() and set upsert for that.
For example,
filter_data = {'uuid': '111'}
new_data = {'$set': {'new_value': 25}}
db.collection.update_one(filter_data, new_data, upsert=True)
This will check if uuid = '111' exists, if not it will create a document; otherwise, update it.

solve E11000 duplicate key error collection: _id_ dup key in pymongo

I am trying to insert a great number of document(+1M) using a bulk_write instruction. In order to do that, I create a list of InsertOne function.
python version = 3.7.4
pymongo version = 3.8.0
Document creation:
document = {
'dictionary': ObjectId(dictionary_id),
'price': price,
'source': source,
'promo': promo,
'date': now_utc,
'updatedAt': now_utc,
'createdAt:': now_utc
}
# add line to debug
if '_id' in document.keys():
print(document)
return document
I create the full list of document by adding a new field from a list of elements and create the query by using InsertOne
bulk = []
for element in list_elements:
for document in documents:
document['new_field'] = element
# add line to debug
if '_id' in document.keys():
print(document)
insert = InsertOne(document)
bulk.append(insert)
return bulk
I do the insert by using bulk_write command
collection.bulk_write(bulk, ordered=False)
I attach the documentation https://api.mongodb.com/python/current/api/pymongo/collection.html#pymongo.collection.Collection.bulk_write
According to the documentation,the _id field is added automatically
Parameter - document: The document to insert. If the document is missing an _id field one will be added.
And somehow it seems that is doing it wrong because some of them have the same value.
Receiving this error(with differents _id of course) for 700k of the 1M documents
'E11000 duplicate key error collection: database.collection index: _id_ dup key: { _id: ObjectId(\'5f5fccb4b6f2a4ede9f6df62\') }'
Seems a bug to me from pymongo, because I used this approach in many situations but I didn't with such size of documents
The _id field has to be unique for sure, but, due to this is done automatically by pymongo, I don't know how to approach to this problem, perhaps using a UpdateOne with upsert True with an impossible filter and hope for the best.
I would appreciate any solution or work around for this problem
It seems that as I was adding the new field of the document and append it into the list, I created similar instances of the same element, so I had the same queries len(list_elements) times and that is why I had the duplicated key error.
to solve the problem, I append to the list a copy of the document
bulk.append(document.copy())
and then create the queries with that list
I would like to thank #Belly Buster for his help in the issue
If any of the documents from your code snippet already contain an _id, a new one won't be added, and you run the risk of getting a duplicate error as you have observed.

How to access the value of one atribute in pymogo

I am new to pymongo and mongo db and created a cluster with the database and the collection I need. I have added data to the Collection but i'm having trouble retrieving the data as individual values. I need to find the only record where the Gender is female and check the Name belonging to that record, Not been able to find help online. Sorry if it's a noob question
import pymongo
client = pymongo.MongoClient('mongodb+srv://#test-v6kig.mongodb.net/admin')
testdb = client['mytestdb']
testcol = testdb['mytestcol']
myquery = { "Gender" : "Female" }
data = testcol.find(myquery)
if "Need to get value of NAME attribute in the data record" == "Sushmit":
print("Y")
else:
print("N")
Any help is appreciated :)
use find_one methods and pass to it the id parameter like this:
single_value= testcol.find_one({'id'})
find_one always returns a single result which is a JSON document. find always returns a cursor (even if there is only a single result or no results). So to get the results from your find query you would do:
cursor=pymongo.find(myquery)
for doc in cursor:
print(doc)

Formatting JSON before inserting to MongoDB

I'm new to MongoDB and I'm still learning so please bear with me.
Let's say I have a document in Mongo with the following form
{ 'objName':
{
'id': 012345678,
'name': 'someName'
}
}
If I insert this JSON using pyMongo it'll take a default Object ID.
What I'd like to do is to set the
_id = 'id'
that is given inside the 'ObjName'.
The problem that I'm facing is that I don't know the value inside 'ObjName' so I need something generic to work regardless of the value inside.
You can assign value to _id
doc = {'objName': {'id': 12345678,'name': 'someName'}}
doc['_id'] = doc['objName']['id']
collection.insert_one(doc)

mongodb: insert if not exists

Every day, I receive a stock of documents (an update). What I want to do is insert each item that does not already exist.
I also want to keep track of the first time I inserted them, and the last time I saw them in an update.
I don't want to have duplicate documents.
I don't want to remove a document which has previously been saved, but is not in my update.
95% (estimated) of the records are unmodified from day to day.
I am using the Python driver (pymongo).
What I currently do is (pseudo-code):
for each document in update:
existing_document = collection.find_one(document)
if not existing_document:
document['insertion_date'] = now
else:
document = existing_document
document['last_update_date'] = now
my_collection.save(document)
My problem is that it is very slow (40 mins for less than 100 000 records, and I have millions of them in the update).
I am pretty sure there is something builtin for doing this, but the document for update() is mmmhhh.... a bit terse.... (http://www.mongodb.org/display/DOCS/Updating )
Can someone advise how to do it faster?
Sounds like you want to do an upsert. MongoDB has built-in support for this. Pass an extra parameter to your update() call: {upsert:true}. For example:
key = {'key':'value'}
data = {'key2':'value2', 'key3':'value3'};
coll.update(key, data, upsert=True); #In python upsert must be passed as a keyword argument
This replaces your if-find-else-update block entirely. It will insert if the key doesn't exist and will update if it does.
Before:
{"key":"value", "key2":"Ohai."}
After:
{"key":"value", "key2":"value2", "key3":"value3"}
You can also specify what data you want to write:
data = {"$set":{"key2":"value2"}}
Now your selected document will update the value of key2 only and leave everything else untouched.
As of MongoDB 2.4, you can use $setOnInsert (http://docs.mongodb.org/manual/reference/operator/setOnInsert/)
Set insertion_date using $setOnInsert and last_update_date using $set in your upsert command.
To turn your pseudocode into a working example:
now = datetime.utcnow()
for document in update:
collection.update_one(
filter={
'_id': document['_id'],
},
update={
'$setOnInsert': {
'insertion_date': now,
},
'$set': {
'last_update_date': now,
},
},
upsert=True,
)
You could always make a unique index, which causes MongoDB to reject a conflicting save. Consider the following done using the mongodb shell:
> db.getCollection("test").insert ({a:1, b:2, c:3})
> db.getCollection("test").find()
{ "_id" : ObjectId("50c8e35adde18a44f284e7ac"), "a" : 1, "b" : 2, "c" : 3 }
> db.getCollection("test").ensureIndex ({"a" : 1}, {unique: true})
> db.getCollection("test").insert({a:2, b:12, c:13}) # This works
> db.getCollection("test").insert({a:1, b:12, c:13}) # This fails
E11000 duplicate key error index: foo.test.$a_1 dup key: { : 1.0 }
You may use Upsert with $setOnInsert operator.
db.Table.update({noExist: true}, {"$setOnInsert": {xxxYourDocumentxxx}}, {upsert: true})
Summary
You have an existing collection of records.
You have a set records that contain updates to the existing records.
Some of the updates don't really update anything, they duplicate what you have already.
All updates contain the same fields that are there already, just possibly different values.
You want to track when a record was last changed, where a value actually changed.
Note, I'm presuming PyMongo, change to suit your language of choice.
Instructions:
Create the collection with an index with unique=true so you don't get duplicate records.
Iterate over your input records, creating batches of them of 15,000 records or so. For each record in the batch, create a dict consisting of the data you want to insert, presuming each one is going to be a new record. Add the 'created' and 'updated' timestamps to these. Issue this as a batch insert command with the 'ContinueOnError' flag=true, so the insert of everything else happens even if there's a duplicate key in there (which it sounds like there will be). THIS WILL HAPPEN VERY FAST. Bulk inserts rock, I've gotten 15k/second performance levels. Further notes on ContinueOnError, see http://docs.mongodb.org/manual/core/write-operations/
Record inserts happen VERY fast, so you'll be done with those inserts in no time. Now, it's time to update the relevant records. Do this with a batch retrieval, much faster than one at a time.
Iterate over all your input records again, creating batches of 15K or so. Extract out the keys (best if there's one key, but can't be helped if there isn't). Retrieve this bunch of records from Mongo with a db.collectionNameBlah.find({ field : { $in : [ 1, 2,3 ...}) query. For each of these records, determine if there's an update, and if so, issue the update, including updating the 'updated' timestamp.
Unfortunately, we should note, MongoDB 2.4 and below do NOT include a bulk update operation. They're working on that.
Key Optimization Points:
The inserts will vastly speed up your operations in bulk.
Retrieving records en masse will speed things up, too.
Individual updates are the only possible route now, but 10Gen is working on it. Presumably, this will be in 2.6, though I'm not sure if it will be finished by then, there's a lot of stuff to do (I've been following their Jira system).
I don't think mongodb supports this type of selective upserting. I have the same problem as LeMiz, and using update(criteria, newObj, upsert, multi) doesn't work right when dealing with both a 'created' and 'updated' timestamp. Given the following upsert statement:
update( { "name": "abc" },
{ $set: { "created": "2010-07-14 11:11:11",
"updated": "2010-07-14 11:11:11" }},
true, true )
Scenario #1 - document with 'name' of 'abc' does not exist:
New document is created with 'name' = 'abc', 'created' = 2010-07-14 11:11:11, and 'updated' = 2010-07-14 11:11:11.
Scenario #2 - document with 'name' of 'abc' already exists with the following:
'name' = 'abc', 'created' = 2010-07-12 09:09:09, and 'updated' = 2010-07-13 10:10:10.
After the upsert, the document would now be the same as the result in scenario #1. There's no way to specify in an upsert which fields be set if inserting, and which fields be left alone if updating.
My solution was to create a unique index on the critera fields, perform an insert, and immediately afterward perform an update just on the 'updated' field.
1. Use Update.
Drawing from Van Nguyen's answer above, use update instead of save. This gives you access to the upsert option.
NOTE: This method overrides the entire document when found (From the docs)
var conditions = { name: 'borne' } , update = { $inc: { visits: 1 }} , options = { multi: true };
Model.update(conditions, update, options, callback);
function callback (err, numAffected) { // numAffected is the number of updated documents })
1.a. Use $set
If you want to update a selection of the document, but not the whole thing, you can use the $set method with update. (again, From the docs)...
So, if you want to set...
var query = { name: 'borne' }; Model.update(query, ***{ name: 'jason borne' }***, options, callback)
Send it as...
Model.update(query, ***{ $set: { name: 'jason borne' }}***, options, callback)
This helps prevent accidentally overwriting all of your document(s) with { name: 'jason borne' }.
In general, using update is better in MongoDB as it will just create the document if it doesn't exist yet, though I'm not sure how to work that with your python adapter.
Second, if you only need to know whether or not that document exists, count() which returns only a number will be a better option than find_one which supposedly transfer the whole document from your MongoDB causing unnecessary traffic.
Method For Pymongo
The Official MongoDB Driver for Python
5% of the times you may want to update and overwrite, while other times you like to insert a new row, this is done with updateOne and upsert
95% (estimated) of the records are unmodified from day to day.
The following solution is taken from this core mongoDB function:
db.collection.updateOne(filter, update, options)
Updates a single document within the collection based on the filter.
This is done with this Pymongo's function update_one(filter, new_values, upsert=True)
Code Example:
# importing pymongo's MongoClient
from pymongo import MongoClient
conn = MongoClient('localhost', 27017)
db = conn.databaseName
# Filter by appliances called laptops
filter = { 'user_id': '4142480', 'question_id': '2801008' }
# Update number of laptops to
new_values = { "$set": { 'votes': 1400 } }
# Using update_one() method for single update with upsert.
db.collectionName.update_one(filter, new_values, upsert=True)
What upsert=True Do?
Creates a new document if no documents match the filter.
Updates a single document that matches the filter.
I do propose the using of await now.

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