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)
I am using pymongo (python module for mongodb).
I want the ObjectID to be created automatically by the server, however it seems to be created by pymongo itself when we don't specify it.
The problem it raises is that I use ObjectID to sort by time (by just sorting by the _id field). However it seems that it is using the time set on each computer so we cannot truly rely on it.
Any idea on how to solve this problem?
If you call save and pass it a document without an _id field, you can force the server to add the _id instead of the client by setting the (enigmatically-named) manipulate option to False:
coll.save({'foo': 'bar'}, manipulate=False)
I'm not Python user but I'm afraid there's no way to generate _id by server. For performance reasons _id is always generated by driver thus when you insert a document you don't need to do another query to get the _id back.
Here's a possible way you can do it by generating a int sequence _id, just like the IDENTITY ID of SqlServer. To do this, you need to keep a record in you certain collection for example in my project there's a seed, which has only one record:
{_id: ObjectId("..."), seqNo: 1 }
The trick is, you have to use findAndModify to keep the find and modify in the same "transaction".
var idSeed = db.seed.findAndModify({
query: {},
sort: {seqNo: 1},
update: { $inc: { seqNo: 1 } },
new: false
});
var id = idSeed.seqNo;
This way you'll have all you instances get a unique sequence# and you can use it to sort the records.
After upgrading MongoEngine from 0.7.9 to 0.8.3, any attempts to save any existing documents in any collection results in a NotUniqueError (user collection shown in example):
Tried to save duplicate unique keys (E11000 duplicate key error index: foo.user.$_id_ dup key: { : ObjectId('xxxxxx') })
I get the same error if I create a new document and save it more than once:
a = Foo()
a.save()
a.save() # results in duplicate error
Mongo by default creates an index on _id which cannot be removed, and I have no other indexes which use _id. Most issues similar to this that I've seen have been on duplicate indexes that aren't _id and can be removed, but this is really odd. I am doing nothing weird with the _id field, just letting Mongo generate it on its own.
Any ideas on what might be causing this to happen?
Thanks!
There was a custom save function which hadn't been migrated to using the new save() arguments, so one of them was caused force_insert to evaluate to true.
So dumb...
Need help in understanding what is happening here and a suggestion to avoid this!
Here is my snippet:
result = [list of dictionary objects(dictionary objects have 2 keys and 2 String values)]
copyResults = list(results);
## Here I try to insert each Dict into MongoDB (Using PyMongo)
for item in copyResults:
dbcollection.save(item) # This is all saving fine in MongoDB.
But when I loop thru that original result list again it shows dictionary objects with a new field added
automatically which is ObjectId from MongoDB!
Later in code I need to transform that original result list to json but this ObjectId is causing issues.No clue why this is getting added to original list.
I have already tried copy or creating new list etc. It still adds up ObjectId in the original list after saving.
Please suggest!
every document saved in mongodb requires '_id' field - which has to be unique among documents in the collection. if you don't provide one, mongodb will automatically create one with ObjectId (bson.objectid.ObjectId for pymongo)
If you need to export documents to json, you have to pop '_id' field before jsonifying it.
Or you could use:
rows['_id'] = str(rows['_id'])
Remember to set it back if you then need to update
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.