I am using MongoDB to track unique views of a resource.
Everytime a user views a specific resource for the first time, a new view is logged in the db.
If that same user views the same resource again, the unique compound index on the collection blocks the insert of the duplicate.
For bulk inserts, with { ordered: false }, Mongo allows the new views through and blocks the duplicates. The return value of the insert is an object with an insertedCount property, telling me how many docs made it past the unique index.
In some cases, I want to know how many docs would be inserted before running the query. Then, based on the dummy insertedCount, I would choose to run the query, or not.
Is there a way to test a query and have it do everything except actually inserting the docs?
I could solve this by running some js serverside to get the answer I need. But I would prefer to let the db do those checks
Collection.InsertOne() returns a *InsertOneResult, which only contains the ID of the inserted document. To get the inserted document, you have to perform another Collection.Find() query. Is there a way to do this in a single step?
A current work around is to use Collection.FindOneAndUpdate() with Upsert set to true, as this returns a *SingleResult that can then be decoded into a struct, and sent back to the client.
If you wish your application to have the complete document:
Generate the _id on client side
Insert the complete document
At that point the document you have is exactly the document that the database has, and returning it from the insert is pointless.
Some other databases generate ids on the server side, but in case of MongoDB each driver implements id generation on the client side such that each document can be completely known prior to the insert.
The save() method seems to replace an entire document/record in my database. If I know the primary key of a document in my database, what is the best way in Grails for me to update a single field on that document without first querying the database using get()?
For instance, I don't like the following code because it executes two queries when all I want to do is to update myField.
def key = "foo"
def doc = MyDomain.get(key) // This is the query I want to eliminate
doc.myField = "bar"
doc.save()
In situations where I know the primary key, I want to simply update a single field, similarly to how Ruby on Rails leverages the ActionModel.update_attribute() method.
Even though my specific database is MongoDB, I think the question is applicable to any database, SQL or NoSQL. If the database supports the ability to update just a single field on one record via one query, using Grails can I avoid the extra get() query to retrieve a full record from the database?
While writing data to mongodb, we are checking if the data is present get the _id and using save update it else using insert add the data. Read save is the best way if you are providing _id in the query while saving it will update/insert based on if the _id is present in the db. Is the save the best method or is there any other way.
If you have all data available to save, just run update() each time but use the upsert functionality. Only one query required:
db.collection.update(
['_id' => $id],
$data,
['upsert' => true]
);
If your _id is generated by mongo you always know there is a record in the database and update is the one to use, but then again you could also save().
If you generated your id's (and thus don't know if it comes from the collection), this will always work without having to run an extra query.
From the documentation
db.collection.save()
Updates an existing document or inserts a new document, depending on its document parameter.
db.collection.insert()
Inserts a document or documents into a collection.
If you use db.collection.insert() in your case you will get duplication key error since it will try to insert new document which has same _id with an existing document. But instead of using save you should use the update method.
I have a collection in which all of my documents have at least these 2 fields, say name and url (where url is unique so I set up a unique index on it). Now if I try to insert a document with a duplicate url, it will give an error and halt the program. I don't want this behavior, but I need something like mysql's insert or ignore, so that mongoDB should not insert the document with duplicate url and continue with the next documents.
Is there some parameter I can pass to the insert command to achieve this behavior? I generally do a batch of inserts using pymongo as:
collection.insert(document_array)
Here collection is a collection and document_array is an array of documents.
So is there some way I can implement the insert or ignore functionality for a multiple document insert?
Set the continue_on_error flag when calling insert(). Note PyMongo driver 2.1 and server version 1.9.1 are required:
continue_on_error (optional): If True, the database will not stop
processing a bulk insert if one fails (e.g. due to duplicate IDs).
This makes bulk insert behave similarly to a series of single inserts,
except lastError will be set if any insert fails, not just the last
one. If multiple errors occur, only the most recent will be reported
by error().
Use insert_many(), and set ordered=False.
This will ensure that all write operations are attempted, even if there are errors:
http://api.mongodb.org/python/current/api/pymongo/collection.html#pymongo.collection.Collection.insert_many
Try this:
try:
coll.insert(
doc_or_docs=doc_array,
continue_on_error=True)
except pymongo.errors.DuplicateKeyError:
pass
The insert operation will still throw an exception if an error occurs in the insert (such as trying to insert a duplicate value for a unique index), but it will not affect the other items in the array. You can then swallow the error as shown above.
Why not just put your call to .insert() inside a try: ... except: block and continue if the insert fails?
In addition, you could also use a regular update() call with the upsert flag. Details here: http://www.mongodb.org/display/DOCS/Updating#Updating-update%28%29
If you have your array of documents already in memory in your python script, why not insert them by iterating through them, and simply catch the ones that fail on insertion due to the unique index?
for doc in docs:
try:
collection.insert(doc)
except pymongo.errors.DuplicateKeyError:
print 'Duplicate url %s' % doc
Where collection is an instance of a collection created from your connection/database instances and docs is the array of dictionaries (documents) you would currently be passing to insert.
You could also decide what to do with the duplicate keys that violate your unique index within the except block.
It is highly recommended to use upsert
stat.update({'location': d['user']['location']}, \
{'$inc': {'count': 1}},upsert = True, safe = True)
Here stat is the collection if visitor location is already present in the collection, count is increased by one, else count is set to 1.
Here is the link for documentation http://www.mongodb.org/display/DOCS/Updating#Updating-UpsertswithModifiers
What I am doing :
Generate array of MongoDB ids I want to insert (hash of some values in my case)
Remove existing IDs (I am using a Redis queue bcoz performance, but you can query mongo)
Insert your cleaned data !
Redis is perfect for that, you can use Memcached or Mysql Memory, according your needs