pymongo sort and find_one issue
Question:
I am trying to sort a collection called user_score using the key position and get the very first document of the result. In this case the collection user_score doesn’t exist and I was hoping to get the result as None, but i was getting a cursor back.
1.
result =
db.user_score.find({'score':'$lt':score}}).sort("position,pymongo.DESCENDING").limit(1)
Now i changed my query like below and did not get anything as expected.
2.
result =
db.user_score.find_one({'score':{'$lt':score}}, sort=[("position", pymongo.DESCENDING)])
What’s the problem with my first query?
Thanks
Answers:
This is the default mongodb behavior on find. Whenever you use find you get a list of the result (in this case an iterable cursor). Only findOne – or it’s PyMongo equivalent find_one will return None if the query has no matches.
In your first query, in the sort function you’re passing one argument ("position,pymongo.DESCENDING"), when you should be passing two arguments ("position", pymongo.DESCENDING).
Be sure to mind your quotation marks.
Use list to convert the value of the cursor into a dict:
list(db.user_score.find({'score':'$lt':score}}).sort("position",pymongo.DESCENDING).limit(1))[0]
A little late in my response but it appears that the current version of PyMongo does support a sort operation on a find_one call.
From the documentation page here:
All arguments to find() are also valid arguments for find_one(),
although any limit argument will be ignored. Returns a single
document, or None if no matching document is found.
Example usage is as follows:
filterdict = {'email' : '[email protected]'}
collection.find_one(filterdict, sort=[('lastseen', 1)])
Hope this helps more recent searchers!
I am trying to sort a collection called user_score using the key position and get the very first document of the result. In this case the collection user_score doesn’t exist and I was hoping to get the result as None, but i was getting a cursor back.
1.
result =
db.user_score.find({'score':'$lt':score}}).sort("position,pymongo.DESCENDING").limit(1)
Now i changed my query like below and did not get anything as expected.
2.
result =
db.user_score.find_one({'score':{'$lt':score}}, sort=[("position", pymongo.DESCENDING)])
What’s the problem with my first query?
Thanks
This is the default mongodb behavior on find. Whenever you use find you get a list of the result (in this case an iterable cursor). Only findOne – or it’s PyMongo equivalent find_one will return None if the query has no matches.
In your first query, in the sort function you’re passing one argument ("position,pymongo.DESCENDING"), when you should be passing two arguments ("position", pymongo.DESCENDING).
Be sure to mind your quotation marks.
Use list to convert the value of the cursor into a dict:
list(db.user_score.find({'score':'$lt':score}}).sort("position",pymongo.DESCENDING).limit(1))[0]
A little late in my response but it appears that the current version of PyMongo does support a sort operation on a find_one call.
From the documentation page here:
All arguments to find() are also valid arguments for find_one(),
although any limit argument will be ignored. Returns a single
document, or None if no matching document is found.
Example usage is as follows:
filterdict = {'email' : '[email protected]'}
collection.find_one(filterdict, sort=[('lastseen', 1)])
Hope this helps more recent searchers!