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pythonmongodbpymongo

pymongo sort and find_one issue

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

Solution

  • 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.