Performing regex queries with PyMongo

Question

If you want to include regular expression options (such as ignore case), try this:

import re
regx = re.compile("^foo", re.IGNORECASE)
db.users.find_one({"files": regx})

Turns out regex searches are done a little differently in pymongo but is just as easy.

Regex is done as follows :

db.collectionname.find({'files':{'$regex':'^File'}})

This will match all documents that have a files property that has a item within that starts with File

To avoid the double compilation you can use the bson regex wrapper that comes with PyMongo:

>>> regx = bson.regex.Regex('^foo')
>>> db.users.find_one({"files": regx})

Regex just stores the string without trying to compile it, so find_one can then detect the argument as a 'Regex' type and form the appropriate Mongo query.

I feel this way is slightly more Pythonic than the other top answer, e.g.:

>>> db.collectionname.find({'files':{'$regex':'^File'}})

It's worth reading up on the bson Regex documentation if you plan to use regex queries because there are some caveats.

The solution of re doesn't use the index at all. You should use commands like:

db.collectionname.find({'files':{'$regex':'^File'}})

( I cannot comment below their replies, so I reply here )