Performing regex queries with PyMongo

I am trying to perform a regex query using PyMongo against a MongoDB server. The document structure is as follows

{
  "files": [
    "File 1",
    "File 2",
    "File 3",
    "File 4"
  ],
  "rootFolder": "/Location/Of/Files"
}

I want to get all the files that match the pattern *File. I tried doing this as such

db.collectionName.find({'files':'/^File/'})

Yet I get nothing back. Am I missing something, because according to the MongoDB docs this should be possible? If I perform the query in the Mongo console it works fine, does this mean the API doesn't support it or am I just using it incorrectly?

Solution 1:

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})

Solution 2:

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

Solution 3:

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.

Solution 4:

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 )