Combining regex with $in operator in mongodb

Question

I'm building regex queries for mongo

        s_list = [f"/^.*{x}.*/i" for x in s_list]
        print(s_list)
        query: dict = resumes_collection.find({
            "first_name": {
                "$in": s_list
            },
        })

This regex is supposed to get all data regardless of case sensitivity etc if we have loop in the database if we enter ^.*OP.*/i or ^.*oO.*/i will return loop in either case Currently this returns nothing. https://www.mongodb.com/docs/manual/reference/operator/query/regex/ Docs for reference. Using motor to interact with mongodb, motor runs pymongo under the hood

You do not need [f"/^.*{x}.*/i" for x in s_list]. Remove the first line and try "first_name": { "$regex": re.compile("|".join(s_list), re.I) }. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Apr 9, 2022 at 9:20

Wiktor Stribiżew · Accepted Answer · 2022-04-09 10:15:26Z

1

You can use

query: dict = resumes_collection.find({
            "first_name": {
                "$regex": re.compile("|".join(s_list), re.I)
            },
        })

Notes:

The [f"/^.*{x}.*/i" for x in s_list] generates a list of strings that are meant to be parsed as regex, but the $in operator will treat them as strings
/.../ regex literal notation is not supported in Python, you need to define a regex object using re.compile
"|".join(s_list) generates a single pattern out of all the s_list strings joining them with a | OR operator. You do not need to match the whole string, adding .* on both sides is not necessary and does more harm in fact.

answered Apr 9, 2022 at 10:15

Wiktor Stribiżew

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