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Home/ Questions/Q 701821
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T03:39:09+00:00 2026-05-14T03:39:09+00:00

I’m currently developing an application to allow students to manage their courses, and I

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I’m currently developing an application to allow students to manage their courses, and I don’t really know how to design the database for a specific feature.
The client wants, a lot like Facebook, that when a student displays the list of people currently in a specific course, the people with the most mutual courses with the logged in user are displayed first. Almost the same as Facebook feature “Friend suggestions” with an additional filter.

As an additional feature, I would like to add a search feature to allow students to search for another one, and displaying first in the search results the people with most mutual courses with the logged in user.

I currently use MySQL, I plan to use Cassandra for some other features, and I also use Memcached for result caching and Sphinx for the search.

Thanks.

—

The app is developed in Python, BTW

And I forgot to mention that the standard approach (using a nice MySQL query to calculate all this with an ORDER BY clause) is wayyyys too slow. So as reads are a lot more frequent than reads, I would like most of the logic to take place once, when the relation people <-> course is added.

I thought about updating a “mutual courses” counter specific to one tuple (user, course) that will be increased for all users of a course when the logged in user joins a new course (or decreased when he leaves it).

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  1. Editorial Team
    Editorial Team
    2026-05-14T03:39:09+00:00Added an answer on May 14, 2026 at 3:39 am

    If you already have your solution, but the problem is just the speed of that query, try doing it sooner. When a user’s friendships change, rerun a job that calculates these things and store all the results away. Don’t runt his as a result of a request, when you need the result so quickly. Do such expensive things only once and do them before a request is ever made.

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