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Home/ Questions/Q 8500541
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T00:58:25+00:00 2026-06-11T00:58:25+00:00

I have a table called categories and a table called business_categories_coupling . In Categories,

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I have a table called categories and a table called business_categories_coupling. In Categories, you have the usual id, name, parent. In the Coupling table, you have business_id and category_id. Each business can have multiple categories, so I store them in that table. It kinda looks like this:

business_id    category_id
73             80
73             81
73             90
74             4
74             10

Right now, my query is just selecting all the categories, doing a foreach and doing a db query in each loop to find how many businesses are in that category. Obviously not the right way to go about it.

Is there a way to do a SQL query that basically selects all the categories, gets the number of times it comes up in the coupling table, and add a count to each category?

SELECT
    C.*
FROM
    CATEGORIES AS C
LEFT JOIN
    BUSINESS_CATEGORIES_COUPLING AS B
ON
    C.id = B.category_id;

Kinda like that, but with a count somewhere. I’ve tried various setups but nothing works like I want. Any suggestions?

EDIT 1

Solution as provided by @phani-rahul, but I added a WHERE clause:

SELECT cat.id AS id, cat.name AS name, cat.slug AS slug, COUNT(cat.id) AS business_count
FROM categories AS cat
LEFT JOIN business_categories_coupling AS coupling ON cat.id=coupling.category_id
WHERE coupling.category_id IS NOT NULL
GROUP BY cat.id
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-11T00:58:27+00:00Added an answer on June 11, 2026 at 12:58 am

    Yes, there is.
    you can use Group by clause:

    select a.id as category, count(a.id) as count_of_category
    from categories a
    left join business_categories_coupling b on a.id=b.category_id
    group by a.id
    

    your result would be something like:

    category   count_of_category
    80          2
    81          5
    90          1
    .           .
    .           .
    .           .
    
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