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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T02:44:00+00:00 2026-05-26T02:44:00+00:00

We have 2500 products on our site, ranked between 60 different categories. Our DB

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We have 2500 products on our site, ranked between 60 different categories. Our DB scheme is 61 columns, labled “product_id”, and then the categories: “category_1”, “category_2″… “category_60”, and 2500 rows, one for each product. If a product is not ranked in a specific cateogry, that corresponding field is marked “0”. If it is ranked, the field is an INT with whatever rank it is: “1” is 1st, “2” is second, etc.

Usually products are only ranked in 2-3 categories, so there are 57+ columns with a “0” in the field. My current query is:

mysql_query("SELECT AVG(category_1 + category_2 + category_3 + category_4 + category_5 + category_6 + category_7 + category_8 + category_9 + category_10 + category_11 + category_12 + category_13 + category_14 + category_15 + category_16 + category_17 + category_18 + category_19 + category_20 + category_21 + category_22 + category_23 + category_24 + category_25 + category_26 + category_27 + category_28 + category_29 + category_30 + category_31 + category_32 + category_33 + category_34 + category_35 + category_36 + category_37 + category_38 + category_39 + category_40 + category_41 + category_42 + category_43 + category_44 + category_45 + category_46 + category_47 + category_48 + category_49 + category_50 + category_51 + category_52 + category_53 + category_54 + category_55 + category_56 + category_57 + category_58 + category_59 + category_60) as 'cat_avg' FROM products.rankings WHERE product_id = '$product_id'");

With this, I’m just getting the sum of the columns, not the AVG. Maybe this has something to do with selecting rows instead of columns, I’m not sure. I tried SUM as well, instead of AVG, same thing.

I’m not really sure where to go from here. What i would like is the Average ranking across all columns for one product, where the column doesn’t equal 0. So if a product_id 123 is ranked 7, 9 and 11, and then the other 57 columns are 0, the average returned would be 9 ((7+9+11)/3), not .45 ((7+9+11+0+0+0….+0))/60)

Note: I did not design this DB, I’m sure there is a better way to design it, but at this point it’s too deeply integrated to change up quickly.

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  1. Editorial Team
    Editorial Team
    2026-05-26T02:44:01+00:00Added an answer on May 26, 2026 at 2:44 am

    Restructured the whole DB… wasn’t as bad as I though, just did a bunch of MySQL queries/updates that got me what I needed. Strained the server for a few hours, but it was well worth it in the end.

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