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Home/ Questions/Q 830739
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T04:05:41+00:00 2026-05-15T04:05:41+00:00

I inherited a mysql database that has a table with columns like this: object_id,

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I inherited a mysql database that has a table with columns like this:

object_id, property, value

It holds data like this:

1,first_name,Jane
1,last_name,Doe
1,age,10
1,color,red
2,first_name,Mike
2,last_name,Smith
2,age,20
2,color,blue
3,first_name,John
3,last_name,Doe
3,age,20
3,color,red
...

Basically what I want to do is treat this table as a regular table. How would I get the id numbers (or all properties) of a person who is age 20 sorted by last and than first name? So far I have:

SELECT object_id FROM table WHERE property = 'age' AND value = '20'
union
SELECT object_id FROM table WHERE property = 'color' AND value = 'red'

But I’m not sure how to go about ordering the data.

Thanks

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  1. Editorial Team
    Editorial Team
    2026-05-15T04:05:41+00:00Added an answer on May 15, 2026 at 4:05 am

    The difficulties of reconstituting rows in an EAV model is exactly what you are seeing.

    You have to pivot the data from rows to columns to get to your “objects” and then query against it.

    Typically a pivot can be done with:

    SELECT object_id
        ,MAX(CASE WHEN property = 'age' THEN value ELSE NULL END) AS age
        ,MAX(CASE WHEN property = 'color' THEN value ELSE NULL END) AS color
    -- ...
    FROM EAVTABLE
    GROUP BY object_id
    

    Then:

    SELECT *
    FROM (
        SELECT object_id
            ,MAX(CASE WHEN property = 'age' THEN value ELSE NULL END) AS age
            ,MAX(CASE WHEN property = 'color' THEN value ELSE NULL END) AS color
        -- ...
        FROM EAVTABLE
        GROUP BY object_id
    ) AS objects
    WHERE age = '20' -- Note you may want to cast from char to proper types above (and these casts can fail).
    ORDER BY last_name, first_name
    
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