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Home/ Questions/Q 7576163
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Editorial Team
  • 0
Editorial Team
Asked: May 30, 20262026-05-30T16:51:21+00:00 2026-05-30T16:51:21+00:00

SELECT upd.*, usr.username AS `username`, usr.profile_picture AS `profile_picture` ,( SELECT COUNT (like.id) FROM likes

  • 0
SELECT 
    upd.*,
    usr.username AS `username`,
    usr.profile_picture AS `profile_picture`
    ,(
        SELECT COUNT (like.id)
        FROM likes as like
        WHERE upd.update_id = like.item_id
           AND like.uid = 118697835834
    ) as liked_update

FROM updates AS upd
LEFT JOIN users AS usr 
    ON upd.uid = usr.uid
WHERE upd.deleted=0
    AND 
    ( upd.uid=118697835834
        OR EXISTS ( SELECT *
                    FROM   subscribers AS sub 
                    WHERE  upd.uid = sub.suid
                    AND  sub.uid = 118697835834
            )
    )
ORDER BY upd.date DESC
LIMIT 0, 15

the subquery in SELECT returns the following error:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL 
    server version for the right syntax to use near 
    'like WHERE upd.update_id = like.item_id AND l' at line 10
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-30T16:51:22+00:00Added an answer on May 30, 2026 at 4:51 pm

    like is a reserved word in SQL; you should use a different alias for your likes table. Change your sub-query from:

    SELECT
        COUNT (like.id)
    FROM
        likes as like
    WHERE
        upd.update_id = like.item_id
        AND like.uid = 118697835834
    

    To something similar to:

    SELECT
        COUNT (l.id)
    FROM
        likes as l
    WHERE
        upd.update_id = l.item_id
        AND l.uid = 118697835834
    
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