php - mysql子查询返回错误

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

SELECT中的子查询返回以下错误：

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

最佳答案

like是SQL中的保留字；您应该为likes表使用不同的别名。将子查询更改为：

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

类似于：

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