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Home/ Questions/Q 1838590
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T06:29:55+00:00 2026-05-17T06:29:55+00:00

Here is the query I am using: SELECT k_id, COUNT(k_id) AS k_count FROM template_keyword_link

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Here is the query I am using:

SELECT k_id, COUNT(k_id) AS k_count 
FROM template_keyword_link 
WHERE k_id IN(1,2,3,4,5)
GROUP BY k_id;

This query returns something like
1 | 6
2 | 1
3 | 4
4 | 1
5 | 9

I want to add something like AND COUNT(k_id) = 1
so I end up with
2 | 1
4 | 1

However I get invalid use a group function.

How would I go about doing this?


Update:

The other part to my question is.

Can this be used as a delete statement?

something like

DELETE FROM 
template_keyword_link tkl
LEFT JOIN keywords k
ON tkl.k_id = k.k_id
WHERE tkl.k_id 
IN(SELECT k_id, COUNT(k_id) AS k_count 
FROM template_keyword_link 
WHERE k_id IN(1,2)
GROUP BY k_id
HAVING k_count = 1);

I get

You have an error in your SQL syntax;


So based on feedback i have altered this to use

DELETE tkl, k FROM 
template_keyword_link tkl
LEFT JOIN keywords k
ON tkl.k_id = k.k_id
WHERE tkl.k_id 
IN(SELECT k_id 
FROM template_keyword_link 
WHERE k_id IN(1,2)
GROUP BY k_id
HAVING COUNT(k_id) = 1);

However now I am getting

You can’t specify target table ‘tkl’ for update in FROM clause

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-17T06:29:55+00:00Added an answer on May 17, 2026 at 6:29 am

    WHERE clause is applied before the COUNT(*) has been calculated, so you need in HAVING, that is applied after.

      SELECT k_id,
             COUNT(k_id) AS k_count 
        FROM template_keyword_link 
       WHERE k_id IN (1, 2, 3, 4, 5)
    GROUP BY k_id
      HAVING k_count = 1
    

    See also: http://dev.mysql.com/doc/refman/5.1/en/select.html

    ** UPD **:

    TIAS 😉 btw, the query syntactically seems fine for me, BUT did not you forget to specify template_keyword_link and keywords join condition clause? Does mysql give you any errors?

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