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Home/ Questions/Q 7409337
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Editorial Team
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Editorial Team
Asked: May 29, 20262026-05-29T06:04:09+00:00 2026-05-29T06:04:09+00:00

All the guys probably will recommend that I read the follow previously question in

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All the guys probably will recommend that I read the follow previously question in this URL http://senta.la/bd08

So I’m just read them carefully and I’m still with my problem. This situation its not equals of the question mentioned in link above.

See below the records I want to found (marked with *):

ID         COURSE       OTHERCOLUMN_X      OTHERCOLUMN_Y       OTHERCOLUMN_Z
123        2111         ...                ...                 ...
*5644      2245         ...                ...                 ...
*5644      2245         ...                ...                 ...
2121       2131         ...                ...                 ...

The previously programmer not thinking use the UNIQUE index to make a compost key to prevent this situation. I know… the good news is that programmer is not a programmer anymore.

This classic query its not provide the result I want…

SELECT id, course FROM mytable
GROUP BY id 
HAVING COUNT(id) > 1

I want to see only this results:

ID         COURSE       OTHERCOLUMN_X      OTHERCOLUMN_Y       OTHERCOLUMN_Z
*5644      2245         ...                ...                 ...
*5644      2245         ...                ...                 ...

I’m just tried a lot of queries but I can find one that show me the results above.

🙁

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-29T06:04:09+00:00Added an answer on May 29, 2026 at 6:04 am

    To find rows with duplicate id:

    SELECT a.*
    FROM mytable AS a
      JOIN
        ( SELECT id 
          FROM mytable
          GROUP BY id 
          HAVING COUNT(*) > 1
        ) AS b
        ON b.id = a.id
    

    To find rows with duplicate (id, course):

    SELECT a.*
    FROM mytable AS a
      JOIN
        ( SELECT id, course 
          FROM mytable
          GROUP BY id, course 
          HAVING COUNT(*) > 1
        ) AS b
        ON  b.id = a.id
        AND b.course = a.course
    
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