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Editorial Team
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Asked: 2026-05-25T23:56:02+00:00

I have a person table which holds person and his manager at the same

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I have a person table which holds person and his manager at the same time.
I’m using SELF JOIN to select managers email but I get a lot of duplicates.

http://imageshack.us/photo/my-images/3/withoutgroupby.png

How can I use GROUP BY with my query

SELECT              P.prs_id AS 'Employee_id', M.prs_id AS 'Manager_id', M.prs_email AS 'Manager_email'
FROM                qrd_prs_person AS P
LEFT OUTER JOIN     qrd_prs_person AS M  
ON                  P.prs_manager_number = M.prs_number

GROUP BY M.prs_id

If I add this line at the end of my query to group by Manager_id, I receive this error

Column ‘qrd_prs_person.prs_id’ is invalid in the select list because
it is not contained in either an aggregate function or the GROUP BY
clause.

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1 Answer

  1. Editorial Team

    I’m not quite sure what you’re trying to achieve?

    If you’re after a query that returns one row per employee, with two optional columns containing the manager’s info, then your original query is correct (without the group by). The relationship is many-to-one, you’re starting with a row per “many” that each has a single (optional) “one”, so there is no need to group by.

    This however is assuming that your data is correct and that prs_number is in fact unique for each employee. If you have two or more managers sharing a prs_number, you will end up with people having multiple managers.

    By making this an outer join you’re also returning people without a manager (i.e. top of the food chain :)), was this your intention?

    EDIT

    If you want only managers returned, then you can’t keep the first column (P.prs_id) and get one row per manager. If you want the list of people that manage one or more people, this will do the trick:

    SELECT              M.prs_id AS 'Manager_id', M.prs_email AS 'Manager_email'
FROM                qrd_prs_person AS P
INNER JOIN          qrd_prs_person AS M  
ON                  P.prs_manager_number = M.prs_number
GROUP BY            M.prs_id, M.prs_email
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