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select count(distinct col1, col2) from table - Microsoft SQL / MS SQL Server

I need to know the number of distinct col1, col2 pairs in a table. The following does not work: select count(distinct col1, col2) from table This one does not work either: select count(*) from (select distinct col1, col2 from table) I do not want to use a view. Does anyone know a solution to this? It should be quite a standard problem....

  1. #1

    Default select count(distinct col1, col2) from table

    I need to know the number of distinct col1, col2 pairs in a table. The
    following does not work:
    select count(distinct col1, col2) from table
    This one does not work either:
    select count(*) from (select distinct col1, col2 from table)

    I do not want to use a view.

    Does anyone know a solution to this? It should be quite a standard problem.


    Mats Guest

  2. #2

    Default Re: select count(distinct col1, col2) from table

    When you select from a sub-select, you need to specify an alias.
    So this should work:
    select count(*) from (select distinct col1, col2 from table) a


    "Mats Olsson" <se> wrote in message
    news:hBL_a.22754$.. 
    problem. 


    Martin Guest

  3. #3

    Default Re: select count(distinct col1, col2) from table

    You almost had it. You need to alias your derived table:


    select count(*) from (select distinct col1, col2 from table) as x

    --
    Tom

    ---------------------------------------------------------------
    Thomas A. Moreau, BSc, PhD, MCSE, MCDBA
    SQL Server MVP
    Columnist, SQL Server Professional
    Toronto, ON Canada
    www.pinnaclepublishing.com/sql


    "Mats Olsson" <se> wrote in message news:hBL_a.22754$..
    I need to know the number of distinct col1, col2 pairs in a table. The
    following does not work:
    select count(distinct col1, col2) from table
    This one does not work either:
    select count(*) from (select distinct col1, col2 from table)

    I do not want to use a view.

    Does anyone know a solution to this? It should be quite a standard problem.



    Tom Guest

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