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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T21:02:18+00:00 2026-05-11T21:02:18+00:00

I’m using SQL Server 2005. With the query below (simplified from my real query):

  • 0

I’m using SQL Server 2005. With the query below (simplified from my real query):

select a,count(distinct b),sum(a) from 
(select 1 a,1 b union all
select 2,2 union all
select 2,null union all
select 3,3 union all
select 3,null union all
select 3,null) a
group by a

Is there any way to do a count distinct without getting

“Warning: Null value is eliminated by an aggregate or other SET operation.”

Here are the alternatives I can think of:

  1. Turning ANSI_WARNINGS off
  2. Separating into two queries, one with count distinct and a where clause to eliminate nulls, one with the sum:

    select t1.a, t1.countdistinctb, t2.suma from
    (
        select a,count(distinct b) countdistinctb from 
        (
            select 1 a,1 b union all
            select 2,2 union all
            select 2,null union all
            select 3,3 union all
            select 3,null union all
            select 3,null
        ) a
        where a.b is not null
        group by a
    ) t1
    left join
    (
        select a,sum(a) suma from 
        (
            select 1 a,1 b union all
            select 2,2 union all
            select 2,null union all
            select 3,3 union all
            select 3,null union all
            select 3,null
        ) a
        group by a
    ) t2 on t1.a=t2.a
    
  3. Ignore the warning in the client

Is there a better way to do this? I’ll probably go down route 2, but don’t like the code duplication.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-11T21:02:19+00:00Added an answer on May 11, 2026 at 9:02 pm
    select a,count(distinct isnull(b,-1))-sum(distinct case when b is null then 1 else 0 end),sum(a) from 
        (select 1 a,1 b union all
        select 2,2 union all
        select 2,null union all
        select 3,3 union all
        select 3,null union all
        select 3,null) a
        group by a
    

    Thanks to Eoin I worked out a way to do this. You can count distinct the values including the nulls and then remove the count due to nulls if there were any using a sum distinct.

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