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:
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
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.
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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