问题
Let's say I have the following table:
category | guid
---------+-----------------------
A | 5BC2...
A | 6A1C...
B | 92A2...
Basically, I want to do the following SQL:
SELECT category, MIN(guid)
FROM myTable
GROUP BY category
It doesn't necessarily have to be MIN. I just want to return one GUID of each category. I don't care which one. Unfortunately, SQL Server does not allow MIN or MAX on GUIDs.
Of course, I could convert the guid into a varchar, or create some nested TOP 1 SQL, but that seems like an ugly workaround. Is there some elegant solution that I've missed?
回答1:
Assuming you're using SQL Server 2005 or later:
;with Numbered as (
select category,guid,ROW_NUMBER() OVER (PARTITION BY category ORDER BY guid) rn
from myTable
)
select * from Numbered where rn=1
回答2:
Just cast it as a BINARY(16)
.
SELECT category, MIN(CAST(guid AS BINARY(16)))
FROM myTable
GROUP BY category
You can cast it back later if necessary.
WITH CategoryValue
AS
(
SELECT category, MIN(CAST(guid AS BINARY(16)))
FROM myTable
GROUP BY category
)
SELECT category, CAST(guid AS UNIQUEIDENTIFIER)
FROM CategoryValue
回答3:
Aggregate functions can be used on Uniqueidentifier columns if SQL Server Version >= 2012
expression
Is a constant, column name, or function, and any combination of arithmetic, bitwise, and string operators. MIN can be used with numeric, char, varchar, uniqueidentifier, or datetime columns, but not with bit columns. Aggregate functions and subqueries are not permitted.
回答4:
declare @T table(category char(1), guid uniqueidentifier)
insert into @T
select 'a', newid() union all
select 'a', newid() union all
select 'b', newid()
select
S.category,
S.guid
from
(
select
T.category,
T.guid,
row_number() over(partition by T.category order by (select 1)) as rn
from @T as T
) as S
where S.rn = 1
If you are on SQL Server 2000 you could to this
select
T1.category,
(select top 1 T2.guid
from @T as T2
where T1.category = T2.category) as guid
from @T as T1
group by T1.category
回答5:
SELECT top 1 category, guid FROM myTable GROUP BY category,guid
来源:https://stackoverflow.com/questions/6069368/aggregate-function-on-uniqueidentifier-guid