SQL -- Selecting Top 1 with Order by? - sql

Question

This was resolved. The statement was in another part of the stored procedure.
The stored procedure I'm writing won't allow me to do this:
declare #dtTopDate datetime
select top 1 #dtTopDate = date_build
from database..table
where database..table.parent = #Parent
and database..table.child = #Child
order by date_build desc
Gives me this error:
Column "database..table.date_build" is
invalid in the ORDER BY clause because
it is not contained in either an
aggregate function or the GROUP BY
clause.
What am I doing wrong?
[Edit] There is no group by statement here. SQL2005.
Here is some more context:
if #Notify = 0
begin
declare #dtTopDate datetime
select top 1 #dtTopDate = date_build
from database..table
where database..table.parent = #Parent
and database..table.child = #Child
order by date_build desc
insert
into database2..table
(parent, child, notification_date, change_date)
values (#Parent, #Child, #dtTopDate, getdate())
return
end

Answer 1

This works for me, but I'm not sure if this is what you are trying to do b/c your example has some errors.
use Test
go
CREATE TABLE [dbo].[MyTable]
(
[MyTableId] [uniqueidentifier] NOT NULL,
[MyDate] [datetime] NOT NULL,
CONSTRAINT [PK_MyTable] PRIMARY KEY CLUSTERED([MyTableId] ASC,[MyDate] ASC)
)
GO
CREATE PROCEDURE ProcTopDate
(
#MyDate datetime OUT
)
AS
BEGIN
SET NOCOUNT ON;
SELECT TOP 1
#MyDate = [MyDate]
FROM [Test].[dbo].[MyTable]
order by MyDate
END
GO
insert into MyTable(MyTableId, MyDate)
values(newid(), getdate())
go
declare #MyDate datetime
exec ProcTopDate #MyDate OUT
print #MyDate

Answer 2

Instead of SELECT TOP 1 ... ORDER BY ...
Why not try SELECT MAX( ..
DECLARE #dtTopDate datetime
SELECT #dtTopDate = MAX(date_build)
from database..table
where database..table.parent = #Parent
and database..table.child = #Child

Answer 3

What version of SQL are you using? It works fine for me on MS SQL Server 2005 (ionce I fix the declaration).

Answer 4

Honestly the only thing I can see wrong is that #dtTopDate =/= #dtLatestDate
Apart from that, there is no GROUP BY clause in your SQL statement.
I just ran this and it worked fine.
declare #OrderDate datetime
select top 1 #OrderDate = OrderDate
from Orders
where Orders.CustomerID = 'ALFKI'
and Orders.EmployeeID = 4
order by OrderDate desc
SELECT #OrderDate

Answer 5

Try qualifying the columns correctly to avoid any ambiguities or x-database schema issue
declare #dtTopDate datetime
select top 1
#dtTopDate = [database]..[table].date_build
FROM
[database]..[table]
where
[database]..[table].parent = #Parent
and [database]..[table].child = #Child
order by
[database]..[table].date_build desc
Or alias it
declare #dtTopDate datetime
select top 1
#dtTopDate = foo.date_build
FROM
[database]..[table] foo
where
foo.parent = #Parent
and foo.child = #Child
order by
foo.date_build desc

Answer 6

The problem was in another part of the stored procedure. I was using a count(*) elsewhere and it required a group by. Thanks for the help.

Answer 7

Try SELECT #dtLatestDate = TOP 1 date_build...

Answer 8

if you want to get really tricky, in T-SQL you can try using the row_number() method and an inner select:
select * from
(
select
db.groupId
, db.date_build
, date_build_rank = row_number() over ( partition by db.groupId order by db.date_build desc)
from
#date_build_tbl db
) as a
where a.date_build_rank < 2;