SQL get data from column(YYY0) with same number as different column(XXXX0) with the maximum date - sql

Question

I am looking for a query to for each row to find the column (YYY.) with the highest/most recent date and would like to find the corresponding column (XXXX.)
Finding the column with the most recent date was possible, but getting the corresponding column left me clueless... All suggestions are welcome!!
So from the table:
| id | XXXX0| YYY0 | XXXX1| YYY1| XXXX9| YYY9|
---------------------------------------------------------------------------------------
| A | 3 | 10-10-2009| 4 |10-10-2010| 1 | 10-10-2011|
| B | 2 | 10-10-2010| 3 |10-10-2012| 6 | 10-10-2011|
| C | 4 | 10-10-2011| 1 |10-10-2010| 7 | 10-10-2012|
| D | 1 | 10-10-2010| 8 |10-10-2013| 9 | 10-10-2012|
I would like to end up with:
| id | LabelX| LabelY|
--------------------------------------
| A | 1 | 10-10-2011|
| B | 3 | 10-10-2012|
| C | 7 | 10-10-2012|
| D | 8 | 10-10-2013|
Added:
This was what I tried to determine the maximum value:
SELECT LTRIM(A) AS A, LTRIM(B) AS B, LTRIM(C)
(Select Max(v)
FROM (VALUES (YYY0), (YYY1), …..(YYY9) AS value(v)) as [MaxDate]
FROM Table

Answer 1

SELECT id,
CASE
WHEN YYYY0 > YYY1 AND YYY0 > YYY2 ... AND YYY0 > YYY9 THEN XXX0
WHEN YYY1 > YYY2 ... AND YYY0 > YYY9 THEN XXX1
...
ELSE XXX9 AS LabelX,
CASE
WHEN YYYY0 > YYY1 AND YYY0 > YYY2 ... AND YYY0 > YYY9 THEN YYY0
WHEN YYY1 > YYY2 ... AND YYY0 > YYY9 THEN YYY1
...
ELSE YYY9 AS LabelY,
...
and replace > by >= depending on which you want to win if they're equal.

Answer 2

If it's a SQL Server 2005 and above you can do it this way (it assumes that dates are unique in each column for specific id):
;with cte as (
select id, xxxx0 as LabelX, yyy0 as LabelY from tab union all
select id, xxxx1, yyy1 from tab union all
select id, xxxx9, yyy9 from tab
)
select t.id, x.LabelX, t.LabelY from (
select t1.id, max(t1.LabelY) as LabelY
from cte t1
group by t1.id
) t
join cte x on t.id = x.id and t.LabelY = x.LabelY
Live SQL Fiddle example

Answer 3

Here's a simplified example for you. I'm using SQL Server 2008, but this SQL is pretty standard and should work fine on most modern implementations (famous last words).
So, given this table schema:
drop table dbo.foobar
go
create table dbo.foobar
(
id char(1) not null primary key ,
X1 int not null , Y1 date not null ,
X2 int not null , Y2 date not null ,
X3 int not null , Y3 date not null ,
)
go
And some sample data:
insert dbo.foobar values ( 'A' , 1 , '1 Jan 2013' , 2 , '1 Feb 2013' , 3 , '1 Mar 2013' )
insert dbo.foobar values ( 'B' , 1 , '1 Mar 2013' , 2 , '1 Jan 2013' , 3 , '1 Feb 2013' )
insert dbo.foobar values ( 'C' , 1 , '1 Feb 2013' , 2 , '1 Mar 2013' , 3 , '1 Jan 2013' )
go
Depending on the nature of your data and the desired semantics of the query and results, either this approach:
--
-- This approach pushes evaluation of the corresponding X to the output column list
--
-- 1. Construct a UNION ALL to normalize the table into a set of id/date pairs
-- 2. Compute max(date) for each id
-- 3. Join back against the original table to recover the source row
-- 4. Use the max(date) value to identify the corresponding X
--
select t.id ,
MaxY = t.y ,
X = case
when t.Y = x.Y1 then x.X1
when t.Y = x.Y2 then x.X2
when t.Y = x.Y3 then x.X3
end
from ( select x.id ,
y = max( x.y )
from ( select id , y=y1 from dbo.foobar
union all select id , y=y2 from dbo.foobar
union all select id , y=y3 from dbo.foobar
) x
group by x.id
) t
join dbo.foobar x on x.id = t.id
order by 1,2,3
go
Or this approach
--
-- This approach looks at each X/Y pair as its own "table" as it were
--
select t.id ,
MaxY = t.y ,
X = coalesce( t1.X1 , t2.X2 , t3.X3 )
from ( select x.id ,
y = max( x.y )
from ( select id , y=Y1 from dbo.foobar
union all select id , y=Y2 from dbo.foobar
union all select id , y=Y3 from dbo.foobar
) x
group by x.id
) t
left join dbo.foobar t1 on t1.id = t.id and t1.y1 = t.Y
left join dbo.foobar t2 on t2.id = t.id and t2.y2 = t.Y
left join dbo.foobar t3 on t3.id = t.id and t3.y3 = t.Y
order by 1,2,3
should work for you. In either event, both queries produce an identical result set:
id MaxY X
-- ---------- -
A 2013-03-01 3
B 2013-03-01 1
C 2013-03-01 2
Good Luck!
[Have you considered normalizing your database design? Third Normal Form makes life a lot easier and usually more efficient.]