Reputation: 1873
MSSQL
Table looks like so
ID 1 | 2 | 3 | 4 | 5
AA1 1 | 1 | 1 | 2 | 1
any clues on how I could make a query to return
ID | MaxNo
AA1 | 4
, usign the above table example? I know I could write a case blah when statement, but I have a feeling there's a much simpler way of doing this
Upvotes: 0
Views: 953
Reputation: 139010
You can use a cross apply where you do max()
over the columns for one row.
select T1.ID,
T2.Value
from YourTable as T1
cross apply
(
select max(T.Value) as Value
from (values (T1.[1]),
(T1.[2]),
(T1.[3]),
(T1.[4]),
(T1.[5])) as T(Value)
) as T2
If you are on SQL Server 2005 you can use union all
in the derived table instead of values()
.
select T1.ID,
T2.Value
from YourTable as T1
cross apply
(
select max(T.Value) as Value
from (select T1.[1] union all
select T1.[2] union all
select T1.[3] union all
select T1.[4] union all
select T1.[5]) as T(Value)
) as T2
Upvotes: 2
Reputation: 239824
You can use UNPIVOT
to get these comparable items, correctly1, into the same column, and then use ROW_NUMBER()
to find the highest valued row2:
declare @t table (ID char(3) not null,[1] int not null,[2] int not null,
[3] int not null,[4] int not null,[5] int not null)
insert into @t (ID,[1],[2],[3],[4],[5]) values
('AA1',1,1,1,2,1)
;With Unpivoted as (
select *,ROW_NUMBER() OVER (ORDER BY Value desc) rn
from @t t UNPIVOT (Value FOR Col in ([1],[2],[3],[4],[5])) u
)
select * from Unpivoted where rn = 1
Result:
ID Value Col rn
---- ----------- ------------------------- --------------------
AA1 2 4 1
1 If you have data from the same "domain" appearing in multiple columns in the same table (such that it even makes sense to compare such values), it's usually a sign of attribute splitting, where part of your data has, incorrectly, been used to form part of a column name.
2 In your question, you say "per row", and yet you've only given a one row sample. If we assume that ID
values are unique for each row, and you want to find the maximum separately for each ID
, you'd write the ROW_NUMBER()
as ROW_NUMBER() OVER (PARTITION BY ID ORDER BY Value desc) rn
, to get (I hope) the result you're looking for.
Upvotes: 3