Reputation: 3
I have rows like this:
( a , #$@$ , $$ , 3 )
( c , ###$ , ## , 0 )
( a , #@$# , !! , 2 )
( b , #@## , $$ , 0 )
If I want to get the result like below
( a , #$@$ , $$ , 3 )
( c , ###$ , ## , 0 )
( b , #@## , $$ , 0 )
Which is based on grouping by column 1 and choose the rows with max value in column 4 independent of other columns (2 & 3).
Instead of creating subquery, is there a way to do this?
Upvotes: 0
Views: 296
Reputation: 1555
Without using subquery, you can use keep dense_rank function (its aggregate version) like below :
with your_table (col1, col2, col3, col4) as (
select 'a', '#$@$' , '$$' , 3 from dual union all
select 'c', '###$' , '##' , 0 from dual union all
select 'a', '#@$#' , '!!' , 2 from dual union all
select 'b', '#@##' , '$$' , 0 from dual
)
select col1
, max(col2)keep(dense_rank first order by col4 desc)col2
, max(col3)keep(dense_rank first order by col4 desc)col3
, max(col4)keep(dense_rank first order by col4 desc)col4
from your_table t
group by col1
;
Upvotes: 1
Reputation: 142710
Why without a subquery? They are designed for such a purpose.
Current table contents:
SQL> select * from test order by col1, col4;
COL1 COL2 COL3 COL4
----- ----- ----- ----------
a #@$# !! 2
a #$@$ $$ 3
b #@$$ $$ 0
c ###$ ## 0
Using analytic function, partition rows by col1
and order them by col4
in descending order; then fetch the "first" row per each group (partition).
SQL> select col1, col2, col3, col4
2 from (select col1, col2, col3, col4,
3 row_number() over (partition by col1 order by col4 desc) rn
4 from test
5 )
6 where rn = 1
7 order by col1;
COL1 COL2 COL3 COL4
----- ----- ----- ----------
a #$@$ $$ 3
b #@$$ $$ 0
c ###$ ## 0
SQL>
Upvotes: 0