Is there a better way to search over a range of values in Oracle than testing against a subquery?
Question
given this table:
x y
-- -
10 a
20 b
30 c
I want the best way to to map values
[10,20) -> a
[20,30) -> b
[30,inf) -> c
Right now I'm using a query like:
select y from foo
where x=(select max(x) from foo
where x<=21);
Is there a better way to do this? Is there an analytic function that might help?
Here's my test case:
create table foo as
select 10 as x ,'a' as y from dual union
select 20,'b' from dual union
select 30,'c' from dual;
-- returns: a,b,b:
select y from foo where x=(select max(x) from foo where x<=19);
select y from foo where x=(select max(x) from foo where x<=20);
select y from foo where x=(select max(x) from foo where x<=21);
Solution
You can rewrite your query to only access the foo table once instead of twice, by using the MAX-KEEP aggregate function.
An example:
SQL> var N number
SQL> exec :N := 19
PL/SQL-procedure is geslaagd.
SQL> select max(y) keep (dense_rank last order by x) y
2 from foo
3 where x <= :N
4 /
Y
-
a
1 rij is geselecteerd.
SQL> exec :N := 20
PL/SQL-procedure is geslaagd.
SQL> select max(y) keep (dense_rank last order by x) y
2 from foo
3 where x <= :N
4 /
Y
-
b
1 rij is geselecteerd.
SQL> exec :N := 21
PL/SQL-procedure is geslaagd.
SQL> select max(y) keep (dense_rank last order by x) y
2 from foo
3 where x <= :N
4 /
Y
-
b
1 rij is geselecteerd.
Also a,b,b as a result. The query plans:
SQL> set serveroutput off
SQL> select /*+ gather_plan_statistics */
2 y
3 from foo
4 where x = (select max(x) from foo where x<=:N)
5 /
Y
-
b
1 rij is geselecteerd.
SQL> select * from table(dbms_xplan.display_cursor(null,null,'predicate -note last'))
2 /
PLAN_TABLE_OUTPUT
-------------------------------------------------------------------------------------------
SQL_ID 3kh85qqnb2phy, child number 0
-------------------------------------
select /*+ gather_plan_statistics */ y from foo where x =
(select max(x) from foo where x<=:N)
Plan hash value: 763646971
----------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
----------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | | | 8 (100)| |
|* 1 | TABLE ACCESS FULL | FOO | 1 | 16 | 4 (0)| 00:00:01 |
| 2 | SORT AGGREGATE | | 1 | 13 | | |
|* 3 | TABLE ACCESS FULL| FOO | 2 | 26 | 4 (0)| 00:00:01 |
----------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - filter("X"=)
3 - filter("X"<=:N)
22 rijen zijn geselecteerd.
SQL> select max(y) keep (dense_rank last order by x) y
2 from foo
3 where x <= :N
4 /
Y
-
b
1 rij is geselecteerd.
SQL> select * from table(dbms_xplan.display_cursor(null,null,'predicate -note last'))
2 /
PLAN_TABLE_OUTPUT
-------------------------------------------------------------------------------------------
SQL_ID avm2zh62c8cwd, child number 0
-------------------------------------
select max(y) keep (dense_rank last order by x) y from foo where x
<= :N
Plan hash value: 3274996510
---------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
---------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | | | 4 (100)| |
| 1 | SORT AGGREGATE | | 1 | 16 | | |
|* 2 | TABLE ACCESS FULL| FOO | 1 | 16 | 4 (0)| 00:00:01 |
---------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
2 - filter("X"<=:N)
20 rijen zijn geselecteerd.
Two full table scans on foo, against one for the new query.
Regards, Rob.
OTHER TIPS
select distinct first_value(y) over (order by x desc) from foo where x<=19;
select distinct first_value(y) over (order by x desc) from foo where x<=20;
select distinct first_value(y) over (order by x desc) from foo where x<=21;
Plus: an index on x will probably be a good idea.
Here's another answer received via usenet. So far this one seems to have the most efficient execution.
select max(y) keep (dense_rank last order by x) from foo where x<=21;
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