ORACLE SQL retrieve n rows without subqueries or derived tables - sql

Question

I'm doing my SQL exercises but I got stuck in one. I need to retrieve the employees with the two highest salaries, but I can't use any type of subquery or derived table. I do it with a subquery like this:
SELECT *
FROM (SELECT * FROM emp ORDER BY sal DESC) new_emp
WHERE ROWNUM < 3;
I also know that this can be achieved using the WITH clause, but I'm wondering if there is any alternative to this.
PS: I'm using Oracle 11.

Answer 1

If you are Oracle version 12.1 or above you can use a row limiting clause. In your case you would just use the subquery plus the row limiting clause like so:
SELECT * FROM emp
ORDER BY sal DESC
FETCH FIRST 5 ROWS ONLY;
Source: https://oracle-base.com/articles/12c/row-limiting-clause-for-top-n-queries-12cr1#top-n

Answer 2

This is actually a pathetic method, in my opinion, but you can use a join:
select e.col1, e.col2, . . .
from emp e left join
emp e2
on e2.salary >= e.salary
group by e.col1, e.col2, . . .
having count(distinct e2.salary) <= 2;
Note: this is really equivalent to a dense_rank(), so if there are ties, you'll get more than two rows. It is easy enough to fix this (assuming you have a unique identifier for each row), but the fix complicates the logic and hides the basic idea.

Answer 3

A good exercise should help to prepare to solve practical problems. So the important thing in this one is not the usage of subquery but to realize that the two highes salaries can have hunderts of employees.
While using the #MT0 view workaround this is the query
CREATE VIEW sal_ordered_emps AS
SELECT e.*,
row_number() over (order by sal desc) as RN
FROM SCOTT.emp e
ORDER BY sal DESC;
select e.* from scott.emp e join
sal_ordered_emps soe on e.sal = soe.sal and rn <= 2
;
result as explained can be more than 2 records
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
---------- ---------- --------- ---------- ------------------- ---------- ---------- ----------
7788 SCOTT ANALYST 7566 19.04.1987 00:00:00 3000 20
7839 KING PRESIDENT 17.11.1981 00:00:00 5000 10
7902 FORD ANALYST 7566 03.12.1981 00:00:00 3000 20

Answer 4

This is cheating but... instead of using a subquery you can define a view:
CREATE VIEW sal_ordered_emps AS
SELECT *
FROM emp
ORDER BY sal DESC;
Then you can do:
SELECT * FROM sal_ordered_emps WHERE ROWNUM < 3;
Alternatively, you can do it with a pipelined function...
CREATE OR REPLACE PACKAGE emp_pkg
AS
TYPE emp_table IS TABLE OF EMP%ROWTYPE;
FUNCTION get_max_sals(
n INT
) RETURN emp_table PIPELINED;
END;
/
CREATE OR REPLACE PACKAGE BODY emp_pkg
AS
FUNCTION get_max_sals(
n INT
) RETURN emp_table PIPELINED
AS
cur SYS_REFCURSOR;
in_rec EMP%ROWTYPE;
i INT := 0;
BEGIN
OPEN cur FOR SELECT * FROM EMP ORDER BY SAL DESC;
LOOP
i := i + 1;
FETCH cur INTO in_rec;
EXIT WHEN cur%NOTFOUND OR i > n;
PIPE ROW(in_rec);
END LOOP;
CLOSE cur;
RETURN;
END;
END;
/
SELECT *
FROM TABLE( emp_pkg.get_max_sals( 2 ) );

Answer 5

This solution use no subquery, but requires three steps:
-- Q1
select max(sal) from scott.emp;
-- Q2
select max(sal) from scott.emp where sal < {result of Q1};
select * from scott.emp where sal in ({result of Q1},{result of Q2});
General you'll need N+1 queries to get top-N salaries emps.