Reputation: 681
I have table called emp
and I a trying to find the lowest paid Clerk. My code select min(sal) as min from emp where job='CLERK';
works fine and i get this:
MIN
----------
800
but I also want to show the name of the clerk which is Smith
. When I run this code select ename, min(sal) as min from emp where job='CLERK' group by name;
it gives me all the Clerks in the table, which is not really want I want. Here is a snippet of my table:
CREATE TABLE EMP
(EMPNO NUMBER(4) NOT NULL,
ENAME VARCHAR2(10),
JOB VARCHAR2(9),
MGR NUMBER(4),
HIREDATE DATE,
SAL NUMBER(7, 2),
COMM NUMBER(7, 2),
DEPTNO NUMBER(2));
INSERT INTO EMP VALUES
(7369, 'SMITH', 'CLERK', 7902,
TO_DATE('17-DEC-1980', 'DD-MON-YYYY'), 800, NULL, 20);
INSERT INTO EMP VALUES
(7499, 'ALLEN', 'SALESMAN', 7698,
TO_DATE('20-FEB-1981', 'DD-MON-YYYY'), 1600, 300, 30);
Upvotes: 0
Views: 75
Reputation: 8178
try this
SELECT * FROM emp
WHERE SAL = (select MIN(SAL) sal from emp WHERE JOB ='CLERK')
and JOB ='CLERK';
Upvotes: 1
Reputation: 1269593
You can do this with a subquery and ranking functions:
select e.*
from (select e.*,
dense_rank() over (partition by job order by salary) as seqnum
from emp
where job = 'CLERK'
) e
where seqnum = 1;
This will return multiple rows if there are ties. If you only want one, you can use row_number()
instead of dense_rank()
.
And, if you want the names of all clerks as a single delimited value, then you can use list_agg()
:
select listagg(e.name, ', ') within group (order by e.name) as names
from (select e.*,
dense_rank() over (partition by job order by salary) as seqnum
from emp
where job = 'CLERK'
) e
where seqnum = 1;
Upvotes: 0
Reputation: 72165
You can use row_number
:
select ename, sal as min
from (
select ename, sal,
row_number() over (order by sal) as rn
from emp
where job='CLERK' ) t
where t.rn = 1
Upvotes: 0