I want to display the name of department with highest SUM of salary.
I am using oracle sql and the table structure is Dept(Deptno,Dname,Loc) and Emp(Empno,Ename,Job,Salary,Deptno).
The query I use was
select Dname
from Dept
where Deptno=
( select Deptno
from Emp
where rownum=1
group by Deptno
order by sum(Salary) Desc
);
This gives an error:
Right parenthesis missing.
When I run the sub-query alone, it successfully returns a Deptno. But with the parent query, I get the above error.
What is the problem and what can be the possible solution?
select dname
from (select dname, rank() over (order by sum(salary) desc) rnk
from dept d
inner join emp e
on e.deptno = d.deptno
group by dname, e.deptno
)
where rnk = 1;
note, in your example putting where rownum=1 where you did is a huge bug. it would mean pick 1 random row and sort it (not really the highest salary row..just any old row)
my solution may get over 1 row if 2 deptartments have the same highest salary. you can use row_number() instead of rank() to just pick one if you want.
Related
The employee table (EMP) specifies groups by department code, the sum of salaries by group, the average (constant treatment), and the number of people in each group's salary, and is presented below, listed in department code order. I would like to modify the following SQL syntax to look up departments whose average salary exceeds 2800000.
SELECT
DEPT
, SUM(SALARY) 합계
, FLOOR(AVG(SALARY)) 평균
, COUNT(*) 인원수
FROM
EMP
GROUP BY
DEPT
ORDER BY DEPT ASC;
question 1. Conditions that need to be modified
question 2. What should I add to the presented code?
I can't read your aliases so I'll just presume what they mean.
If query - you posted in the question - works OK, then use it as a CTE and select desired data from it:
with data as
(select dept,
sum(salary) sumsal,
floor(avg(salary)) avgsal,
count(*) cnt
from emp
group by dept
)
select *
from data
where avgsal > 2800000;
you can use below sql:
SELECT
DEPT
, SUM(SALARY) 합계
, FLOOR(AVG(SALARY)) 평균
, COUNT(*) 인원수
FROM
EMP
GROUP BY
DEPT having FLOOR(AVG(SALARY) > 2800000
ORDER BY DEPT ASC;
You can filter aggregated result using having
SELECT
DEPT
, SUM(SALARY) 합계
, FLOOR(AVG(SALARY)) 평균
, COUNT(*) 인원수
FROM
EMP
GROUP BY DEPT
HAVING AVG(SALARY) >2800000
ORDER BY DEPT ASC;
This question already has answers here:
How to find the employee with the second highest salary?
(5 answers)
Closed 3 years ago.
I have to get the name of employee with the second highest salary the table name from where I am fetching is emp. I know the query for second highest salary which is
select max(sal)
from emp
where sal < (select max(sal) from emp)
it works and it returns the right answer.
But I have to get the name of the employee as well. I simply tried
select name, max(sal)
from emp
where sal < (select max(sal) from emp)
I get this error:
ORA-00937: not a single-group group function
how can i remove the error in order to get the name and salary both.
thank you to anyone who helps.
You can use
select name,sal from emp where sal = (select max(sal) from emp where sal < (select max(sal) from emp));
use this :
with cte (
select ROW_NUMBER() over(order by sal desc)rnum ,name,sal from emp )
select * from cte where rnum = 2
You can get this easily with a window function. Try something like this:
SELECT name, sal
FROM emp
QUALIFY RANK OVER(ORDER BY sal DESC) = 2
This will order your rows by Salary and then give each row a ranking. Then it will return the rows with ranking = 2.
If you want to ensure you only get one row back, change RANK to ROW_NUMBER.
I have tried checking all the duplicate answers provided in this site but couldn't able to come up with rectifying my error could you guys help me out on this?
Getting this error
ORA-00923: FROM keyword not found where expected
SELECT TOP 1 sal FROM emp WHERE sal in
(SELECT DISTINCT TOP 3 sal FROM emp ORDER BY sal DESC)
ORDER BY sal ASC;
I am solving this using oracle apex site
You appear to want to find thew record with the third-lowest salary.
You can use a subquery to assign an analytic rank to each salary, and then filter on that:
select * from (
select e.*, dense_rank() over (order by sal desc) as rnk
from emp e
)
where rnk = 3;
From 12c you can use the offset and fetch syntax to do the same thing:
select * from emp
order by sal desc
offset 3 rows
fetch first row only;
You need to decide how to deal with ties though - if more than one employee shares that third-lowest salary; or even if the lowest and second lowest are awarded to more than one person. Depending on what you want to happen you can look at variations of the 12c row-limiting syntax, and other analytic functions - row_number(), dense_rank() - and their windowing options.
I was able to get my query now
SELECT A.* FROM emp A where 3 =(SELECT COUNT(DISTINCT sal) FROM emp B WHERE A.sal <= B.sal);
Yet another option, using analytic function:
WITH ranks
AS (SELECT empno, DENSE_RANK () OVER (ORDER BY sal DESC) rnk FROM emp)
SELECT e.*
FROM emp e, ranks r
WHERE e.empno = r.empno
AND r.rnk = 3;
I have a table emp (ename,eid,did,sal,sex) where column did is foreign key with another table dept (did,dname).
I want to get the max sal of the company along with name and did of the person who is getting it.
I am executing following query
select did,ename ,max(sal) from emp;
But the Result is :
ORA-00937: not a single-group group function
so My question is can't I have more than 1 columns ?
You can't use MAX aggregate function like this. Using MAX without a GROUP BY clause will just return one record with the maximum sal value. You are not allowed to place any more non-aggregated fields in the SELECT clause.
If you want to get the record having the maximum sal value, then you have to do a self-join:
select e1.*
from emp as e1
inner join (
select max(sal) as max_sal
from emp
) as e2 on e1.sal = e2.max_sal
Note: The above query will return more than one records from table emp, in case more than one records share the same maximum sal value.
Edit:
If you want to get the maximum salary per department then you have to include a GROUP BY clause in the derived table used:
select e1.*
from emp as e1
inner join (
select did, max(sal) as max_sal
from emp
group by did
) as e2 on e1.did = e2.did and e1.sal = e2.max_sal
You would have to group the results by the sal column:
select did,ename,max(sal) from emp group by sal
MAX operates over the group specified in a GROUP BY clause and finds the maximum value for each group. When not specifying a set of columns to group by, it finds the max for the entire results set.
You need to find the max sal, then you can query the rows that match:
select did, ename, sal
from emp
where sal = (select max(sal) from emp)
I want to Display department number, names and salaries of employees who are earning max salary in their departments which are in the same table.
I am using oracle sql. The Table structure I am using is
Emp(Empno,Ename,Job,Salary,Deptno)
I have read about this and I think that this can be done by the use of correlated sub-queries. The query I fired was
select E1.Ename,E1.Ename,E1.Salary
from Emp E1
where E1.Empno=(
select Empno
from Emp E2
where Salary=(
select max(Salary)
from Emp
where Deptno=E1.Deptno
)
);
This gives an error saying "single-row subquery returns more than one row".
What am I doing wrong? What should do to correct it?
SELECT EmpNo, Ename,Job,Salary,Deptno
FROM
(
SELECT EmpNo, Ename,Job,Salary,Deptno,
DENSE_RANK() OVER (PARTITION BY DeptNo
ORDER BY Salary DESC ) rn
FROM Emp
) a
WHERE a.rn = 1
DENSE_RANK
or by using MAX
SELECT a.*
FROM Emp a
INNER JOIN
(
SELECT DeptNo, MAX(Salary) Max_sal
FROM Emp
GROUP BY DeptNo
) b ON a.DeptNo = b.DeptNo AND
a.Salary = b.Max_SAL