I'm having an assignment for my sql class which i can't seem to figure out. This is the description of the select wanted:
Show all employees of which the salary is higher than the average of the colleagues in their department, only for the departments with at least 4 employees.
I've been able to find parts of the query like
select ename
from emp
where sal > any (select avg(sal)
from emp
group by
deptno);
to get the names of the employees which earn more than the avg.
or
select count(deptno)
from emp
group by
deptno having count(deptno) > 4;
to getthe number of employees in each department.
But somehow it doesn't work linking them together. Maybe someone can help me shine a light on this on.
Just put your second query in with an AND clause:
select ename
from emp
where sal > any (select avg(sal)
from emp
group by
deptno)
and deptno in (select deptno
from emp
group by
deptno having count(deptno) > 4);
You can use Having Clause in Conjunction with Group By
select ename
from emp
where sal > any (select avg(sal)
from emp
group by
deptno)
having count(*)>4;
select ename
from (
select deptno, count(deptno)
from emp
group by deptno
having count(deptno) > 4) valid_depts join emp ON emp.deptno=valid_depts.deptno
where sal > any (select avg(sal)
from emp
group deptno);
Related
I have written a piece of code in Oracle database 19c to extract the salary of employees who earns more than the average salary of their respective departments but the query shows "missing right parenthesis". Can anybody help me with it, like where and what the error is all about?
select ename
from emp
where sal > (select round(avg(sal)) as avg_sal, deptno
from emp
group by deptno
order by 2);
You can use these queries to get job done.
select ename
from emp e1
where sal > (select round(avg(sal)) as avg_sal
from emp e2
where e2.deptno = e1.deptno
);
or
select e1.ename
from emp e1,
(
select round(avg(sal)) as avg_sal, deptno
from emp
group by deptno
) e2
where e1.deptno = e2.deptno
and e1.sal > avg_sal
;
Now the output of your subquery is 2 columns and multiple rows, which is against the business logic you described.
To gain the desired result you need to modify the query a bit:
remove the second column in subquery;
remove GROUP BY and ORDER BY clauses from the subquery.
Should be as follows:
SELECT ename FROM emp WHERE sal > (SELECT round(AVG(sal)) AS avg_sal FROM emp);
Now subquery returns only one values which is valuated with each salary in the main query and returns employees' names of ones, that have salary more that average.
The "missing right parenthesis" error is that the ORDER BY clause is not allowed in the sub-query so the SQL parser expects the query to end after the GROUP BY clause.
Once you fix that you get a further error that the sub-query will return multiple rows:
select ename
from emp
where sal > (select round(avg(sal)) as avg_sal, deptno
from emp
group by deptno);
Outputs a different error message:
ORA-00913: too many values
To fix that you need to restrict the sub-query to a single row and correlate the outer-query to the sub-query:
select ename
from emp e
where sal > (select round(avg(sal)) as avg_sal
from emp a
WHERE e.deptno = a.deptno);
Which, for the sample data:
CREATE TABLE emp (ename, sal, deptno) AS
SELECT 'Alice', 100, 1 FROM DUAL UNION ALL
SELECT 'Beryl', 200, 1 FROM DUAL UNION ALL
SELECT 'Carol', 300, 1 FROM DUAL UNION ALL
SELECT 'Debra', 100, 2 FROM DUAL UNION ALL
SELECT 'Ester', 200, 2 FROM DUAL;
Outputs:
ENAME
Carol
Ester
If you want to do it in a single table-scan then you can use analytic functions:
SELECT ename
FROM (
SELECT ename,
sal,
ROUND(AVG(sal) OVER (PARTITION BY deptno)) AS avg_sal
FROM emp
)
WHERE sal > avg_sal;
db<>fiddle here
The schema of the table is:
Emp (Empno, Ename, Job, Mgr, Hiredate, Sal, Comm, Deptno)
Question:
Display the job, deptno and average salary of employees belonging to
department 10 or 20 and their salary is more than 2000 and average
salary is more than 2500.
My Query:
SELECT Job, Deptno, AVG(Sal) "AVGSALARY"
FROM Emp
WHERE Sal > 2000 AND Deptno IN(10, 20)
GROUP BY Job
HAVING AVG(Sal) > 2500;
Error:
ORA-00979: not a GROUP BY expression
I cannot find what is wrong I am doing here that makes this error appear.
I am using the same column name in the GROUP BY statement that I have mentioned in the SELECT statement.
Also, the aggregate function I used in the HAVING statement is also present in the SELECT statement.
My question is 'how can I remove this error from this statement?'.
Any performance improvement query is also welcome.
You need to include all the non-aggregated columns in the GROUP BY statement. So include Deptno in the GROUP BY as well:
SELECT Job, Deptno, AVG(Sal) "AVGSALARY"
FROM Emp
WHERE Sal > 2000 AND Deptno IN(10, 20)
GROUP BY Job, Deptno
HAVING AVG(Sal) > 2500;
This question already has answers here:
Oracle show all employees with greater than average salary of their department
(4 answers)
Closed 3 years ago.
I'm trying to display the employeeno#, employee name, department, and salary for all employees who earn more than the average salary in their own department.
I've only tried this:
Select empno, ename, deptno,sal As Salary
From emp
Where sal >(SELECT AVG (sal) FROM emp);
For this structure, you can use a correlation condition. And that requires introducing table aliases:
select e.empno, e.ename, e.deptno, e.sal As Salary
from emp e
where e.sal > (select avg(e2.sal) from emp e2 where e2.deptno = e.deptno);
It is more efficient to join the table to the average salaries of each deptno under the condition that the salary of the employee is more than the average:
select e.empno, e.ename, e.deptno, e.sal
from emp e inner join (
select deptno, avg(sal) sal from emp group by deptno
) t
on t.deptno = e.deptno and t.sal < e.sal
In Oracle Database the most efficient query would be using analytic functions:
select empno, ename, deptno, sal from
(select emp.*, AVG(sal) over (partition by deptno) avgsal from emp)
where sal > avgsal;
Just one table scan and filtering on the view.
I have a table Employee with those columns
empid, empname, job, hiredate, sal, comm, deptno
I want to verify if this is correct:
Number of employees in each department
How many people there are in each type of job in each department
Display the department and number of employees in department with fewer than 6 employees
Find the employee name and its salary who is earning maximum salary in dept 20
Here is what I have tried:
Query #1:
select DEPTNO, count(*) AS NO_OF_PERSONS
from EMP
group by DEPTNO;
Query #2:
select job, count(*) AS NO_OF_PERSONS
from EMP
group by job;
Query #3:
update EMPLOYEE
set sal = sal + 1000
where com > 2500;
And I am unable to do the 4th part.
I hope that below queries will help you.
no of employees in each dept?
SELECT DEPTNO,
count(*) AS NO_OF_PERSONS
FROM EMP
GROUP BY DEPTNO;
how many people are there in each type of job in each department?
SELECT job,
deptno,
count(*) AS NO_OF_PERSONS
FROM EMP
GROUP BY job,
deptno;
display the department and no of employees in department with fewer than 6 employee.
SELECT deptno,
count(*)
FROM emp
GROUP BY deptno
HAVING count(*) < 6;
find the employee name and its salary who is earning maximum salary in dept 20.
SELECT Max(salary_amount),
empid
FROM EMP
WHERE deptno = 20
GROUP BY empid;
You can start learning Basic SQL here
select e1.name, e1.sal
from EMP e1
where e1.DEPTNO = 20
and not exists(select *
from EMP e2
where e2.DEPTNO = 20
and e2.sal > e1.sal);
Practicing some SQL, we have to get the name of the employees whose salary is the greatest of his department. But if in any department there were more than one employer with the greatest salary, we would not have to consider that department.
We got the first part but not the second one (because there are two employees with the same greatest salary (3,000) in the same department (20)).
This is what we did:
SQL> SELECT ename, sal, deptno FROM emp a
WHERE sal >= ALL (SELECT sal FROM emp WHERE deptno=a.deptno)
ORDER BY sal;
And this is what we got:
ENAME SAL DEPTNO
---------- ------- ------
BLAKE 2,850 30
FORD 3,000 20
SCOTT 3,000 20
KING 5,000 10
4 filas seleccionadas.
Any help will be useful, thank you!
SELECT ename, sal, deptno
FROM emp a
WHERE not exists (
SELECT *
FROM emp
WHERE deptno=a.deptno
and sal >= a.sal
and ename != a.ename)
ORDER BY sal;
with cte as
( SELECT ename, sal, deptno
, row_number() over (partition by deptno order by sal desc) as rn
FROM emp
)
select ename, sal, deptno from cte where rn = 1
except
select ename, sal, deptno from cte where rn = 2
order by sal
if this does not work in oracle - it used to be also tagged mssql
You can have what you need with some analytic functions:
select ename,
deptno,
sal
from (
select ename,
deptno,
sal,
row_number() over(partition by deptno order by sal desc) AS num,
count(1) over(partition by deptno, sal) AS count
from emp
)
where num = 1
and count = 1
The inner query orders by salary and counts the number of employees with the same salary in the same department; the outer one simply filters for employees with the maximum salary, where only one employee has that salary in the department.
With a different approach, simply modifying your query, you can try:
SELECT ename, sal, deptno FROM emp a
WHERE sal >= ALL (SELECT sal FROM emp WHERE deptno=a.deptno)
and (select count(1) from emp b where a.deptno = b.deptno and a.sal = b.sal) = 1
The first way gives better performance, with a single table scan, while the second one needs a nested query, thus being less efficient
try to use GROUP BY column_name`, it will show the record without duplicate.