i have three columns employee_id,e_name,Sal and i want the sum of Sal group by employee_id and i want all the columns like employee_id,e_name,Sal,sum as out put in oracle.
in/put= OP=
employee_id,e_name,Sal employee_id,e_name,sal,sum
select employee_id,e_name,sal, sum(Sal) over()
from YOUR_TABLE
you will have the same value of sum for each employee
Related
When I try it without displaying department_id it works fine as :
SQL> SELECT MAX(AVG(SALARY)) FROM EMPLOYEE GROUP BY DEPARTMENT_ID;
MAX(AVG(SALARY))
----------------
800000
But when I want to display the department_id's too, it gives me error as follows:
SQL> SELECT DEPARTMENT_ID, MAX(AVG(SALARY)) FROM EMPLOYEE GROUP BY DEPARTMENT_ID;
SELECT DEPARTMENT_ID, MAX(AVG(SALARY)) FROM EMPLOYEE GROUP BY DEPARTMENT_ID
*
ERROR at line 1:
ORA-00937: not a single-group group function
Is there any explanation for this? What am I doing wrong? I went through answers of previous questions like this and tried their solutions but got same or some other error. Any help would be appreciated.
I suggest you use
SELECT department_id, avg_salary
FROM ( SELECT DEPARTMENT_ID,
AVG (SALARY) avg_salary,
RANK () OVER (ORDER BY AVG (salary) DESC) rnk
FROM EMPLOYEE
GROUP BY DEPARTMENT_ID)
WHERE rnk = 1;
i.e.
use your first query as a "source"
additionally, rank average salaries in descending order (using the rank analytic function)
select row which ranks as highest
I have the table EMPLOYEES with columns EMPLOYEE_ID, NAME, SALARY, COMMISSION_PCT.
I have to calculate the average salary of employees taking into account commissions.
How can I do that? Thank you!
Just like you said:
select avg(sal + nvl(comm, 0)) avg_sal_comm
from emp
If commission part is expressed as percentage, then
select avg(sal + sal * nvl(comm, 0) / 100) avg_sal_comm
from emp;
NVL is here because not all employees have comm column value.
I don't want rank column but still, want the data in the same format by applying dense rank.
select ename,position,deptno,dense_rank() over(partition by deptno order by ename asc) as rank from emp where deptno in ('10','30');
Why do you need the rank just order by deptno first then ename.
SELECT ename,position,deptno,
FROM emp
WHERE deptno in ('10','30')
ORDER BY DeptNo, Ename
Using the analytical function two options derived table or CTE
Derived table/inline view.
SELECT ename,position,deptno
FROM (select ename,position,deptno,dense_rank() over(partition by deptno order by ename asc) as rank
from emp
where deptno in ('10','30')) Z
ORDER BY deptNo, rank
Common Table Expression (CTE):
with Z AS (SELECT ename,position,deptno
, dense_rank() over(partition by deptno order by ename asc) as rank
FROM emp
WHERE deptno in ('10','30'))
SELECT ename,position,deptno
FROM z
ORDER BY deptno, rank
Both these last 2 techniques simply avoid exposing the rank function to the outer query in which the results are returned. They are "Tricks" and sub-optimal execution time. unless there's a specific reason to have the rank data; I'd not use it.
I have a table with the following data:
SCORE ROW_ID NAME
0.4 1011 ABC
0.95 1011 DEF
0.4 501 GHI
0.95 501 XYZ
At any point of time, i only need single row of data with maximum score, if there has more than 1 records, take the one with minimum row_id.
Is it possible to achieve by using RANK or DENSE_RANK function? How about partition by?
MAX(score) keep(dense_rank first order by row_id)
You are looking for max score, one row, so use row_number():
select score, row_id, name
from (select t.*, row_number() over (order by score desc, row_id) rn from t)
where rn = 1
demo
You can use rank and dense_rank in your example, but they can return more than one row, for instance when you add row (0.95, 501, 'PQR') to your data.
keep dense_rank is typically used when searched value is other than search criteria, for instance if we look for salary of employee who works the longest:
max(salary) keep (dense_rank first order by sysdate - hiredate desc)
max in this case means that if there are two or more employees who works longest, but exactly the same number of days than we take highest salary.
max(salary)
keep (dense_rank first order by sysdate - hiredate desc)
over (partition by deptno)
This is the same as above, but salary of longest working employees is shown for each department separately. You can even use empty over() to show salary of longest working employee in separate column except other data like name, salary, hire_date.
You dont need to use dense_rank. This would help
SELECT * FROM (
SELECT
SCORE,
ROW_ID
NAME
FROM T
ORDER BY SCORE DESC, ROW_ID DESC
)
WHERE ROWNUM = 1;
So I have two tables salary and emp whose definition is shown as below
[
I am tring to create a query that Find the employee who draws the maximum salary and Display the employee details along with the nationality.
I created this query
select empcode,
max(basic) as "Highest Sal"
from salary
join emp on empcode;
Please help with this
Your query uses a simple aggregate max(basic) which would find the highest salary. Except you need to join to the EMP table to display other details. This means you can't use aggregation, because we need to GROUP BY the non-aggregated columns, which would make a nonsense of the query.
Fortunately we can solve the problem with an analytic function. The subquery selects all the relevant information and ranks each employee by salary, with a rank of 1 being the highest paid. We use rank() here because that will handle ties: two employees with the same basic will be in the same rank.
select empcode
, empname
, nationality
, "Highest Sal"
from (
select emp.empcode
, emp.empname
, emp.nationality
, salary.basic as "Highest Sal"
, rank() over (order by salary.basic desc ) as rnk
from salary join emp on emp.empcode = salary.empcode
)
where rnk = 1;
Find the employee who draws the maximum salary
An employee can have multiple salaries in your datamodel. An employee's (total) salary hence is the sum of these. You want to find the maximum salary per employee and show the employee(s) earning that much.
You can use MAX OVER to find the maximum sum:
select e.*, s.total_salary
from emp e
join
(
select
empcode,
sum(basic) as total_salary,
max(sum(basic)) over () as max_total_salary
from salary
) s on s.empcode = e.empcode and s.total_salary = s.max_total_salary
order by e.empcode;
Try this:
SELECT * FROM
(SELECT E.EmpCode, E.EmpName, E.DOB, E.DOJ, E.DeptCode, E.DesgCode, E.PhNo,
E.Qualification, E.Nationality, S.Basic, S.HRA, S.TA, S.UTA, S.OTRate
FROM EMP AS E JOIN SALARY AS S ON (E.EmpCode = S.EmpCode) order by S.Basic desc)
WHERE rownum = 1