I have data of two tables. In the first table I have employee id's and their names. In the second table, I have employee details (such as salary, location, join date, projects handled, designation).
I am trying to write a query where I can bring the data of maximum and minimum salary group by location and designation. This is what I have so far:
SELECT E.EMPID,I.EMPID, E.EMPNAME, I.EMPNAME,
B.[MAXIMUM SALARY],F.[MINIMUM SALARY],
B.DESIG,F.DESIG,B.LOCATION,F.LOCATION
FROM
((SELECT MAX (SALARY) [MAXIMUM SALARY], LOCATION, DESIG
FROM
EMPDETAILS
GROUP BY LOCATION, DESIG) B
JOIN
(SELECT C.EMPID,C.SALARY,C.GENDER,C.DESIG,
C.JOINON,C.LOCATION,C.PROJECTS,D.EMPNAME
FROM EMPDETAILS C
JOIN
EMPLOYEE D
ON C.EMPID=D.EMPID) E
ON E.DESIG=B.DESIG AND E.SALARY=B.[MAXIMUM SALARY] AND E.LOCATION =
B.LOCATION)
FULL OUTER JOIN[enter image description here][1]
((SELECT MIN(SALARY) AS [MINIMUM SALARY], LOCATION, DESIG
FROM EMPDETAILS
GROUP BY LOCATION, DESIG) F
JOIN
(SELECT G.EMPID,H.EMPNAME,G.DESIG,G.GENDER,
G.JOINON,G.LOCATION,G.PROJECTS,G.SALARY
FROM EMPDETAILS G
JOIN EMPLOYEE H
ON G.EMPID=H.EMPID) I
ON I.DESIG=F.DESIG AND F.[MINIMUM SALARY]=I.SALARY AND
I.LOCATION=F.LOCATION)
ON I.DESIG=E.DESIG
ORDER BY B.LOCATION, f.location
Giving me:
no clue what you really want/need, but if you want employeeID-level results, you'll need something like:
SELECT
d.*,
min_salary,
max_salary
FROM
EMPDETAILS d JOIN
EMPLOYEE e ON
d.EMPID=e.EMPID JOIN
(SELECT
empid,
MIN(salary) AS min_salary,
MAX(salary) AS max_salary
FROM
EMPDETAILS
GROUP BY
empid) m ON
e.empid = m.empid
if you want min/max salaries grouped by LOCATION, DESIG as in your code, you'll need to return those results separately from employee-level details:
SELECT
LOCATION, DESIG,
MIN(salary) AS min_salary,
MAX(salary) AS max_salary
FROM
EMPDETAILS
GROUP BY
LOCATION, DESIG
you don't need separate subqueries for your min and max groupings, you don't need to join your detail tables multiple times, and you don't need a full outer join.
Related
Sql query to find the active employees. sort the output in descending order based on employee id column and fetch only the first 15 records.
Display employee id, employee name, city, state, country,salary, active date, status, department name, manager name.
employee name: concatenate firstname & lastname with space character and fetch only first 10 letters from employee name
manager name: display firstname and lastname(separated by space)
Tables:
EMPLOYEE--dept_id,emp_activefrom, emp_dob, emp_fname, emp_lname, emp_sal, emp_status, emp_terminationdate, loc_id, mgr_id
LOCATION-- city, country, loc_id, state
DEPARTMENT--dept_head,dept_id, dept_name
the below given is the code which i used,
SELECT b.* FROM (SELECT e.emp_id, SUBSTR(e.emp_fname||' '||e.emp_lname,1,10) AS emp_name, l.city, l.state, l.country, e.emp_sal, e.emp_activefrom, e.emp_status,d.dept_name, d.dept_head FROM employee e, location l, department d WHERe e.dept_id=d.dept_id AND e.loc_id=l.loc_id AND e.emp_status='Active' ORDER BY 1 desc) b WHERE rownum<=15;
I dont know what is wrong here. please help me out to solve this.
It's always better use proper join for better understanding. To get the manager name self join (employee e inner jion employee m) is used here.
SELECT e.emp_id, SUBSTR(e.emp_fname||' '||e.emp_lname,1,10) AS emp_name, l.city, l.state, l.country, e.emp_sal, e.emp_activefrom,
e.emp_status,d.dept_name, d.dept_head , m.emp_fname||' '||m.emp_lname as manager_name
FROM employee e
inner join department d on e.dept_id=d.dept_id
inner join location l on e.loc_id=l.loc_id
inner join employee m on e.Mgrid=m.EmpId
where e.emp_status='Active' and
rownum<=15
order by e.EmpId desc
Kazi is correct that you want to do this using JOINs. However you need to be careful about employees that have no managers when doing the JOIN. In addition, Oracle is finicky about limiting the number of rows -- and I assume you are using Oracle based on rownum comparisons. The correct method uses the fetch first clause.
So, you seem to be describing something like this:
select e.emp_id, substr(e.emp_fname || ' ' || e.emp_lname,1,10) AS emp_name,
l.city, l.state, l.country,
e.emp_sal, e.emp_activefrom, e.emp_status,
d.dept_name, d.dept_head,
(m.emp_fname || ' ' || m.emp_lname) as manager_name
from employee e join
department d
on e.dept_id = d.dept_id join
location l
on e.loc_id = l.loc_id left join
employee m
on e.Mgrid = m.EmpId
where e.emp_status = 'Active'
order by e.EmpId desc
fetch first 15 rows only;
Department wise maximum salary in a table it's working:
select * from employee where (emp_dept, emp_sal)
in (select emp_dept, max(emp_sal) from employee group by emp_dept);
but my concern is : i want emp_id, emp_name, emp_sal, emp_dept column with department wise maximnum salary
customer table : cust_id, cust_name, emp_id
employee table : emp_id, emp_name, emp_sal, emp_dep
I think you want :
select e.emp_id, e.emp_name, e.emp_sal, e.emp_dept
from employee e
where e.emp_sal = (select max(e1.emp_sal) from employee e1 where e1.emp_dept = e.emp_dept);
Assuming you want all the records from the employee table with the maximum salary added, then you could use something like:
Select e.emp_id, e.emp_name, e.emp_sal, e.emp_dept, d.max_salary
from employee e
left join
(select emp_dept,max(emp_sal) as max_salary
from employee
group by emp_dept) d
ON e.emp_dept=d.emp_dept
The sub query calculates the maximum salary for each department and then using the left join you can get it into the final table.
If you want the employee details of the employee with the maximum salary, then:
Select e.emp_id, e.emp_name, e.emp_sal, e.emp_dept, d.max_salary
from employee e
inner join
(select emp_dept,max(emp_sal) as max_salary
from employee
group by emp_dept) d
ON e.emp_dept=d.emp_dept and e.emp_sal=d.max_salary
In this case the join is based on the salary component as well, thereby returning only the employees with the maximum salary. Hope this helps.
Just use a window function:
select e.*,
max(emp_sal) over (partition by emp_dept) as dept_max_sal
from employee ;
I found a couple of SQL tasks on Hacker News today, however I am stuck on solving the second task in Postgres, which I'll describe here:
You have the following, simple table structure:
List the employees who have the biggest salary in their respective departments.
I set up an SQL Fiddle here for you to play with. It should return Terry Robinson, Laura White. Along with their names it should have their salary and department name.
Furthermore, I'd be curious to know of a query which would return Terry Robinsons (maximum salary from the Sales department) and Laura White (maximum salary in the Marketing department) and an empty row for the IT department, with null as the employee; explicitly stating that there are no employees (thus nobody with the highest salary) in that department.
Return one employee with the highest salary per dept.
Use DISTINCT ON for a much simpler and faster query that does all you are asking for:
SELECT DISTINCT ON (d.id)
d.id AS department_id, d.name AS department
,e.id AS employee_id, e.name AS employee, e.salary
FROM departments d
LEFT JOIN employees e ON e.department_id = d.id
ORDER BY d.id, e.salary DESC;
->SQLfiddle (for Postgres).
Also note the LEFT [OUTER] JOIN that keeps departments with no employees in the result.
This picks only one employee per department. If there are multiple sharing the highest salary, you can add more ORDER BY items to pick one in particular. Else, an arbitrary one is picked from peers.
If there are no employees, the department is still listed, with NULL values for employee columns.
You can simply add any columns you need in the SELECT list.
Find a detailed explanation, links and a benchmark for the technique in this related answer:
Select first row in each GROUP BY group?
Aside: It is an anti-pattern to use non-descriptive column names like name or id. Should be employee_id, employee etc.
Return all employees with the highest salary per dept.
Use the window function rank() (like #Scotch already posted, just simpler and faster):
SELECT d.name AS department, e.employee, e.salary
FROM departments d
LEFT JOIN (
SELECT name AS employee, salary, department_id
,rank() OVER (PARTITION BY department_id ORDER BY salary DESC) AS rnk
FROM employees e
) e ON e.department_id = d.department_id AND e.rnk = 1;
Same result as with the above query with your example (which has no ties), just a bit slower.
This is with reference to your fiddle:
SELECT * -- or whatever is your columns list.
FROM employees e JOIN departments d ON e.Department_ID = d.id
WHERE (e.Department_ID, e.Salary) IN (SELECT Department_ID, MAX(Salary)
FROM employees
GROUP BY Department_ID)
EDIT :
As mentioned in a comment below, if you want to see the IT department also, with all NULL for the employee records, you can use the RIGHT JOIN and put the filter condition in the joining clause itself as follows:
SELECT e.name, e.salary, d.name -- or whatever is your columns list.
FROM employees e RIGHT JOIN departments d ON e.Department_ID = d.id
AND (e.Department_ID, e.Salary) IN (SELECT Department_ID, MAX(Salary)
FROM employees
GROUP BY Department_ID)
This is basically what you want. Rank() Over
SELECT ename ,
departments.name
FROM ( SELECT ename ,
dname
FROM ( SELECT employees.name as ename ,
departments.name as dname ,
rank() over (
PARTITION BY employees.department_id
ORDER BY employees.salary DESC
)
FROM Employees
JOIN Departments on employees.department_id = departments.id
) t
WHERE rank = 1
) s
RIGHT JOIN departments on s.dname = departments.name
Good old classic sql:
select e1.name, e1.salary, e1.department_id
from employees e1
where e1.salary=
(select maxsalary=max(e.salary) --, e. department_id
from employees e
where e.department_id = e1.department_id
group by e.department_id
)
Table1 is emp - empno, ename, sal, deptno
Table2 is dept - deptno, dname.
Query could be (includes ties & runs on 11.2g):
select e1.empno, e1.ename, e1.sal, e1.deptno as department
from emp e1
where e1.sal in
(SELECT max(sal) from emp e, dept d where e.deptno = d.deptno group by d.dname)
order by e1.deptno asc;
SELECT
e.first_name, d.department_name, e.salary
FROM
employees e
JOIN
departments d
ON
(e.department_id = d.department_id)
WHERE
e.first_name
IN
(SELECT TOP 2
first_name
FROM
employees
WHERE
department_id = d.department_id);
`select d.Name, e.Name, e.Salary from Employees e, Departments d,
(select DepartmentId as DeptId, max(Salary) as Salary
from Employees e
group by DepartmentId) m
where m.Salary = e.Salary
and m.DeptId = e.DepartmentId
and e.DepartmentId = d.DepartmentId`
The max salary of each department is computed in inner query using GROUP BY. And then select employees who satisfy those constraints.
Assuming Postgres
Return highest salary with employee details, assuming table name emp having employees department with dept_id
select e1.* from emp e1 inner join (select max(sal) avg_sal,dept_id from emp group by dept_id) as e2 on e1.dept_id=e2.dept_id and e1.sal=e2.avg_sal
Returns one or more people for each department with the highest salary:
SELECT result.Name Department, Employee2.Name Employee, result.salary Salary
FROM ( SELECT dept.name, dept.department_id, max(Employee1.salary) salary
FROM Departments dept
JOIN Employees Employee1 ON Employee1.department_id = dept.department_id
GROUP BY dept.name, dept.department_id ) result
JOIN Employees Employee2 ON Employee2.department_id = result.department_id
WHERE Employee2.salary = result.salary
SQL query:
select d.name,e.name,e.salary
from employees e, depts d
where e.dept_id = d.id
and (d.id,e.salary) in
(select dept_id,max(salary) from employees group by dept_id);
Take look at this solution
SELECT
MAX(E.SALARY),
E.NAME,
D.NAME as Department
FROM employees E
INNER JOIN DEPARTMENTS D ON D.ID = E.DEPARTMENT_ID
GROUP BY D.NAME
T1: employee [id, salary]
T2: department [name, employeeid]
(employeeid is a foreign key to T1's id)
Problem: Write a query to fetch the name of the department which receives the maximum salary.
My Solution:
SELECT DISTINCT name
FROM department AS a
INNER JOIN employee AS b ON a.employeeid = b.id
AND b.salary
IN (
SELECT max( salary )
FROM employee AS c
)
Edit: The problem statement is accurate, and we're not trying to find out the employee who has the highest salary. It says "....Department which receives.....", not "...employee who receives....".
Is this ok? Or can this be optimized?
GROUP BY the name of the department and order by SUM(salary).
SELECT department.name
FROM department
JOIN employee ON department.employeeid = employee.id
GROUP BY department.name
ORDER BY SUM(salary) DESC
LIMIT 1
How about:
SELECT employee.id, employee.salary, department.name
FROM department, employee
where
employee.id = department.employeeid and
employee.salary = (select max(salary) from employee)
When using the SQL MIN() function, along with GROUP BY, will any additional columns (not the MIN column, or one of the GROUP BY columns) match the data in the matching MIN row?
For example, given a table with department names, employee names, and salary:
SELECT MIN(e.salary), e.* FROM employee e GROUP BY department
Obviously I'll get two good columns, the minimum salary and the department. Will the employee name (and any other employee fields) be from the same row? Namely the row with the MIN(salary)?
I know there could very possibly be two employees with the same (and lowest) salary, but all I'm concerned with (now) is getting all the information on the (or a single) cheapest employee.
Would this select the cheapest salesman?
SELECT min(salary), e.* FROM employee e WHERE department = 'sales'
Essentially, can I be sure that the data returned along with the MIN() function will matches the (or a single) record with that minimum value?
If the database matters, I'm working with MySql.
If you wanted to get the "cheapest" employee in each department you would have two choices off the top of my head:
SELECT
E.* -- Don't actually use *, list out all of your columns
FROM
Employees E
INNER JOIN
(
SELECT
department,
MIN(salary) AS min_salary
FROM
Employees
GROUP BY
department
) AS SQ ON
SQ.department = E.department AND
SQ.min_salary = E.salary
Or you can use:
SELECT
E.*
FROM
Employees E1
LEFT OUTER JOIN Employees E2 ON
E2.department = E1.department AND
E2.salary < E1.salary
WHERE
E2.employee_id IS NULL -- You can use any NOT NULL column here
The second statement works by effectively saying, show me all employees where you can't find another employee in the same department with a lower salary.
In both cases, if two or more employees have equal salaries that are the minimum you will get them both (all).
SELECT e.*
FROM employee e
WHERE e.id =
(
SELECT id
FROM employee ei
WHERE ei.department = 'sales'
ORDER BY
e.salary
LIMIT 1
)
To get values for each department, use:
SELECT e.*
FROM department d
LEFT JOIN
employee e
ON e.id =
(
SELECT id
FROM employee ei
WHERE ei.department = d.id
ORDER BY
e.salary
LIMIT 1
)
To get values only for those departments that have employees, use:
SELECT e.*
FROM (
SELECT DISTINCT eo.department
FROM employee eo
) d
JOIN
employee e
ON e.id =
(
SELECT id
FROM employee ei
WHERE ei.department = d.department
ORDER BY
e.salary
LIMIT 1
)
Of course, having an index on (department, salary) will greatly improve all three queries.
The fastest solution:
SET #dep := '';
SELECT * FROM (
SELECT * FROM `employee` ORDER BY `department`, `salary`
) AS t WHERE IF ( #dep = t.`department`, FALSE, ( #dep := t.`department` ) OR TRUE );
Another approach can be using Analytical functions. Here is the query using analytical and ROW_NUM functions
select first_name, salary from (select first_name,salary, Row_NUMBER() over (PARTITION BY DEPARTMENT_ID ORDER BY salary ASC) as row_count from employees) where row_count=1;