There is such a task: By joining the tables HR.DEPARTMENTS and HR.EMPLOYEES, display complete data on departments in which the minimum salary is below 5000.
I tried to do this, but it gives an error
select distinct d.department_id,department_name,
d.manager_id, location_id
from hr.departments d
left join hr.employees e on e.department_id = d.department_id
where min(e.salary) < 5000
order by 1
Error: group function is not allowed here
This is what hr.employees looks like:
EMPLOYEE_ID FIRST_NAME LAST_NAME EMAIL PHONE_NUMBER HIRE_DATE JOB_ID SALARY COMMISSION_PCT MANAGER_ID DEPARTMENT_ID
100 Steven King SKING 515.123.4567 17-JUN-03 AD_PRES 24000 - - 90
hr.departments:
DEPARTMENT_ID DEPARTMENT_NAME MANAGER_ID LOCATION_ID
10 Administration 200 1700
You cannot use MIN in the WHERE clause, because MIN is an aggregation result over many rows, but in a WHERE clause you look at single rows (before any aggregation takes place).
The task to get the departments in question by joining the tables is a bit weird, because this is not how this should be done in SQL. If you must do it this way, then you only need a slight change to your query: Change the join into an inner join and check the rows' salary.
select distinct
d.department_id, department_name, d.manager_id, location_id
from hr.departments d
join hr.employees e on e.department_id = d.department_id
where e.salary < 5000
order by d.department_id;
The proper solution would use EXISTS or IN instead, so as not to create an unnecessarily large intermediate result that you must get rid of with DISTINCT:
select *
from hr.departments
where department_id in (select department_id from employees where salary < 5000)
order by department_id;
or
select *
from hr.departments d
where exists
(
select null
from employees e
where e.salary < 5000
and e.department_id = d.department_id
)
order by department_id;
This works for your solution, where is use for row filtering like gender = 'Male' while having is for aggregating filtering functions like min(salary) < 5000 but for having you need to group by with something like department.
SELECT
*
FROM
DimEmployee
WHERE
EmployeeID IN (
SELECT
EmployeeID
FROM
DimEmployee
GROUP BY
EmployeeID
HAVING
MIN(Salary) < 5000
)
First of all, don't use distinct, unless you have to. Secondly, you can't use group functions like that.
In order to solve this, you need to break the task into steps, breaking down your sentences.
"...the tables"
So we have this:
SELECT * FROM hr.departments;
... and ...
SELECT * FROM hr.employees;
"HR.DEPARTMENTS and HR.EMPLOYEES"
As you pointed our, the FK is the department.
(we first test the join, then add what we need)
(the 1 is just a placeholder; you can use EMPLOYEE_ID or COUNT(1), it's irrelevant)
SELECT 1
FROM hr.employees e
LEFT JOIN hr.departments d on e.department_id = d.department_id;
"display complete data on departments"
Well, this is simple, you just enumerate the columns you need or use d.*. We'll do this later.
"which the minimum salary is below 5000"
Now we get to the blocking issue. Let's list the records.
SELECT d.*
FROM hr.employees e, hr.departments d
WHERE e.department_id = d.department_id
AND EXISTS (SELECT 1 FROM hr.employees m WHERE m.department_id = d.department_id GROUP BY m.department_id HAVING min(m.salary) < 5000);
But what's this? We get a line for every employee of that department. Well, we can either use DISTINCT, but that is bad practice or we can fix the query.
We'll just remove the employees from the join.
SELECT d.*
FROM hr.departments d
WHERE EXISTS (SELECT 1 FROM hr.employees e WHERE e.department_id = d.department_id GROUP BY e.department_id HAVING min(e.salary) < 5000);
UPDATE:
To respect the task "By joining the tables"
So we have this:
SELECT d.*
FROM hr.departments d,
(
SELECT e.department_id
FROM hr.employees e
GROUP BY e.department_id
HAVING min(salary) < 5000
) e
WHERE e.department_id = d.department_id;
I am stuck on a school project. I need the second-highest salary per department, but without having the third, fourth etc. I am using Oracle SQL. and I need to do it with subquery.
This is the only code I can come up with so far. I can filter all salaries after the highest, but I cant filter only the second highest. I have two statements, that produce the same result at the end.
select e.department_id, d.department_name, e.first_name, e.last_name, e.salary
from employees e inner join departments d
on (e.department_id = d.department_id)
where e.salary < (select max(salary)
from employees e
where e.department_id = d.department_id
order by d.department_id
offset 1 row fetch next 2 row only)
order by e.department_id;
Or
select department_id, first_name, last_name, salary
from employees
where salary < any (select max(salary)
from employees
group by department_id)
order by department_id;
I have 2 tables, one table storing details of staff (columns are staff_id, staff_name, department_id) and another table storing details of department (columns are department_id, department_name, department_block_num).
I need to write a query to display names of department and staff count in each department, if staff not exist display count as 0.
Here is code
Select department_name,
case department.department_id
when department.department_id=staff.department_id then count(staff_name)
else 0 end staff_count
From department, staff
Group by department_name
order by department_name;
You are looking for a LEFT JOIN and an aggregation query:
select d.department_id, d.department_name,
count(s.department_id) as staff_count
from department d left join
staff s
on d.department_id = s.department_id
group by d.department_id, d.department_name
So i have two tables
EMPLOYEE- Contains columns including EMPLOYEE_NAME, DEPARTMENT_ID and SALARY
DEPARTMENTS - Contains columns including DEPARTMENT_NAME, and DEPARTMENT_ID
I need to display the department name and the average slary for each department and order it by the average salaries.
I am new to DBs and am having trouble.
I try to do a subquery in the from field ( this subquery returns exactly what i need minus the department name which requires me to then join the departments table to the results) all the data in the subquery is in one table- employees. while department name is in the departments table.
here is what i tried.
SELECT D.DEPARTMENT_NAME, T.PERDEPT
FROM
(
SELECT DEPARTMENT_ID, AVG(SALARY) AS PERDEPT
FROM EMPLOYEE
GROUP BY DEPARTMENT_ID
ORDER BY PERDEPT
) AS TEST T
JOIN DEPARTMENTS
ON D.DEPARTMENT_ID=T.DEPARTMENT_ID;
This returns a
SQL command not properly terminated
on the line with the AS TEST T
any and all help is greatly appreciated
many thanks
This query should do what you ask:
select d.department_name, avg(e.salary) as avg_salary
from salary_department d
left join employee e on e.department_id = d.department_id
group by d.department_name
order by avg(e.salary)
Simply correct your table aliases as you seem to have two aliases for subquery (TEST and T) and no assignment for D. Adjust SQL with one alias for each table/query reference:
...
(
SELECT ...
) AS T
JOIN DEPARTMENTS D
With that said, you do not even need the subquery as aggregate query with JOIN should suffice, assuming DEPARTMENT_ID is unique in DEPARTMENTS table to not double count the aggregate.
SELECT D.DEPARTMENT_NAME,
AVG(E.SALARY) AS PERDEPT
FROM EMPLOYEE E
JOIN DEPARTMENTS D
ON E.DEPARTMENT_ID = D.DEPARTMENT_ID
GROUP BY E.DEPARTMENT_ID,
D.DEPARTMENT_NAME
ORDER BY AVG(SALARY)
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