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
this is the schema Write a query to display the name of the department that has the maximum student count.
this is what is tried.
select d.department_name,count(s.student_id)
from department d left join student s
on d.department_id=s.department_id
group by d.department_name,d.department_id
order by d.department_name;
and i think there is something missing in my code
You're almost there.
Order the result in descending order on the number of students and then take the first row:
SELECT department_name
FROM
(
SELECT d.department_name,
COUNT(*) AS nr_students
FROM department d
JOIN student s
ON d.department_id = s.department_id
GROUP BY d.department_name
ORDER BY nr_students DESC
)
WHERE ROWNUM <= 1;
Based on the schema mentioned, you would have to make a join (INNER JOIN) to the department table from the staff table to get the name of the department.
If the name of the department is not desired and the counts can just be based on the department_id, then a join is not required.
The queries for both the scenarios is mentioned below.
Oracle SQL query for result with the department name, i.e. with INNER JOIN
SELECT D.DEPARTMENT_NAME, COUNT(S.DEPARTMENT_ID) AS STAFF_COUNT FROM **DEPARTMENT D, STAFF S** --INDICATES INNER JOIN IN ORACLE SQL
WHERE D.DEPARTMENT_ID = S.DEPARTMENT_ID
GROUP BY D.DEPARTMENT_NAME
ORDER BY STAFF_COUNT DESC
Oracle SQL query for result without the department name, just the department_id
SELECT S.DEPARTMENT_ID,COUNT(S.DEPARTMENT_ID) AS STAFF_COUNT FROM STAFF S
GROUP BY S.DEPARTMENT_ID
ORDER BY STAFF_COUNT DESC
Hope this helps. Cheers.
I tried this and it worked.
select department_name
from department d inner join student s
on s.department_id=d.department_id
having count(*) in (
select max(count(student_id))
from student s join department d
on d.department_id=s.department_id
group by d.department_id)
group by d.department_id,department_name;
Select * from (
SELECT D.DEPARTMENT_NAME, COUNT(S.DEPARTMENT_ID) AS STAFF_COUNT
FROM DEPARTMENT D, STAFF S
WHERE D.DEPARTMENT_ID = S.DEPARTMENT_ID
GROUP BY D.DEPARTMENT_NAME
ORDER BY STAFF_COUNT DESC)
where rownum=1;
This query will give department name that has maximum number of student count
An employee belongs to a department (foreign key = D_ID). An employee has a SSN (primary key), name, salary and D_ID.
A department has multiple employees (primary key = ID)
I want a query that returns Department name| Name of Highest Paid Employee of that department | His Salary | Average salary of employees working in the same department.
I know how to select the first part:
SELECT
D.name, E.name, E.salary
FROM
Employee E, Department D
WHERE
salary IN (SELECT MAX(E.salary)
FROM Employee E
GROUP BY E.D_ID)
AND E.D_ID = D.ID
I know also how to select the last part:
SELECT AVG(E.salary)
FROM Employee E
GROUP BY E.D_ID
How do I put these together in a single query?
You can use window functions for that:
select department_name, employee_name, salary, avg_dept_salary
from (
select e.name as employee_name,
d.name as department_name,
e.salary,
max(e.salary) over (partition by d.id) as max_dept_salary,
avg(e.salary) over (partition by d.id) as avg_dept_salary
from Employee E
join Department D on e.d_id = d.id
) t
where salary = max_dept_salary
order by department_name;
The above is standard ANSI SQL and runs on all modern DBMS.
I would do something like this:
SELECT d.name
, e.name
, e.salary
, n.avg_salary
FROM Department d
JOIN ( SELECT m.d_id
, MAX(m.salary) AS max_salary
, AVG(m.salary) AS avg_salary
FROM Employee m
GROUP BY m.d_id
) n
ON n.d_id = d.id
JOIN Employee E
ON e.d_id = d.id
AND e.salary = n.max_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
So I'm trying to figure this question out my professor gave me and I cannot seem to get the code right no matter what I try. The employee table has the employee information, like salary, while the workon table has the information about the hours for the project The code I have now is
select e.name
from employee e, workon w
where e.empid = w.empid
and e.name in
(select name
from employee
having salary < avg (salary)
and empid in
(select empid
from workon
having sum (hours) > 100))
group by e.name
Maybe something like this:
select
e.name
from
employee e
inner join workon wo on e.employee = wo.employee
where
e.salary < (select avg(salary) from employee)
and sum(wo.hours) > 100
group by
e.name;
Try this one:
SELECT name FROM employee WHERE salary < (SELECT AVG(salary) FROM employee) having sum(hours) > 100 GROUP BY name;