I'm trying to figure a way to retrieve employees working in two different departments.
I have 3 simple tables:
employee (employee_id, employee_name)
department (department_id, department_name)
working (eid, did, work_time)
So I have tried to write a SQL query:
select employee_name
from employee, working,department
where eid = employee_id
and did = department_id
and department_name = 'software'
and dname = 'hardware';
But it doesn't work, what is my problem?
The problem is that you are requiring department to be both 'software' and 'hardware'. Also, dname is not a field.
Correcting your query:
select employee_name
from employee, working, department
where eid = employee_id and did = department_id
and (department_name = 'software' or department_name = 'hardware');
But I would prefer this kind of query:
SELECT DISTINCT e.employee_name
FROM employee e
JOIN working w ON w.eid = e.employee_id
JOIN department d ON d.department_id = w.did
WHERE d.department_name IN ('software', 'hardware');
That is to get employees that work in any of the two departments (or both).
If you want only employees that work in both departments, try this:
SELECT e.employee_id, e.employee_name
FROM employee e
JOIN working w ON w.eid = e.employee_id
JOIN department d ON d.department_id = w.did
WHERE d.department_name IN ('software', 'hardware')
GROUP BY e.employee_id HAVING COUNT(DISTINCT d.department_id) = 2;
Would something like this work for you?
SELECT
count(*) as cnt,
employee.employee_name
FROM
employee
JOIN working ON working.eid = employee.employee_id
JOIN department ON department.department_id = working.did
WHERE
department.department_name = 'software' or department.department_name = 'hardware'
GROUP BY employee.employee_name
HAVING cnt > 1
This would count each employee who is linked both software or hardware department. Or you can leave WHERE clause away to get all employees working more than one departments.
What is my problem?
There is no dname column in your tables.
You can simplify the problem as you don't need the department table since the working table contains the department id in the did column.
Then you need to GROUP BY each employee and find those HAVING a COUNT of two DISTINCT department ids:
SELECT MAX(e.employee_name)
FROM employee e
INNER JOIN working w
ON e.employee_id = w.eid
GROUP BY e.employee_id
HAVING COUNT(DISTINCT w.did) = 2
If you want to consider only the software and hardware departments then:
SELECT MAX(e.employee_name)
FROM employee e
INNER JOIN working w
ON e.employee_id = w.eid
INNER JOIN department d
ON w.did = d.department_id
WHERE d.department_name IN ('software', 'hardware')
GROUP BY e.employee_id
HAVING COUNT(DISTINCT w.did) = 2
You can easily obtain employees who work in one specific department:
select *
from Employee e inner join
Working w on e.employee_id = w.eid inner join
Department d on w.did = d.department_id
where d.name = 'software'
Now ambiguity cames. If you want to get all employees work either in software or in hardware:
-- Employees who work at either software or hardware or both departments
select *
from Employee e inner join
Working w on e.employee_id = w.eid inner join
Department d on w.did = d.department_id
where d.name = in ('software', 'hardware')
If you want to get employees who works in both software and hardware departments:
-- Employees who work in both hardware and software deparments simultaneously
select *
from Employee e inner join
Working w on e.employee_id = w.eid inner join
Department d on w.did = d.department_id
where d.name = 'software'
intersect
select *
from Employee e inner join
Working w on e.employee_id = w.eid inner join
Department d on w.did = d.department_id
where d.name = 'hardware'
Related
I am trying to figure out how to "find all employees' name who has a manager that lives in the same city as them." For this problem, we have two tables. We need to make a query.
"employee"
The employee table that we can refer to has both normal employees and managers
employeeid
name
projectid
city
1
jeff
1
new york
2
larry
1
new york
3
Linda
2
detroit
4
tom
2
LA
"Managertable"
Our manager table which we can refer to with mangerid = employeeid
projectid
mangerid
1
2
2
3
Right now I have found a way to get just the employees and filter out the managers, but now I am trying to figure out the next step to get to the comparison of managers and employees. Would this just be another subquery?
SELECT name
FROM employee e
WHERE employeeid not in(
SELECT mangerid
FROM Managertable pm
INNER JOIN employee e
ON pm.mangerid= e.employeeid);
Expected result :
employee name
jeff
I think the easient way to achieve this would be like this:
SELECT
e.*
FROM employee e
inner join Managertable mt on e.projectid = mt.projectid
inner join employee manager on mt.mangerid = manager.employeeid
WHERE
e.city = manager.city
and e.employeeid <> manager.employeeid;
One approach is a correlated subquery in which we look up the employee's manager's city.
select e.name
from employee e
where city =
(
select m.city
from managertable mt
join employee m on m.employeeid = mt.managerid
where mt.projectid = e.projectid
and m.employeeid <> e.employeeid
);
The same thing can be written with an EXISTS clause, if you like that better.
Based off the table structure you're showing, something like this might work
First find the employee ids of employees who have managers in the same city, then join it back on employee to retrieve all data from the table
;WITH same_city AS (
SELECT DISTINCT e.employeeid
FROM employee AS e
INNER JOIN managertable AS mt ON e.projectid = mt.projectid
INNER JOIN employee AS m ON mt.managerid = e.employeeid
WHERE e.city = m.city
)
SELECT e.*
FROM employee
INNER JOIN same_city AS sc ON e.employeeid = sc.employeeid
I don't see how projectid is relevant in your question because you didn't mention that as a requirement or restriction. Here's a method using a CTE to get the managers and their cities, then join to it to find employees who live in the same city as a manager.
with all_managers as (
select distinct m.managerid, e.city
from manager m
join employee e
on m.managerid = e.employeeid
)
select e.name
from employee e
join all_managers a
on e.city = a.city
and e.employeeid <> a.managerid;
name
jeff
But it you want us to assume that an employee reports to only that manager as listed in the projectid, then here's a modification to ensure that is met:
with all_managers as (
select distinct m.managerid, e.city, e.projectid
from manager m
join employee e
on m.managerid = e.employeeid
)
select e.name
from employee e
join all_managers a
on e.city = a.city
and e.projectid = a.projectid
and e.employeeid <> a.managerid;
View on DB Fiddle
You just need two joins:
one between "managers" and "employees" to gather managers information
one between "managers" and "employees" to gather employees information with respect to the manager's projectid and city.
SELECT employees.name
FROM managers
INNER JOIN employees managers_info
ON managers.mangerid = managers_info.employeeid
INNER JOIN employees
ON managers.projectid = employees.projectid
AND managers_info.employeeid <> employees.employeeid
AND managers_info.city = employees.city
This gets me the number of employees per department
SELECT department, COUNT(idEmployees) empCount
FROM employees
GROUP BY department;
You could join on departments to get the buildingID, and then group by that:
SELECT buildingID, COUNT(*)
FROM employees e
JOIN department d ON e.department = d.departmentID
GROUP BY buildingID
i think this is the solution
select d.buildingID, count(e.ID)
from department d
inner join employees e
on d.departmentID = e.department
group by d.buildingID
This question can be answered in two ways. Firstly, if all employees of a department must be attached in a building. Secondly few employees of a department are attached to a building.
For one where employees are attached to a building it's mandatory
SELECT d.buildingId
, COUNT(e.id) count_employee
FROM departments d
INNER JOIN employees e
ON d.departmentid = e.department
GROUP BY d.buildingId
For second where employees are attached to a building but it's optional. In that case LEFT JOIN is used.
SELECT b.buildingId
, COALESCE(t.count_employee, 0) count_employee
FROM building b
LEFT JOIN (SELECT d.buildingId
, COUNT(e.id) count_employee
FROM departments d
INNER JOIN employees e
ON d.departmentid = e.department
GROUP BY d.buildingId) t
ON b.buildingId = t.buildingId
If a building is attached with multiple departments and one employee is assigned with multiple departments then count building wise same employee is only onetime not multiple times. In that case DISTINCT keyword is used inside COUNT().
SELECT d.buildingId
, COUNT(DISTINCT e.id) count_employee
FROM departments d
INNER JOIN employees e
ON d.departmentid = e.department
GROUP BY d.buildingId
Is it possible to retrieve the number of departments and the number of employees working in that department if given the location Id? Location_id column is in the departments table. Employees and Department share department_id column
Cannot get the correct result with this query:
select
count(E.employee_id), count(D.DEPARTMENT_ID)
from
employees e
join
departments d on (e.department_id = d.department_id)
where
D.LOCATION_ID = 1700;
I suspect you need distinct when counting the departments:
SELECT
COUNT( E.employee_id )
, COUNT(DISTINCT d.department_id )
FROM employees e
JOIN departments d ON e.department_id = d.department_id
WHERE d.location_id = 1700;
or, you need to group by department:
SELECT
COUNT( E.employee_id )
, d.department_id
FROM employees e
JOIN departments d ON e.department_id = d.department_id
WHERE d.location_id = 1700
GROUP BY d.department_id;
I have a table named employee. It includes managers and employees. There is also birthdate column and i need to find "older employees than THEIR managers". How can I do it? Could you give me some clue?
You could use an inner join to the same table:
select e.*
from employees e
inner join employees m
on e.managerid = m.id
and e.birthdate < m.birthdate
or exists() in the where:
select e.*
from employees e
where exists (
select 1
from employees m
where e.managerid = m.id
and e.birthdate < m.birthdate
)
If your employees structure is something like:
id,
name,
birthdate,
manager_id
This could be a solution:
select id, name
from employees e
where
exists
(select *
from employees m
where
m.id = e.manager_id and
m.birthdate > e.birthdate);
Problem:
Create a list of department names, the manager id,
manager name (employee last name) of that department, and the average salary in each
department.
SELECT d.department_name, d.manager_id, AVG(e.salary)
FROM employees e
INNER JOIN departments d ON (e.department_id = d.department_id)
GROUP BY d.department_name, d.manager_id;
And it works nice, but when I add the e.last_name, I get all the last names from employees table.
I do believe the answer to be out here and not quite far, although out of my reach at this point.
In order to pull the name of the manager, you need to join employees again, this time on d.manager_id:
SELECT d.department_name, d.manager_id, m.name, AVG(e.salary)
FROM employees e
INNER JOIN departments d ON (e.department_id = d.department_id)
LEFT OUTER JOIN employees m ON (m.employee_id = d.manager_id)
GROUP BY d.department_name, d.manager_id, m.name;
The kind of join (inner or outer) is not essential here, because you group by d.manager_id.
It looks like you need to join d.manager_id to employees again to get the managers last_name:
SELECT d.department_name, d.manager_id, e2.last_name, AVG(e.salary)
FROM employees e
INNER JOIN departments d ON e.department_id = d.department_id
INNER JOIN employees e2 ON d.manager_id = e2.employee_id
GROUP BY d.department_name, d.manager_id, e2.last_name