I need to find employee with lowest salary- first and last name, gender and department.
I tried with this but I have problems adding department name.
SELECT first_name, last_name, gender, salary FROM salaries
JOIN employees using(emp_no) where order by salary desc limit 0,1;
Any help?
schema-link
This should work, you are missing a couple of joins among other things (like the department name in the field list):
SELECT e.first_name, e.last_name, e.gender, s.salary, d.dept_name
FROM salaries AS s
INNER JOIN employees AS e ON s.emp_no=e.emp_no
INNER JOIN dept_emp AS de ON de.emp_no=e.emp_no
INNER JOIN departments AS d ON d.dept_no=de.dept_no
ORDER BY s.salary ASC limit 0, 1
Related
I have three different tables salaries, dept_managers(Department managers), and departments
the first two tables are connected through emp_no(Employee number), While the other two tables are connected through dept_no(the department number),
I want to have a query that gives me the average amount of salaries that each department has.
I tried this query but It did not work.
SELECTd.dept_name,dm.dept_noFROM departments d
JOIN dept_manager dm ON dm.dept_no = d.dept_no
JOIN(
SELECT AVG(salary) AS Average_salary
FROM salaries s
GROUP BY salary)
ON d.emp_no = dm.emp_no
;
maybe I'm missing something here but wouldn't you just join department to employees and then join salaries to employees and get the avg salary grouped by department?
select d.dept_name, avg(s.salary)
from departments d join employees e on (e.dept_no = d.dept_no)
join salaries s on (s.emp_no = e.emp_no)
group by d.dept_name
I have 2 tables in my database.
Table 1. employee
id
name
department_id
Table 2. department
id
name
What will be the query to fetch all employees with their department?
So I have written this query
SELECT employee.name
, department.name
FROM employee
JOIN department
ON employee.department_id = department.id
And this seems to be correct but I am not able to write a query if I want to fetch only the department that has the highest number of employees. How can I achieve this?
To guarantee just one department...
SELECT
*
FROM
department
WHERE
id = (SELECT department_id
FROM employee
GROUP BY department_id
ORDER BY COUNT(*) DESC
LIMIT 1
)
Note, if two departments are tied with joint maximum employees, this will still only select One of them (arbitrarily chosen, potentially different each time).
To handle ties, you could do the following...
SELECT *
FROM department
WHERE id IN (SELECT department_id
FROM employee
GROUP BY department_id
HAVING COUNT(*) = (SELECT COUNT(*)
FROM employee
GROUP BY department_id
ORDER BY COUNT(*) DESC
LIMIT 1
)
)
This is a real pain to handle in MySQL. Here is one option:
SELECT d1.id, d1.name
FROM department d1
INNER JOIN employee e1
ON d1.id = e1.department_id
GROUP BY d1.id, d1.name
HAVING COUNT(*) = (SELECT COUNT(*) FROM department d2 INNER JOIN employee e2
ON d2.id = e2.department_id
GROUP BY d2.id ORDER BY COUNT(*) DESC LIMIT 1);
Note that if you were using a database with analytic function support, such as SQL Server, then the problem gets much easier:
SELECT id, name
FROM
(
SELECT d.id, d.name, DENSE_RANK() OVER (ORDER BY COUNT(*) DESC) dr
FROM department d
INNER JOIN employee e
ON d.id = e.department_id
GROUP BY d.id, d.name
) t
WHERE t.dr = 1;
This question can be solved in multiple ways:
Using sub query
SELECT name FROM department
WHERE id IN
(SELECT department_id FROM employee HAVING COUNT(department_id)
IN
(SELECT MAX(COUNT(department_id)) FROM employee) GROUP BY department_id)
Using Join
SELECT name FROM employee e
INNER JOIN
department d ON e.department_id = d.id
HAVING COUNT(e.department_id)
IN
(SELECT MAX(COUNT(department_id)) from employee) group by department_id)
first check the column related two types have same name, same data type and the use subquery
SELECT name
FROM department
WHERE id IN (
SELECT department_id
FROM employees
HAVING COUNT(department_id) IN (
SELECT MAX(COUNT(dept_id))
FROM employees
)
GROUP BY department_id
)
Your query should work for the first question.
for the second, You can use this. The sub query would give you the dept Id for the most employees, which the outer query would give additional details for.
select * from department where department_id in
(select limit 1 Employee.department_id from Employee group by department_id
order by count(Employee.name) desc)
I have 2 tables employees(id, first_name, last_name, salary, department_id_ and department(id, name) and I want to show number of employees in each department.
I have this question here:
SELECT department.name, COUNT(*) AS 'employees_number'
FROM department
LEFT JOIN employees
ON employees.department_id = department.id
GROUP BY department.id, department.name;
But for some reason, in departments where I have no people, it shows a number of employees as 1. Any idea why this is happening?
With an outer join you still get a result row when no match in the outer table is found. Only all employee column values are null then.
So rather than count the records, you want to count matched records, i.e. where an employee was found and its data is not null. So Count a column in the employee table (nulls are not counted, when counting a column or expression). E.g. use COUNT(e.department_id) or COUNT(e.id):
SELECT d.name, COUNT(e.id) AS employees_number
FROM department d
LEFT JOIN employees e ON e.department_id = d.id
GROUP BY d.id, d.name;
What I prefer though, is to aggregate/count before joining. The query looks a bit more complicated, but is less prone to errors on future query changes:
SELECT d.name, COALESCE(e.how_many, 0) AS employees_number
FROM department d
LEFT JOIN
(
SELECT department_id, COUNT(*) AS how_many
FROM employees
GROUP BY department_id
) e ON e.department_id = d.id;
As it's one aggregated column only you want, you can move the subquery to your SELECT clause and get thus a simpler query:
SELECT
d.name,
(
SELECT COUNT(*)
FROM employees e
WHERE e.department_id = d.id
) AS employees_number
FROM department d;
Using SUM instead of COUNT also can give you what you want:
SELECT
department.name,
SUM(CASE WHEN employees.id IS NOT NULL THEN 1 ELSE 0 END) AS 'employees_number'
FROM department
LEFT JOIN employees
ON employees.department_id = department.id
GROUP BY department.id, department.name;
SQL Fiddle:
http://sqlfiddle.com/#!9/8b8976/1
select department.name, count(employee.id) as co from
department left join employee on
department.id = employee.dept_id group by department.name
order by co desc, department.name asc
how to write a query to display the name of the departments that have the least student count. Sort the result based on department name in ascending order
select d.department_name from
(select dd.department_name, count(di.department_id) as id from student di
join department dd on di.department_id=dd.department_id group by dd.department_name) d,
(select min(count(*)) as new from student group by department_id) d2
where d.id=d2.new;
select d.department_name from Department d, Student s where
d.department_id = s.department_id
group by d.department_name
having count(s.student_id)<=all
(select count(s.student_id) from Department d, Student s where
d.department_id = s.department_id
group by d.department_name)
order by department_name;
Try this.
select d.department_id, d.department_name
from Department d
join Student s on d.department_id = s.department_id
group by d.department_id
having count(s.student_id) = (select min(count(s2.student_id))
from student s2
join department d2
on s2.department_id = d2.department_id
group by d2.department_id)
order by d.department_name
You must join the 2 tables to have the needed information.
You'll also have to group them by the selected information such that you can count the students.
And lastly, you place the condition. Needing a subquery to retrieve the minimum number of students.
select department_name
from Department
join Student
on Department.department_id=Student.department_id
having count(*) in
( select min(count(*)) from Student group by department_id)
group by Department.department_id,department_name
order by department_name asc;
Requirements: using one query only to show each department with their employee counts (two tables, department and employee); order by number of employees decreasing;(for departments with no employees, still show them); for departments with the same size, order by department name alphabetically. My solution not showing the departments with no employees, also, it is not showing alphabetically order of departments when same number of employees:
SELECT d.DEPT_ID,
DEPT_NAME, COUNT(s.STUDENT_ID) as numStudents
FROM Departments d, Students s
WHERE d.DEPT_ID = s.DEPT_ID
GROUP BY d.DEPT_ID
ORDER BY numStudents DESC;
Use a LEFT JOIN instead of the cross join you are currently doing.
SELECT d.DEPT_ID,
d.DEPT_NAME, COALESCE(COUNT(s.STUDENT_ID),0) as numStudents
FROM Departments d,
LEFT JOIN Students s
ON s.dept_id = d.dept_id
GROUP BY d.DEPT_ID, d.dept_name
ORDER BY numStudents DESC, DEPT_NAME;
In order to show departments with no students, you have to use LEFT JOIN, not a cross product. And to order by name when they have the same number of students, you need to add that to your ORDER BY clause.
SELECT d.DEPT_ID, DEPT_NAME, IFNULL(COUNT(s.STUDENT_ID), 0) AS numStudents
FROM Departments AS d
LEFT JOIN Students AS s ON d.DEPT_ID = s.DEPT_ID
GROUP BY d.DEPT_ID
ORDER BY numStudents DESC, DEPT_NAME