AdventureWorks Database: Find the departments that each have more than 15 employees
SQL Query - AdventureWorks: Exercise-94 with Solution
94. From the following tables write a SQL query to return the departments of a company that each have more than 15 employees.
Sample table: humanresources.employeebusinessentityid|nationalidnumber|loginid |jobtitle |birthdate |maritalstatus|gender|hiredate |salariedflag|vacationhours|sickleavehours|currentflag|rowguid |modifieddate |organizationnode| ----------------+----------------+----------------------------+----------------------------------------+----------+-------------+------+----------+------------+-------------+--------------+-----------+------------------------------------+-----------------------+----------------+ 1|295847284 |adventure-works\ken0 |Chief Executive Officer |1969-01-29|S |M |2009-01-14|true | 99| 69|true |f01251e5-96a3-448d-981e-0f99d789110d|2014-06-30 00:00:00.000|/ | 2|245797967 |adventure-works\terri0 |Vice President of Engineering |1971-08-01|S |F |2008-01-31|true | 1| 20|true |45e8f437-670d-4409-93cb-f9424a40d6ee|2014-06-30 00:00:00.000|/1/ | 3|509647174 |adventure-works\roberto0 |Engineering Manager |1974-11-12|M |M |2007-11-11|true | 2| 21|true |9bbbfb2c-efbb-4217-9ab7-f97689328841|2014-06-30 00:00:00.000|/1/1/ | 4|112457891 |adventure-works\rob0 |Senior Tool Designer |1974-12-23|S |M |2007-12-05|false | 48| 80|true |59747955-87b8-443f-8ed4-f8ad3afdf3a9|2014-06-30 00:00:00.000|/1/1/1/ | 5|695256908 |adventure-works\gail0 |Design Engineer |1952-09-27|M |F |2008-01-06|true | 5| 22|true |ec84ae09-f9b8-4a15-b4a9-6ccbab919b08|2014-06-30 00:00:00.000|/1/1/2/ | 6|998320692 |adventure-works\jossef0 |Design Engineer |1959-03-11|M |M |2008-01-24|true | 6| 23|true |e39056f1-9cd5-478d-8945-14aca7fbdcdd|2014-06-30 00:00:00.000|/1/1/3/ | 7|134969118 |adventure-works\dylan0 |Research and Development Manager |1987-02-24|M |M |2009-02-08|true | 61| 50|true |4f46deca-ef01-41fd-9829-0adab368e431|2014-06-30 00:00:00.000|/1/1/4/ | 8|811994146 |adventure-works\diane1 |Research and Development Engineer |1986-06-05|S |F |2008-12-29|true | 62| 51|true |31112635-663b-4018-b4a2-a685c0bf48a4|2014-06-30 00:00:00.000|/1/1/4/1/ | 9|658797903 |adventure-works\gigi0 |Research and Development Engineer |1979-01-21|M |F |2009-01-16|true | 63| 51|true |50b6cdc6-7570-47ef-9570-48a64b5f2ecf|2014-06-30 00:00:00.000|/1/1/4/2/ | 10|879342154 |adventure-works\michael6 |Research and Development Manager |1984-11-30|M |M |2009-05-03|true | 16| 64|true |eaa43680-5571-40cb-ab1a-3bf68f04459e|2014-06-30 00:00:00.000|/1/1/4/3/ | -- more --
Sample Solution:
-- Selecting the job title and counting employees in each designation
SELECT jobtitle,
COUNT(businessentityid) AS EmployeesInDesig
-- From the employee table in the HumanResources schema
FROM humanresources.employee e
-- Grouping the results by job title
GROUP BY jobtitle
-- Filtering the results to include only those with more than 15 employees
HAVING COUNT(businessentityid) > 15;
Explanation:
- This SQL query retrieves the job titles and counts the number of employees in each job title designation from the employee table in the HumanResources schema.
- The COUNT(businessentityid) function is used to count the number of employees for each job title group.
- The results are grouped by the job title using the GROUP BY clause.
- The HAVING clause is used to filter the results to include only those job titles with more than 15 employees.
Sample Output:
jobtitle |employeesindesig| ----------------------------+----------------+ Production Technician - WC30| 25| Production Technician - WC60| 26| Production Technician - WC50| 26| Production Technician - WC20| 22| Production Technician - WC10| 17| Production Technician - WC40| 26|
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