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Databases SQL Reporting by Example Day 3: Getting Good at Grouping Day 3: Final Exam

Sujal khatiwada
Sujal khatiwada
1,525 Points
Posted by Sujal khatiwada
Sujal khatiwada
1,525 Points

Quiz problem

ELECT MANAGERS.FIRST_NAME, MANAGERS.LAST_NAME FROM MANAGERS JOIN EMPLOYEES ON EMPLOYEES.MANAGER_ID = MANAGERS.ID GROUP BY MANAGERS.ID .........................

1 Answer

Brendan Whiting
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Brendan Whiting
Front End Web Development Techdegree Graduate 84,738 Points
Brendan Whiting
.a{fill-rule:evenodd;}techdegree seal-36
Brendan Whiting
Front End Web Development Techdegree Graduate 84,738 Points 
SELECT MANAGERS.FIRST_NAME, MANAGERS.LAST_NAME
FROM MANAGERS
JOIN EMPLOYEES ON EMPLOYEES.MANAGER_ID = MANAGERS.ID
GROUP BY MANAGERS.ID HAVING COUNT(1) > 5;

"HAVING" is like "WHERE" but to use with aggregate functions. COUNT is the aggregate function. It's a little bit confusing the order of the clauses. We GROUP BY by MANAGERS.ID column from the EMPLOYEES table, so it will count the number of employees that have the same manager and only show the records where the count is greater than 5.