SQL Exercise: Departments where more than two employees are working
SQL JOINS: Exercise-29 with Solution
From the following tables write a SQL query to find the names of departments where more than two employees are employed. Return dpt_name.
Sample table: emp_details
SELECT emp_department.dpt_name FROM emp_details INNER JOIN emp_department ON emp_dept =dpt_code GROUP BY emp_department.dpt_name HAVING COUNT(*) > 2;
Output of the Query:
dpt_name Finance IT HR
The said SQL query is selecting the 'dpt_name' column from the emp_department table.
This code is joining the emp_details table with the emp_department table on the matching value in the 'emp_dept' column of the emp_details table and the 'dpt_code' column of the emp_department table.
The query also use GROUP BY clause that groups the result-set by one or more columns.
The HAVING clause is used in conjunction with the GROUP BY clause. It is used to filter groups based on a specified condition, in this case, the condition is COUNT(*) > 2, which means it will only return the groups that have more than 2 rows.
It returns the name of the department whose count of employees is more than 2.
Relational Algebra Expression:
Relational Algebra Tree:
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Previous SQL Exercise: Find employees and departments with a given budget.
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SQL: Tips of the Day
Grouped LIMIT in PostgreSQL: Show the first N rows for each group?
db=# SELECT * FROM xxx; id | section_id | name ----+------------+------ 1 | 1 | A 2 | 1 | B 3 | 1 | C 4 | 1 | D 5 | 2 | E 6 | 2 | F 7 | 3 | G 8 | 2 | H (8 rows)
I need the first 2 rows (ordered by name) for each section_id, i.e. a result similar to:
id | section_id | name ----+------------+------ 1 | 1 | A 2 | 1 | B 5 | 2 | E 6 | 2 | F 7 | 3 | G (5 rows)
PostgreSQL v9.3 you can do a lateral join
select distinct t_outer.section_id, t_top.id, t_top.name from t t_outer join lateral ( select * from t t_inner where t_inner.section_id = t_outer.section_id order by t_inner.name limit 2 ) t_top on true order by t_outer.section_id;
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