SQL Exercise: List the employees at a given place for over 10 years
SQL employee Database: Exercise-60 with Solution
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60. From the following table, write a SQL query to search for employees working in PERTH or MELBOURNE and month part of their achieved experience is more than 10. Return employee ID, employee name, department ID, salary, and department location.
Sample table: employees
Sample table: department
SELECT e.emp_id, e.emp_name, e.dep_id, e.salary, d.dep_location FROM employees e, department d WHERE e.dep_id = d.dep_id AND d.dep_location IN ('PERTH', 'MELBOURNE') AND EXTRACT(MONTH FROM age(CURRENT_DATE, hire_date)) > 10;
emp_id | emp_name | dep_id | salary | dep_location --------+----------+--------+---------+-------------- 64989 | ADELYN | 3001 | 1700.00 | PERTH 65271 | WADE | 3001 | 1350.00 | PERTH (2 rows)
The given query in SQL that selects the employee ID, name, department ID, salary, and department location of employees who work in the departments located in either 'PERTH' or 'MELBOURNE' and whose hire date is more than 10 months ago from the current date.
The statement joins the 'employees' and 'department' tables based on the common column dep_id.
The "WHERE" clause restricts the results to employees whose department location is either 'PERTH' or 'MELBOURNE' using the "IN" operator, and it further filters the results to employees whose hire date is more than 10 months ago from the current date using the "age" and "EXTRACT" functions. The "age" function calculates the difference between the current date and the hire date, and the "EXTRACT" function extracts the month value from the result of the "age" function.
Sample Database: employee
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Previous SQL Exercise: Analysts or managers with a salary range, no commission.
Next SQL Exercise: Employees with location, salary range, and joined in 91.
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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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