SQL Exercise: List the employees who joined before 1991
SQL employee Database: Exercise-12 with Solution
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12. From the following table, write a SQL query to find those employees who joined before 1991. Return complete information about the employees.
Sample table: employees
Pictorial Presentation:

Sample Solution:
SELECT *
FROM employees
WHERE hire_date<('1991-1-1');
Sample Output:
emp_id | emp_name | job_name | manager_id | hire_date | salary | commission | dep_id --------+----------+----------+------------+------------+--------+------------+-------- 63679 | SANDRINE | CLERK | 69062 | 1990-12-18 | 900.00 | | 2001 (1 row)
Explanation:
The said query in SQL that selects all columns from the 'employees' table where the hire_date is before January 1, 1991.
The "WHERE" clause filters the results for all employees whose hire date is earlier than January 1, 1991.
Relational Algebra Expression:

Relational Algebra Tree:

Practice Online
Sample Database: employee

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Previous SQL Exercise: List the employees not in department 2001.
Next SQL Exercise: Average salaries of the employees works as ANALYST.
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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;
Database: PostgreSQL
Ref: https://bit.ly/3AfYwZI
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