SQL Exercise: List the employee with their grade for the grade 4
SQL employee Database: Exercise-64 with Solution
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64. From the following tables, write a SQL query to determine which employees have a grade of 4 and a salary between the minimum and maximum. Return all information of each employees and their grade and salary related details.
Pictorial Presentation:

Sample table: employees
Sample table: salary_grade
Sample Solution:
SELECT *
FROM employees e,
salary_grade s
WHERE e.salary BETWEEN s.min_sal AND s.max_sal
AND s.grade = 4;
Sample Output:
emp_id | emp_name | job_name | manager_id | hire_date | salary | commission | dep_id | grade | min_sal | max_sal --------+----------+----------+------------+------------+---------+------------+--------+-------+---------+--------- 66928 | BLAZE | MANAGER | 68319 | 1991-05-01 | 2750.00 | | 3001 | 4 | 2101 | 3100 67832 | CLARE | MANAGER | 68319 | 1991-06-09 | 2550.00 | | 1001 | 4 | 2101 | 3100 65646 | JONAS | MANAGER | 68319 | 1991-04-02 | 2957.00 | | 2001 | 4 | 2101 | 3100 67858 | SCARLET | ANALYST | 65646 | 1997-04-19 | 3100.00 | | 2001 | 4 | 2101 | 3100 69062 | FRANK | ANALYST | 65646 | 1991-12-03 | 3100.00 | | 2001 | 4 | 2101 | 3100 (5 rows)
Explanation:
The said query in SQL that retrieves all the employees whose salary falls within the salary range of grade 4 and displays all columns for each matching row from two tables, 'employees' and 'salary_grade' .
The WHERE clause first checks whether the employee's salary falls within the range of minimum and maximum salaries defined for grade 4 in the "salary_grade" table.
It then retrieve data only for employees who belong to grade 4.
Relational Algebra Expression:

Relational Algebra Tree:

Practice Online
Sample Database: employee

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Previous SQL Exercise: List the employees who are senior to their own manager.
Next SQL Exercise: List the employees in given department, joined after 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;
Database: PostgreSQL
Ref: https://bit.ly/3AfYwZI
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