SQL Exercise: Employees who earn more than 100 as daily salary
SQL employee Database: Exercise-29 with Solution
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29. From the following table, write a SQL query to find out which employees earn more than 100 per day as a salary. Return employee ID, employee name, salary, and experience.
Sample table: employees
Pictorial Presentation:

Sample Solution:
SELECT emp_id,
emp_name,
salary,
age(CURRENT_DATE, hire_date) "Experience"
FROM employees
WHERE (salary/30)>100;
Sample Output:
emp_id | emp_name | salary | Experience --------+----------+---------+------------------------- 68319 | KAYLING | 6000.00 | 26 years 2 mons 12 days 67858 | SCARLET | 3100.00 | 20 years 9 mons 11 days 69062 | FRANK | 3100.00 | 26 years 1 mon 27 days (3 rows)
Explanation:
The said query in SQL that selects the "emp_id", "emp_name", "salary", and experience (in years and months) from the 'employees' table and filters the results based on a calculation using the "salary" column.
The experience of the employee in years and months, calculated using the "age" function applied to the "hire_date" column.
The WHERE clause filters the results to include only those employees whose daily salary (calculated by dividing the monthly salary by 30) is greater than 100.
Practice Online
Sample Database: employee

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Previous SQL Exercise: Experiences of all employees working with Manger 68319.
Next SQL Exercise: Employees retiring after 31-Dec-99 after 8 years.
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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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