﻿ SQL: List employees id, salary, and commission

# SQL Exercise: List employees id, salary, and commission

## SQL employee Database: Exercise-9 with Solution

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9. From the following table, write a SQL query to find the employee ID, salary, and commission of all the employees.

Sample table: employees

Pictorial Presentation:

Sample Solution:

``````SELECT emp_id,
salary,
commission
FROM employees;
``````

Sample Output:

``` emp_id | salary  | commission
--------+---------+------------
68319 | 6000.00 |
66928 | 2750.00 |
67832 | 2550.00 |
65646 | 2957.00 |
67858 | 3100.00 |
69062 | 3100.00 |
63679 |  900.00 |
64989 | 1700.00 |     400.00
65271 | 1350.00 |     600.00
66564 | 1350.00 |    1500.00
68454 | 1600.00 |       0.00
68736 | 1200.00 |
69000 | 1050.00 |
69324 | 1400.00 |
(14 rows)
```

Explanation:

The given query in SQL that selects the emp_id, salary, and commission columns from the 'employees' table.

Relational Algebra Expression:

Relational Algebra Tree:

## Practice Online

Sample Database: employee

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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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