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SQL Exercise: Display all the information of the employees

SQL employee Database: Exercise-1 with Solution

[An editor is available at the bottom of the page to write and execute the scripts.]

1. From the following table return complete information about the employees.

Sample table: employees


Pictorial Presentation:

SQL exercises on employee Database: Display all the information of the employees

Sample Solution:

SELECT * FROM employees;

Sample Output:

 emp_id | emp_name | job_name  | manager_id | hire_date  | salary  | commission | dep_id
--------+----------+-----------+------------+------------+---------+------------+--------
  68319 | KAYLING  | PRESIDENT |            | 1991-11-18 | 6000.00 |            |   1001
  66928 | BLAZE    | MANAGER   |      68319 | 1991-05-01 | 2750.00 |            |   3001
  67832 | CLARE    | MANAGER   |      68319 | 1991-06-09 | 2550.00 |            |   1001
  65646 | JONAS    | MANAGER   |      68319 | 1991-04-02 | 2957.00 |            |   2001
  67858 | SCARLET  | ANALYST   |      65646 | 1997-04-19 | 3100.00 |            |   2001
  69062 | FRANK    | ANALYST   |      65646 | 1991-12-03 | 3100.00 |            |   2001
  63679 | SANDRINE | CLERK     |      69062 | 1990-12-18 |  900.00 |            |   2001
  64989 | ADELYN   | SALESMAN  |      66928 | 1991-02-20 | 1700.00 |     400.00 |   3001
  65271 | WADE     | SALESMAN  |      66928 | 1991-02-22 | 1350.00 |     600.00 |   3001
  66564 | MADDEN   | SALESMAN  |      66928 | 1991-09-28 | 1350.00 |    1500.00 |   3001
  68454 | TUCKER   | SALESMAN  |      66928 | 1991-09-08 | 1600.00 |       0.00 |   3001
  68736 | ADNRES   | CLERK     |      67858 | 1997-05-23 | 1200.00 |            |   2001
  69000 | JULIUS   | CLERK     |      66928 | 1991-12-03 | 1050.00 |            |   3001
  69324 | MARKER   | CLERK     |      67832 | 1992-01-23 | 1400.00 |            |   1001
(14 rows)

Explanation:

The said query in SQL that retrieves all the data from the 'employees' table. The asterisk (*) represents a wildcard character that include all the columns of the table in the result set.

Relational Algebra Expression:

Relational Algebra Expression: Display all the information of the employees.

Relational Algebra Tree:

Relational Algebra Tree: Display all the information of the employees.

Practice Online


Sample Database: employees

employee database structure

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Previous SQL Exercise: SQL Exercises of Employee Database
Next SQL Exercise: Find the salaries of all employees.

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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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