SQL Exercise: Employees earn over 11000 or 7th digit in their phone
SQL SORTING and FILTERING on HR Database: Exercise-18 with Solution
18. From the following table, write a SQL query to find those employees who earn above 11000 or the seventh character in their phone number is 3. Sort the result-set in descending order by first name. Return full name (first name and last name), hire date, commission percentage, email, and telephone separated by '-', and salary.
Sample table : employees
Sample Solution:
SELECT first_name ||' '||last_name AS Full_Name, hire_date ,
commission_pct, email ||' - '||phone_number AS Contact_Details, salary
FROM employees
WHERE salary > 11000
OR phone_number LIKE '______3%'
ORDER BY first_name DESC;
Sample Output:
full_name | hire_date | commission_pct | contact_details | salary
-------------------+------------+----------------+--------------------------------+----------
William Gietz | 2002-06-07 | 0.00 | WGIETZ - 515.123.8181 | 8300.00
Valli Pataballa | 2006-02-05 | 0.00 | VPATABAL - 590.423.4560 | 4800.00
Susan Mavris | 2002-06-07 | 0.00 | SMAVRIS - 515.123.7777 | 6500.00
Steven King | 2003-06-17 | 0.00 | SKING - 515.123.4567 | 24000.00
Shelley Higgins | 2002-06-07 | 0.00 | SHIGGINS - 515.123.8080 | 12000.00
Shanta Vollman | 2005-10-10 | 0.00 | SVOLLMAN - 650.123.4234 | 6500.00
Payam Kaufling | 2003-05-01 | 0.00 | PKAUFLIN - 650.123.3234 | 7900.00
Pat Fay | 2005-08-17 | 0.00 | PFAY - 603.123.6666 | 6000.00
Neena Kochhar | 2005-09-21 | 0.00 | NKOCHHAR - 515.123.4568 | 17000.00
Nancy Greenberg | 2002-08-17 | 0.00 | NGREENBE - 515.124.4569 | 12000.00
Michael Hartstein | 2004-02-17 | 0.00 | MHARTSTE - 515.123.5555 | 13000.00
Matthew Weiss | 2004-07-18 | 0.00 | MWEISS - 650.123.1234 | 8000.00
Lisa Ozer | 2005-03-11 | 0.25 | LOZER - 011.44.1343.929268 | 11500.00
Lex De Haan | 2001-01-13 | 0.00 | LDEHAAN - 515.123.4569 | 17000.00
Kevin Mourgos | 2007-11-16 | 0.00 | KMOURGOS - 650.123.5234 | 5800.00
Karen Partners | 2005-01-05 | 0.30 | KPARTNER - 011.44.1344.467268 | 13500.00
John Russell | 2004-10-01 | 0.40 | JRUSSEL - 011.44.1344.429268 | 14000.00
Jennifer Whalen | 2003-09-17 | 0.00 | JWHALEN - 515.123.4444 | 4400.00
Hermann Baer | 2002-06-07 | 0.00 | HBAER - 515.123.8888 | 10000.00
Diana Lorentz | 2007-02-07 | 0.00 | DLORENTZ - 590.423.5567 | 4200.00
David Austin | 2005-06-25 | 0.00 | DAUSTIN - 590.423.4569 | 4800.00
Bruce Ernst | 2007-05-21 | 0.00 | BERNST - 590.423.4568 | 6000.00
Alexander Hunold | 2006-01-03 | 0.00 | AHUNOLD - 590.423.4567 | 9000.00
Alberto Errazuriz | 2005-03-10 | 0.30 | AERRAZUR - 011.44.1344.429278 | 12000.00
Adam Fripp | 2005-04-10 | 0.00 | AFRIPP - 650.123.2234 | 8200.00
(25 rows)
Code Explanation:
The said query in SQL that retrieves the first name and last name concatenated into one column, "Full_Name", using the "||" operator, the hire date, commission percentage, email and phone number concatenated into one column "Contact_Details", and the salary from the 'employees' table.
The result is filtered to only include rows where the value in the "salary" column is greater than 11000 or where the value in the "phone_number" column starts with a 6-digit number ending with "3".
It is sorting the result based on the "first_name" column values that are used to sort the results in descending order.
Practice Online
Query Visualization:
Duration:
Rows:
Cost:
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SQL: Tips of the Day
Difference between natural join and inner join
One significant difference between INNER JOIN and NATURAL JOIN is the number of columns returned-
Consider:
TableA TableB +------------+----------+ +--------------------+ |Column1 | Column2 | |Column1 | Column3 | +-----------------------+ +--------------------+ | 1 | 2 | | 1 | 3 | +------------+----------+ +---------+----------+
The INNER JOIN of TableA and TableB on Column1 will return
SELECT * FROM TableA AS a INNER JOIN TableB AS b USING (Column1); SELECT * FROM TableA AS a INNER JOIN TableB AS b ON a.Column1 = b.Column1;
+------------+-----------+---------------------+ | a.Column1 | a.Column2 | b.Column1| b.Column3| +------------------------+---------------------+ | 1 | 2 | 1 | 3 | +------------+-----------+----------+----------+
The NATURAL JOIN of TableA and TableB on Column1 will return:
SELECT * FROM TableA NATURAL JOIN TableB +------------+----------+----------+ |Column1 | Column2 | Column3 | +-----------------------+----------+ | 1 | 2 | 3 | +------------+----------+----------+
Ref: https://bit.ly/3AG5CId
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