SQL Exercises, Practice, Solution - Wildcard and Special operators
SQL [22 exercises with solution]
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1. From the following table, write a SQL query to find the details of those salespeople who come from the 'Paris' City or 'Rome' City. Return salesman_id, name, city, commission.
Sample table: salesman
salesman_id | name | city | commission
-------------+------------+----------+------------
5001 | James Hoog | New York | 0.15
5002 | Nail Knite | Paris | 0.13
5005 | Pit Alex | London | 0.11
5006 | Mc Lyon | Paris | 0.14
5007 | Paul Adam | Rome | 0.13
5003 | Lauson Hen | San Jose | 0.12
Sample Output:
salesman_id name city commission 5002 Nail Knite Paris 0.13 5006 Mc Lyon Paris 0.14 5007 Paul Adam Rome 0.13
Click me to see the solution with visual presentation
2. From the following table, write a SQL query to find the details of the salespeople who come from either 'Paris' or 'Rome'. Return salesman_id, name, city, commission.
Sample table: salesman
salesman_id | name | city | commission
-------------+------------+----------+------------
5001 | James Hoog | New York | 0.15
5002 | Nail Knite | Paris | 0.13
5005 | Pit Alex | London | 0.11
5006 | Mc Lyon | Paris | 0.14
5007 | Paul Adam | Rome | 0.13
5003 | Lauson Hen | San Jose | 0.12
Sample Output:
salesman_id name city commission 5002 Nail Knite Paris 0.13 5006 Mc Lyon Paris 0.14 5007 Paul Adam Rome 0.13
Click me to see the solution with visual presentation
3. From the following table, write a SQL query to find the details of those salespeople who live in cities other than Paris and Rome. Return salesman_id, name, city, commission.
Sample table: salesman
salesman_id | name | city | commission
-------------+------------+----------+------------
5001 | James Hoog | New York | 0.15
5002 | Nail Knite | Paris | 0.13
5005 | Pit Alex | London | 0.11
5006 | Mc Lyon | Paris | 0.14
5007 | Paul Adam | Rome | 0.13
5003 | Lauson Hen | San Jose | 0.12
Sample Output:
salesman_id name city commission 5001 James Hoog New York 0.15 5005 Pit Alex London 0.11 5003 Lauson Hen San Jose 0.12
Click me to see the solution with visual presentation
4. From the following table, write a SQL query to retrieve the details of all customers whose ID belongs to any of the values 3007, 3008 or 3009. Return customer_id, cust_name, city, grade, and salesman_id.
Sample table: customer
customer_id | cust_name | city | grade | salesman_id
-------------+----------------+------------+-------+-------------
3002 | Nick Rimando | New York | 100 | 5001
3007 | Brad Davis | New York | 200 | 5001
3005 | Graham Zusi | California | 200 | 5002
3008 | Julian Green | London | 300 | 5002
3004 | Fabian Johnson | Paris | 300 | 5006
3009 | Geoff Cameron | Berlin | 100 | 5003
3003 | Jozy Altidor | Moscow | 200 | 5007
3001 | Brad Guzan | London | | 5005
Sample Output:
customer_id cust_name city grade salesman_id 3007 Brad Davis New York 200 5001 3008 Julian Green London 300 5002 3009 Geoff Cameron Berlin 100 5003
Click me to see the solution with visual presentation
5. From the following table, write a SQL query to find salespeople who receive commissions between 0.12 and 0.14 (begin and end values are included). Return salesman_id, name, city, and commission.
Sample table: salesman
salesman_id | name | city | commission
-------------+------------+----------+------------
5001 | James Hoog | New York | 0.15
5002 | Nail Knite | Paris | 0.13
5005 | Pit Alex | London | 0.11
5006 | Mc Lyon | Paris | 0.14
5007 | Paul Adam | Rome | 0.13
5003 | Lauson Hen | San Jose | 0.12
Sample Output:
salesman_id name city commission 5002 Nail Knite Paris 0.13 5006 Mc Lyon Paris 0.14 5007 Paul Adam Rome 0.13 5003 Lauson Hen San Jose0.12
Click me to see the solution with visual presentation
6. From the following table, write a SQL query to select orders between 500 and 4000 (begin and end values are included). Exclude orders amount 948.50 and 1983.43. Return ord_no, purch_amt, ord_date, customer_id, and salesman_id.
Sample table: orders
ord_no purch_amt ord_date customer_id salesman_id ---------- ---------- ---------- ----------- ----------- 70001 150.5 2012-10-05 3005 5002 70009 270.65 2012-09-10 3001 5005 70002 65.26 2012-10-05 3002 5001 70004 110.5 2012-08-17 3009 5003 70007 948.5 2012-09-10 3005 5002 70005 2400.6 2012-07-27 3007 5001 70008 5760 2012-09-10 3002 5001 70010 1983.43 2012-10-10 3004 5006 70003 2480.4 2012-10-10 3009 5003 70012 250.45 2012-06-27 3008 5002 70011 75.29 2012-08-17 3003 5007 70013 3045.6 2012-04-25 3002 5001
Sample Output:
ord_no purch_amt ord_date customer_id salesman_id 70005 2400.60 2012-07-27 3007 5001 70003 2480.40 2012-10-10 3009 5003 70013 3045.60 2012-04-25 3002 5001
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7. From the following table, write a SQL query to retrieve the details of the salespeople whose names begin with any letter between 'A' and 'L' (not inclusive). Return salesman_id, name, city, commission.
Sample table: salesman
salesman_id | name | city | commission
-------------+------------+----------+------------
5001 | James Hoog | New York | 0.15
5002 | Nail Knite | Paris | 0.13
5005 | Pit Alex | London | 0.11
5006 | Mc Lyon | Paris | 0.14
5007 | Paul Adam | Rome | 0.13
5003 | Lauson Hen | San Jose | 0.12
Sample Output:
salesman_id name city commission 5001 James Hoog New York 0.15
Click me to see the solution with visual presentation
8. From the following table, write a SQL query to find the details of all salespeople except those whose names begin with any letter between 'A' and 'M'. Return salesman_id, name, city, commission.
Sample table: salesman
salesman_id | name | city | commission
-------------+------------+----------+------------
5001 | James Hoog | New York | 0.15
5002 | Nail Knite | Paris | 0.13
5005 | Pit Alex | London | 0.11
5006 | Mc Lyon | Paris | 0.14
5007 | Paul Adam | Rome | 0.13
5003 | Lauson Hen | San Jose | 0.12
Sample Output:
salesman_id name city commission 5002 Nail Knite Paris 0.13 5005 Pit Alex London 0.11 5006 Mc Lyon Paris 0.14 5007 Paul Adam Rome 0.13 5003 Lauson Hen San Jose 0.12
Click me to see the solution with visual presentation
9. From the following table, write a SQL query to retrieve the details of the customers whose names begins with the letter 'B'. Return customer_id, cust_name, city, grade, salesman_id..
Sample table: customer
customer_id | cust_name | city | grade | salesman_id
-------------+----------------+------------+-------+-------------
3002 | Nick Rimando | New York | 100 | 5001
3007 | Brad Davis | New York | 200 | 5001
3005 | Graham Zusi | California | 200 | 5002
3008 | Julian Green | London | 300 | 5002
3004 | Fabian Johnson | Paris | 300 | 5006
3009 | Geoff Cameron | Berlin | 100 | 5003
3003 | Jozy Altidor | Moscow | 200 | 5007
3001 | Brad Guzan | London | | 5005
Sample Output:
customer_id cust_name city grade salesman_id 3007 Brad Davis New York 200 5001 3001 Brad Guzan London 5005
Click me to see the solution with visual presentation
10. From the following table, write a SQL query to find the details of the customers whose names end with the letter 'n'. Return customer_id, cust_name, city, grade, salesman_id.
Sample table: customer
customer_id | cust_name | city | grade | salesman_id
-------------+----------------+------------+-------+-------------
3002 | Nick Rimando | New York | 100 | 5001
3007 | Brad Davis | New York | 200 | 5001
3005 | Graham Zusi | California | 200 | 5002
3008 | Julian Green | London | 300 | 5002
3004 | Fabian Johnson | Paris | 300 | 5006
3009 | Geoff Cameron | Berlin | 100 | 5003
3003 | Jozy Altidor | Moscow | 200 | 5007
3001 | Brad Guzan | London | | 5005
Sample Output:
customer_id cust_name city grade salesman_id 3008 Julian Green London 300 5002 3004 Fabian Johnson Paris 300 5006 3009 Geoff Cameron Berlin 100 5003 3001 Brad Guzan London 5005
Click me to see the solution with visual presentation
11. From the following table, write a SQL query to find the details of those salespeople whose names begin with ‘N’ and the fourth character is 'l'. Rests may be any character. Return salesman_id, name, city, commission.
Sample table : salesman
salesman_id | name | city | commission
-------------+------------+----------+------------
5001 | James Hoog | New York | 0.15
5002 | Nail Knite | Paris | 0.13
5005 | Pit Alex | London | 0.11
5006 | Mc Lyon | Paris | 0.14
5007 | Paul Adam | Rome | 0.13
5003 | Lauson Hen | San Jose | 0.12
Sample Output:
salesman_id name city commission 5002 Nail Knite Paris 0.13
Click me to see the solution with visual presentation
12. From the following table, write a SQL query to find those rows where col1 contains the escape character underscore ( _ ). Return col1.
Sample table: testtable
col1 -------------------------- A001/DJ-402\44_/100/2015 A001_\DJ-402\44_/100/2015 A001_DJ-402-2014-2015 A002_DJ-401-2014-2015 A001/DJ_401 A001/DJ_402\44 A001/DJ_402\44\2015 A001/DJ-402%45\2015/200 A001/DJ_402\45\2015%100 A001/DJ_402%45\2015/300 A001/DJ-402\44
Sample Output:
col1 A001/DJ-402\44_/100/2015 A001_\DJ-402\44_/100/2015 A001_DJ-402-2014-2015 A002_DJ-401-2014-2015 A001/DJ_401 A001/DJ_402\44 A001/DJ_402\44\2015 A001/DJ_402\45\2015%100 A001/DJ_402%45\2015/300
Click me to see the solution with visual presentation
13. From the following table, write a SQL query to identify those rows where col1 does not contain the escape character underscore ( _ ). Return col1.
Sample table: testtable
col1 -------------------------- A001/DJ-402\44_/100/2015 A001_\DJ-402\44_/100/2015 A001_DJ-402-2014-2015 A002_DJ-401-2014-2015 A001/DJ_401 A001/DJ_402\44 A001/DJ_402\44\2015 A001/DJ-402%45\2015/200 A001/DJ_402\45\2015%100 A001/DJ_402%45\2015/300 A001/DJ-402\44
Sample Output:
col1 A001/DJ-402%45\2015/200 A001/DJ-402\44
Click me to see the solution with visual presentation
14. From the following table, write a SQL query to find rows in which col1 contains the forward slash character ( / ). Return col1.
Sample table: testtable
col1 -------------------------- A001/DJ-402\44_/100/2015 A001_\DJ-402\44_/100/2015 A001_DJ-402-2014-2015 A002_DJ-401-2014-2015 A001/DJ_401 A001/DJ_402\44 A001/DJ_402\44\2015 A001/DJ-402%45\2015/200 A001/DJ_402\45\2015%100 A001/DJ_402%45\2015/300 A001/DJ-402\44
Sample Output:
col1 A001/DJ-402\44_/100/2015 A001_\DJ-402\44_/100/2015 A001/DJ_401 A001/DJ_402\44 A001/DJ_402\44\2015 A001/DJ-402%45\2015/200 A001/DJ_402\45\2015%100 A001/DJ_402%45\2015/300 A001/DJ-402\44
Click me to see the solution with visual presentation
15. From the following table, write a SQL query to identify those rows where col1 does not contain the forward slash character ( / ). Return col1.
Sample table: testtable
col1 -------------------------- A001/DJ-402\44_/100/2015 A001_\DJ-402\44_/100/2015 A001_DJ-402-2014-2015 A002_DJ-401-2014-2015 A001/DJ_401 A001/DJ_402\44 A001/DJ_402\44\2015 A001/DJ-402%45\2015/200 A001/DJ_402\45\2015%100 A001/DJ_402%45\2015/300 A001/DJ-402\44
Sample Output:
col1 A001_DJ-402-2014-2015 A002_DJ-401-2014-2015
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16. From the following table, write a SQL query to find those rows where col1 contains the string ( _/ ). Return col1.
Sample table: testtable
col1 -------------------------- A001/DJ-402\44_/100/2015 A001_\DJ-402\44_/100/2015 A001_DJ-402-2014-2015 A002_DJ-401-2014-2015 A001/DJ_401 A001/DJ_402\44 A001/DJ_402\44\2015 A001/DJ-402%45\2015/200 A001/DJ_402\45\2015%100 A001/DJ_402%45\2015/300 A001/DJ-402\44
Sample Output:
col1 A001/DJ-402\44_/100/2015 A001_\DJ-402\44_/100/2015
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17. From the following table, write a SQL query to find those rows where col1 does not contain the string ( _/ ). Return col1.
Sample table: testtable
col1 -------------------------- A001/DJ-402\44_/100/2015 A001_\DJ-402\44_/100/2015 A001_DJ-402-2014-2015 A002_DJ-401-2014-2015 A001/DJ_401 A001/DJ_402\44 A001/DJ_402\44\2015 A001/DJ-402%45\2015/200 A001/DJ_402\45\2015%100 A001/DJ_402%45\2015/300 A001/DJ-402\44
Sample Output:
col1 A001_DJ-402-2014-2015 A002_DJ-401-2014-2015 A001/DJ_401 A001/DJ_402\44 A001/DJ_402\44\2015 A001/DJ-402%45\2015/200 A001/DJ_402\45\2015%100 A001/DJ_402%45\2015/300 A001/DJ-402\44
Click me to see the solution with visual presentation
18. From the following table, write a SQL query to find those rows where col1 contains the character percent ( % ). Return col1.
Sample table: testtable
col1 -------------------------- A001/DJ-402\44_/100/2015 A001_\DJ-402\44_/100/2015 A001_DJ-402-2014-2015 A002_DJ-401-2014-2015 A001/DJ_401 A001/DJ_402\44 A001/DJ_402\44\2015 A001/DJ-402%45\2015/200 A001/DJ_402\45\2015%100 A001/DJ_402%45\2015/300 A001/DJ-402\44
Sample Output:
col1 A001/DJ-402%45\2015/200 A001/DJ_402\45\2015%100 A001/DJ_402%45\2015/300
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19. From the following table, write a SQL query to find those rows where col1 does not contain the character percent ( % ). Return col1.
Sample table: testtable
col1 -------------------------- A001/DJ-402\44_/100/2015 A001_\DJ-402\44_/100/2015 A001_DJ-402-2014-2015 A002_DJ-401-2014-2015 A001/DJ_401 A001/DJ_402\44 A001/DJ_402\44\2015 A001/DJ-402%45\2015/200 A001/DJ_402\45\2015%100 A001/DJ_402%45\2015/300 A001/DJ-402\44
Sample Output:
col1 A001/DJ-402\44_/100/2015 A001_\DJ-402\44_/100/2015 A001_DJ-402-2014-2015 A002_DJ-401-2014-2015 A001/DJ_401 A001/DJ_402\44 A001/DJ_402\44\2015 A001/DJ-402\44
Click me to see the solution with visual presentation
20. From the following table, write a SQL query to find all those customers who does not have any grade. Return customer_id, cust_name, city, grade, salesman_id.
Sample table: customer
customer_id | cust_name | city | grade | salesman_id
-------------+----------------+------------+-------+-------------
3002 | Nick Rimando | New York | 100 | 5001
3007 | Brad Davis | New York | 200 | 5001
3005 | Graham Zusi | California | 200 | 5002
3008 | Julian Green | London | 300 | 5002
3004 | Fabian Johnson | Paris | 300 | 5006
3009 | Geoff Cameron | Berlin | 100 | 5003
3003 | Jozy Altidor | Moscow | 200 | 5007
3001 | Brad Guzan | London | | 5005
Sample Output:
customer_id cust_name city grade salesman_id 3001 Brad Guzan London 5005
Click me to see the solution with visual presentation
21. From the following table, write a SQL query to locate all customers with a grade value. Return customer_id, cust_name,city, grade, salesman_id.
Sample table: customer
customer_id | cust_name | city | grade | salesman_id
-------------+----------------+------------+-------+-------------
3002 | Nick Rimando | New York | 100 | 5001
3007 | Brad Davis | New York | 200 | 5001
3005 | Graham Zusi | California | 200 | 5002
3008 | Julian Green | London | 300 | 5002
3004 | Fabian Johnson | Paris | 300 | 5006
3009 | Geoff Cameron | Berlin | 100 | 5003
3003 | Jozy Altidor | Moscow | 200 | 5007
3001 | Brad Guzan | London | | 5005
Sample Output:
customer_id cust_name city grade salesman_id 3002 Nick Rimando New York 100 5001 3007 Brad Davis New York 200 5001 3005 Graham Zusi California 200 5002 3008 Julian Green London 300 5002 3004 Fabian Johnson Paris 300 5006 3009 Geoff Cameron Berlin 100 5003 3003 Jozy Altidor Moscow 200 5007
Click me to see the solution with visual presentation
22. From the following table, write a SQL query to locate the employees whose last name begins with the letter 'D'. Return emp_idno, emp_fname, emp_lname and emp_dept. Go to the editor
Sample table: emp_details
EMP_IDNO EMP_FNAME EMP_LNAME EMP_DEPT --------- --------------- --------------- ---------- 127323 Michale Robbin 57 526689 Carlos Snares 63 843795 Enric Dosio 57 328717 Jhon Snares 63 444527 Joseph Dosni 47 659831 Zanifer Emily 47 847674 Kuleswar Sitaraman 57 748681 Henrey Gabriel 47 555935 Alex Manuel 57 539569 George Mardy 27 733843 Mario Saule 63 631548 Alan Snappy 27 839139 Maria Foster 57
Sample Output:
emp_idno emp_fname emp_lname emp_dept 843795 Enric Dosio 57 444527 Joseph Dosni 47
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