﻿ SQL Exercises - Wildcard and Special operators - w3resource

# 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
```

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
```

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
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
```

Click me to see the solution with visual presentation

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
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
```

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
```

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
```

Click me to see the solution with visual presentation

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
```

Click me to see the solution with visual presentation

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
```

Click me to see the solution with visual presentation

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
```

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
```

Click me to see the solution with result

## Practice Online

More to Come !

Query visualizations are generated using Postgres Explain Visualizer (pev).

Do not submit any solution of the above exercises at here, if you want to contribute go to the appropriate exercise page.

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