Errata

In the example last entry has a apostrophe (') whereas in the solution this apostrophe is on purpose missing to produce the error.
So the SQL-statement in the question/exercise does not produce any error as expected.

In the example last entry has a apostrophe (') whereas in the solution this apostrophe is on purpose missing to produce the error.
So the SQL-statement in the question/exercise does not produce any error as expected.

Location 1076 of 10324 in kindle edition.

The solution is "SELECT drink_name, color, ice FROM drink_info WHERE cost >= 3.8;

There are 3 drinks that meet this criteria.

Soda and It @ 3.8
Greyhound @ 4
Kiss on the Lips @ 5.5

If you left out the '=', it would work.

The original query is:
mysql> select * from easy_drinks where main IN ('peach nectar', 'soda');

For me that returns:
+-------------+
| drink_name |
+-------------+
| Blue Moon |
| Oh My Gosh |
| Hot Gold |
| Lone Tree |
| Soda and It |
| Greyhound |
+-------------+

The answer in the book is:
select drink_name from easy_drinks where main between 'p' and 't';
+-------------+
| drink_name |
+-------------+
| Blue Moon |
| Oh My Gosh |
| Lime Fizz |
| Hot Gold |
| Lone Tree |
| Soda and It |
| Greyhound |
+-------------+

Because Lime Fizz contains Sprite.

The correct answer is:
select drink_name from easy_drinks where main between 'p' and 'sp';
+-------------+
| drink_name |
+-------------+
| Blue Moon |
| Oh My Gosh |
| Hot Gold |
| Lone Tree |
| Soda and It |
| Greyhound |
+-------------+

The (recent, corrected) solution still does not work. There are two problems with it.

1) MySQL does not support FIFTH, SIXTH etc in an ALTER TABLE statement. It only supports FIRST and AFTER. See the ALTER TABLE syntax reference here: http://dev.mysql.com/doc/refman/5.5/en/alter-table.html
Note that this affects page 201 and 202 also.

2) Even when using FIRST or AFTER in an ALTER TABLE statement, you need to specify the data type.

See this transcript (which includes my corrected solution):
--------------------------------
mysql> CREATE TABLE `hooptie` (
-> `color` varchar(20) default NULL,
-> `year` varchar(4) default NULL,
-> `make` varchar(20) default NULL,
-> `mo` varchar(20) default NULL,
-> `howmuch` float(10,3) default NULL
-> ) ENGINE=MyISAM DEFAULT CHARSET=latin1;
Query OK, 0 rows affected (0.03 sec)

mysql>
mysql> INSERT INTO `hooptie` (`color`,`year`,`make`,`mo`,`howmuch`) VALUES ('sil
ver','1998','Porsche','Boxter','17992.539');
Query OK, 1 row affected (0.00 sec)

mysql> INSERT INTO `hooptie` (`color`,`year`,`make`,`mo`,`howmuch`) VALUES (NULL
,'2000','Jaguar','XJ','15995.000');
Query OK, 1 row affected (0.00 sec)

mysql> INSERT INTO `hooptie` (`color`,`year`,`make`,`mo`,`howmuch`) VALUES ('red
','2002','Cadillac','Escalade','40215.898');
Query OK, 1 row affected (0.00 sec)

mysql> ALTER TABLE hooptie
-> RENAME TO car_table,
-> ADD COLUMN car_id INT NOT NULL AUTO_INCREMENT FIRST,
-> ADD PRIMARY KEY (car_id),
-> ADD COLUMN VIN VARCHAR(17) AFTER car_id,
-> CHANGE COLUMN mo model VARCHAR(20),
-> MODIFY COLUMN color AFTER model,
-> MODIFY COLUMN year SIXTH,
-> CHANGE COLUMN howmuch price DECIMAL(7,2);
ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that
corresponds to your MySQL server version for the right syntax to use near 'AFTER
model,
MODIFY COLUMN year SIXTH,
CHANGE COLUMN howmuch price DECIMAL(7,2)' at line 7

mysql> ALTER TABLE hooptie
-> RENAME TO car_table,
-> ADD COLUMN car_id INT NOT NULL AUTO_INCREMENT FIRST,
-> ADD PRIMARY KEY (car_id),
-> ADD COLUMN VIN VARCHAR(17) AFTER car_id,
-> CHANGE COLUMN mo model VARCHAR(20),
-> MODIFY COLUMN color varchar(20) AFTER model,
-> MODIFY COLUMN year varchar(4) AFTER color,
-> CHANGE COLUMN howmuch price DECIMAL(7,2);
Query OK, 3 rows affected, 2 warnings (0.03 sec)
Records: 3 Duplicates: 0 Warnings: 2

mysql> select * from car_table;
+--------+------+----------+----------+--------+------+----------+
| car_id | VIN | make | model | color | year | price |
+--------+------+----------+----------+--------+------+----------+
| 1 | NULL | Porsche | Boxter | silver | 1998 | 17992.54 |
| 2 | NULL | Jaguar | XJ | NULL | 2000 | 15995.00 |
| 3 | NULL | Cadillac | Escalade | red | 2002 | 40215.90 |
+--------+------+----------+----------+--------+------+----------+
3 rows in set (0.00 sec)

When running the code from the solutions against the table, existing values in the 'category' column are erased where no 'WHEN' statement was provided (tested with MySQL 5.5.27).
Adding 'ELSE category' solved the issue for me (solution 2 and 3).

The answer states that the CHECK constraint is not available in MySQL. This was true at the time of publishing, however as of MySQL 8.0.16 it is now available. See https://mysqlserverteam.com/mysql-8-0-16-introducing-check-constraint/

Last line of SELECT statement reads:

...ON boys.toy_id = toys.toy_id;

This line caused an error in a PHP trial. I believe it should read,

...ON boys.boy_id = toys.toy_id;

I made the change in the code, and it worked fine. The boys table does not have a toy_id field.

In Left Outer Join The Explanation says that in query after from in left table and After ANY JOIN FLAVOUR is Right Table

But On 426 Page When Explaining right join the book points , after the from tab1 as right and after RIGHT OUTER JOIN as left table.

The description presented is not that of an updatable view, but of an insertable view. Failing to include all of the NOT NULL columns in a view does not prevent it from being used to UPDATE or DELETE columns. The final paragraph on the page does point out conditions that prevent a view from being updatable, such as aggregate operators.

Furthermore, whether or not INSERT can be used actually depends on whether the view contains all of the columns for which a default value is not defined. It is possible to use INSERT on a view that does not contain all of the NOT NULL columns, provided those columns have a default value other than NULL.