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BOOK: Beginning PHP 6, Apache, MySQL 6 Web Development ISBN: 9780470391143
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Old September 20th, 2009, 07:39 PM
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Default Page 87

Ok so I am having the same problem as the poster below me

Here is the error i am getting
Quote:
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 'INTERGER UNSIGNED NOT NULL DEFAULT 0, PRIMARY KEY (movie_id), KEY movie_t' at line 7
the code is
Code:
<?php
//database creation and connection.
//connect to the database
$db = mysql_connect('moviesite.mattanders.com', 'bp6amm', 'bp6ampass')or die ('Unable to connect. Check your connection parameters.');

// creates the database if it is not there 
$query = 'CREATE DATABASE IF NOT EXISTS mattanders_movies';
mysql_query($query, $db) or die(mysql_error($db));// these mysql_ variables are inherent to mysql. though it is also being supplied the die variable from the $db variable

//make sure tthe created database is the active one 
mysql_select_db('mattanders_movies', $db) or die(mysql_error($db));

//create the movie table
$query = 'CREATE TABLE movie (movie_id		INTEGER UNSIGNED	NOT NULL AUTO_INCREMENT,
		movie_name  	VARCHAR(255)     	NOT NULL,
		movie_type		TINYINT			 	NOT NULL DEFAULT 0,
		movie_year 		SMALLINT UNSIGNED	NOT NULL DEFAULT 0,
		movie_leadactor	INTEGER UNSIGNED	NOT NULL DEFAULT 0,
		movie_director	INTERGER UNSIGNED	NOT NULL DEFAULT 0,
		
		PRIMARY KEY (movie_id),
		KEY movie_type (movie_type, movie_year)
		)
		ENGINE=MyISAM';
	mysql_query($query, $db)
 or die (mysql_error($db));
 
 // create the movietype table
 $query = 'CREATE TABLE movietype (
 		 movietype_id 	TINYINT UNSIGNED	NOT NULL AUTO_INCREMENT,
		 movietype_label	VARCHAR(100)	NOT NULL,
		 PRIMARY KEY (movietype_id)
	)
	ENGINE=MyISAM';
mysql_query($query, $db) or die(mysql_error($db));

echo 'Movie database successfully created!';
?>
Whats wrong?

According to the code editor im using that section of code is on line 14.

Last edited by Golarin; September 20th, 2009 at 07:45 PM.. Reason: question
 
Old September 21st, 2009, 04:03 PM
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Default Changed

I changed the password to 123456
 
Old September 21st, 2009, 04:57 PM
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Default

well i fixed this error, but now i am having a problem with

Code:
<?php
$db = mysql_connect('moviesite.mattanders.com', 'bp6amm', '123456') or die('Unable to connect. Check your connection parameters.');
mysql_select_db('mattanders_movies', $db) or die(mysql_error($db));

// select movie titles and thier genre after 1990
$query = 'SELECT
		movie.movie_name, movietype.movietype_label
	FROM
		movie, movietype
	WHERE
		movie.movie_type = movietype.movietype_id AND 
		movie_year > 1990
	ORDER BY 
		movie_type';
$result = mysql_query($query, $db) or die(mysql_error($db));

// Show the results
echo '<table border="1">';
while ($row = mysql_fetch_assoc($result)) {
	echo '<tr>';
	foreach ($row as $value) {
	    echo '<td>' . $value . '</td>';
	}
	echo '</tr>';
}
echo '</table>';
?>
Nothing shows on the page when i try running it... so i think i messed something up


And i tried it with the code from the book and still got nothing to show on my page

Last edited by Golarin; September 21st, 2009 at 05:00 PM..
 
Old September 27th, 2009, 03:58 AM
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Default

hmmm... according to the problem u state:
Quote:
Nothing shows on the page when i try running it....
This means the browser catch no ERROR on your code? so it might be your DB value which is blank, or your the problem lies with the ECHOs.

To check the Database you can chk with MySQL Query Browser on page:102 they give tips abt it.

for the problem on the echos u might want to try alternatives way to echo out the results:

// Show the results
echo '<table border="1">';
while ($row = mysql_fetch_assoc($result)) {
echo '<tr>';
extract ($row);
echo '<td>' . $movie_name . '</td>';
echo '<td>' . $movietype_label . '</td>';
echo '</tr>';
}
echo '</table>';
?>

but then again, if you can successfully complete page 96. it means your foreach to create table wasn't wrg.

Another thing you can check is your browser's settings.





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