Cannot execute this PHP code...

[anuj_web]

<?php

$dbcnx = @mysql_connect('localhost', 'username', 'password');
if (!$dbcnx) {
echo( '<p>Unable to connect to the ' . 'database server at this time.</p>' );
exit();
}
mysql_select_db('anujweb_donor', $dbcnx);

$sql = 'CREATE TABLE anuj(donorID INT NOT NULL AUTO_INCREMENT PRIMARY KEY,name text)';
if ( @mysql_query($sql) ) {
echo('<p>donor table successfully created!</p>');
} else {
die('<p>Error creating Jokes table: ' .mysql_error() . '</p>');
}
?>



It returns an error no database selected

PS:usename password intentionaly hidden

 

[LHVWB]

You shouldn't need to pass the database reference to the mysql_select_db() function, try it without it.
mysql_select_db('anujweb_donor');

 

[anuj_web]

thanks man
it worked
thanks a lot
Edit:
btw how many tables can be made in a single databse ??

 

[marshian]

there's no limit as far as i know

 

[anuj_web]

k thanks