Hello,
I'm starting to learn how to manage databases with
MySQL/PHP. The program is very basic but I have a
strange error: "Resource id #3".
I read many things on Internet and understood that
others have this error because they are trying to echo
the pointer of the query and not the result itself.
The message that you receive is not an error, it
is a description of the object $result after the
query is done.
So, you will not get anywhere just by echo $result,
you have to process that information using one
of the mysql_fetch_array(),mysql_fetch_assoc(),
mysql_result().
In your case try :
> $value)
{
echo $key . " :: ". $value;
}
}
//the query didn't work
else echo mysql_error();
}
else echo mysql_error();
}
Bastien
From: Marc Henri <[EMAIL PROTECTED]>
To: php-db@lists.php.net
Subject: [PHP-DB] Error: "
Hello,
Thank you very much to Bastien Koert and Firan
Corneliu. You both solved my problem.
Shame on me!
_
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