"char" as a table name causing problems in PHP code...

L33tN3k0 · #1 (**permalink**) 08-22-04, 04:23

PHP Code:


			
$result = mysql_query("SELECT * FROM char",$db);

Every time I execute my code, it returns the error

Quote:

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in C:\htdocs\index.php on line 37

However when I change "char" to "login", it works fine...

All I can figure is causing it is that the table is named "char" which, I'm assuming, is a restricted word...

I searched all over Google and couldn't figure out what to look for, then tried replacing char with a variable ".$chardb." which is defined as

PHP Code:


			
$chardb="char"

... Which gave the same compilation error.

So, my question is, besides changing the name of the table, what can I do to make this work?

Thanks in advance

~Neko