Trouble checking if a name already exists

[coolgamer48]

Hey, so I'm making this registration page for a site (it's not really registering a user, but it's like the same thing). I actually have a sign in page with simmilar code that works fine. But for some reason it won't work when I'm doing it now. Can anyone find a problem with this code:

		$sql = "CREATE TABLE kingdoms
	(
		name varchar(15),
		PRIMARY KEY(name),
		date_est date(),
		acreage int(1000),
		money int(20),
		user varchar(15)
	)";
	if (mysql_query($sql,$con) == false)
	{
		echo("We're sorry, there was an error with the Kingdom database.");
	}
	$sql = "SELECT * FROM kingdoms WHERE name=\"".mysql_real_escape_string($_POST["kingdom_name"])."\"";
	$kingdoms_list = mysql_query($sql,$con);
	if (mysql_num_rows($kingdoms_list) > 0)
	{
		echo "Sorry, that Kingdom name is already in use. You can<br />
			<ul>
				<li>Try a different name if you have something in mind</li>
				<li>Consider something like ".$_POST["kingdom_name"]." of ".$_COOKIE["username"]."</li>
			</ul>";
	}

 

I'm pretty sure the problem is with creating the table, but I'm not sure. Please and thank you!

[kts]

It will error when you try to create, because it already exists, try removing that from your code

[revraz]

Could put a Else statement here

 

if (mysql_query($sql,$con) == false)

{

echo("We're sorry, there was an error with the Kingdom database.");

}

else {

//then do your code

 

The way you have it, if there is an errror, the rest of your code still executes.

[revraz]

This is probably better

 

$tablename="Kingdoms";

$res = mysql_query("show table status like '$tablename'") or die(mysql_error());

$table_exists = mysql_num_rows($res) == 1;