Why this:
Notice: Undefined variable: email_error in c:\easyphp1-7\www\join.php on line 65
if the following two queries define the variable?:
if(!$_POST['email_address']){
$errors .= "Missing Email address\n";
$email_error = true ;
}
if(!$_POST['email_address2']){
$errors .= "Missing Email address".
"Verification\n";
$email_error = true ;
}
//If both emails were posted, validate they match.
(line65) if( $email_error == false){
if($_POST['email_address'] !=
$_POST['email_address2']){
$error = true;
$errors .= "Email addresses do not match!\n\n";
$email_error = true;
}
AND
why this:
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in
c:\easyphp1-7\www\join.php on line 102
from this:
//Verify if username already exists
$_ucount = mysql_result(mysql_query("SELECT_COUNT(*)
AS ucount FROM book_mydb.members
WHERE members.username =
(line 102) '{$_POST['username']}' "),0);
Joseph (I'm not *that* far advanced, these are from a tutorial; but after aplying
what little knowledge I've gained I'm getting the same errors. And the servers *are*
turned on!)
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