Quoting "Joseph, Smile-Poet" <[EMAIL PROTECTED]>:
> 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 ;
> }
>
This error is coming up because, if you don't have a missing email address,
$email_error isn't being defined at all. What you might want to do is add a line
above the two if blocks that defines $email_error = false, then if the email is
missing, it will be redefined as true then. Make sense?
>
> //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!)
For this one I'm not sure - what I usually do is create my SELECT statement as a
variable (eg. $select = "SELECT_COUNT(*)...";) and include that variable in the
mysql_query function. This makes it easy to echo out the $select variable to see
what is being passed to the function and to find out why the query isn't working.
--
Robin Hastings
[EMAIL PROTECTED]
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