[wdvltalk] Re: $_POST and access questions

[Sheila Fenelon]

Hi Joseph,

Joseph, Smile-Poet wrote:
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;
        }
$email_error needs an initial value. Add this line
         $email_error = false;
before the first 'if' statement.

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in c:\easyphp1-7\www\join.php on line 102

This usually means there is something wrong with the SQL statement.

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);
Break this up into several steps so you can find the problem,

$sql = "SELECT_COUNT(*) AS ucount FROM book_mydb.members 
'{$_POST['username']}' ";

$result = mysql_query($sql);

if ($result) {
  $_ucount = mysql_result($result);
} else {
  echo mysql_errno() . ": " . mysql_error(). "<br>\n";
  echo "SQL: $sql<br>\n";
}


Sheila
--
Sheila Fenelon
http://www.shefen.com/


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