How can I pass a variable from my function to another part of my file?

[Conjurer]

I am trying to pass a variable out of a function and reuse it in the page code. 

 

On the web page they see I would like to report to them the email value that is in the database and since that is pulled in the first part of the function I thought maybe I could just put some lines in the html using the $email variable, but when I do it gives me an error message of Undefined variable: email

 

Is there a way I can keep the variable live outside of the functions so I can reuse it instead of having to look up the value again?

 

Here is the code where I am trying to access it:

$username = $_POST['username'];
  echo 'This is the username you submitted: ' .$username .'<br>';
  try
    {
    $password = reset_password($username);
    notify_password($username, $password);   <<<This is the function that finds the email I am trying to pass
    echo "Your new password has been e-mailed to you at [b]$email[/b].<br />";
echo "If you no longer have access to the [b]$email [/b]e-mail account, please contact our webmaster at [email protected] for assistance.<br />";
    echo ' <div id="nav"><a href="http://www.mydomain.com/">mydomain Home</a> <a href="index.php5">Logout</a></div>';  
    }
    catch (Exception $e)
    {
    echo 'Your password could not be reset - please try again later.';
  echo $e;  //added to see error
    }

 

 

And here is the function that looks up the email address I want to pass into the rest of the web page:

//---------------------------------------------------------------------------
// function to notify user of new password
//---------------------------------------------------------------------------

function notify_password($username, $password)
// notify the user that their password has been changed
{
    $conn = db_connect();
    $result = $conn->query("select email from directory 
                            where user_id=(select user_id from users where username = '$username')");[/b]
    if (!$result)
    {
      throw new Exception('Could not find email address.');  
    }
    else if ($result->num_rows==0)
    {
      throw new Exception('Could not find email address.');   // username not in db
    }
    else
    {
      $row = $result->fetch_object();
      $email = $row->email;
      $from = "From: [email protected] \r\n";
      $mesg = "Your password for the directory has been changed. \r\n"
  				."You can login to the membership directory at http://www.mydomain.com/Directory/index.php5 \r\n\r\n"
				."Your username is: $username \r\n"
				."Your new password is: $password\r\n\r\n"
              ."You can either keep this password or change it to one you preferr the next time you log in. "
		  ."To change it, once you log in click on the \"Change Password\" button.\r\n";
      
      
      if (mail($email, 'Mydomain login information', $mesg, $from))
        return true;      
      else
        throw new Exception('Could not send email. Please contact [email protected] for assistance!');
    }
} 

 

Any ideas would be greatly appreciated.

[DarkerAngel]

When you call the function

 

call it like

 

 

$email = notify_password($username, $password)

 

and at the end of the function

 

      if (mail($email, 'Mydomain login information', $mesg, $from))

        return $email;     

      else

        throw new Exception('Could not send email. Please contact [email protected] for assistance!');

    }