$email = $cheatinf[8]; $emaillen = strpos($email, "@"); $email2 = substr($email, 0, $emaillen);
Makes $email2 come up as null. $cheatinf[8] returns the e-mail address. zx_storm@yahoo.com I'm trying to get it to cut off everything past the @. Can someone help me?
Syntax seems right. Perhaps some other problem? All I can suggest: $email = $cheatinf[8]; echo("$email<BR>"); $emaillen = strpos($email, "@"); echo("$emaillen<BR>"); $email2 = substr($email, 0, $emaillen); echo("$email2<BR>");