Fatal error: Cannot use isset, use null expression instead

Question

I get this error when trying to submit this form.

Fatal error: Cannot use isset() on the result of an expression (you can use "null !== expression" instead) in C:\Users\USER\XAMPP\htdocs\contactform\webform.php on line 3

This is line 3:

if(isset($userEmail = $_POST["email"]) && $userEmail = $_POST["email"] != "")

This is the entire part of code:

<?php
    $message_sent = false;
    if(isset($userEmail = $_POST["email"]) && $userEmail = $_POST["email"] != ""){
        if(filter_var($_POST["email"], FILTER_VALIDATE_EMAIL)){
            $userName = $_POST["name"];
            $userEmail = $_POST["email"];
            $messageSubject = $_POST["subject"];
            $message = $_POST["message"];
            $to = "smartvishwa132@gmail.com";
            $body = "";
            $body .= "From: ".$userName. "\r\n";
            $body .= "Email: ".$userEmail. "\r\n";
            $body .= "Message: ".$message. "\r\n";
            mail($to,$messageSubject,$body);
            $message_sent = true;
        }
    }
?>

How do I fix this?

score 1 · Accepted Answer · answered Dec 29 '20 at 03:25

if(isset($userEmail = $_POST["email"]) && $userEmail = $_POST["email"] != ""){

Here, you are assigning it to a variable and then checking isset all in a single call. Split it up into separate calls:

check if it is set
check if it is not an empty string
validate it against the built-in email validator
assign it to a variable

Likely should be

if(isset($_POST["email"]) && $_POST["email"] != "") {
    if(filter_var($_POST["email"], FILTER_VALIDATE_EMAIL)) {   
        ...
        $userEmail = $_POST["email"];
        ...

Thank you, it works and I am able to get the contact form working. — Vishwajith S, Dec 29 '20 at 15:43

Fatal error: Cannot use isset, use null expression instead

1 Answers1