When I use the following code, it doesn't print out "else if test" until after I refresh the page.
unset($_SESSION["product"]);
if ( !empty($_SESSION["product"]) && ($_SESSION["product"] != NULL)) {
echo "if test";
} else if {
echo "else if test";
}
When I use:
var_dump($_SESSION["product"]);
It shows:
array(0) { }
(Added some code markup ~ Mod)
A few things...
!empty($anything) will not be true if $anything is null, so there's no need for the separate NULL check.else {, not else if { since you don't have an if condition?Anyway, this works for me from the terminal:
21:54 $ php -a
Interactive shell
php > $_SESSION = ['product' => 'something'];
php > unset($_SESSION['product']);
php > var_export($_SESSION);
array (
)
php > if(!empty($_SESSION['product'])) {
php { echo "It's not empty";
php { } else {
php { echo "It's empty.";
php { }
It's empty.
It should be else not else if.
When I paste $var_export($_SESSION); into my code, I get the following error:
Notice: Undefined variable: var_export in C:\xampp\htdocs\TopView\cart.php on line 82
Fatal error: Uncaught Error: Function name must be a string in C:\xampp\htdocs\TopView\cart.php:82 Stack trace: #0 {main} thrown in C:\xampp\htdocs\TopView\cart.php on line 82
It's not $var_export (where are you pasting it from?), it's var_export.
The code didn't actually make the call to unset the variables until after the not empty conditional was asked. I was misled because var_dump still printed out an empty session array. Once I changed the order of my code, it worked.
