PHP session array created from database and want to print to screen only one variable. I'm using this guide.
The array is created by :
$user = $libUsr->check($_POST['email'], $_POST['password']);
if ($user===false) { echo "ERR"; }
else {
$_SESSION['user'] = [
"name" => $user['user_name'],
"email" => $user['user_email'],
"status" => $user['user_status'],
"data" => $user['user_data']
];
echo "OK";
}
break;
The header of the signedin page after login is below, with start_session() called in 2a-config.php
require __DIR__ . DIRECTORY_SEPARATOR . "lib" . DIRECTORY_SEPARATOR . "2a-
config.php";
if (!is_array($_SESSION['user'])) {
header("Location: index.php");
die();
}
I've tried all combinations such as these (with the php tags of course):
echo $user["name"];
echo $user[1]["name"];
print_r(array_column($user, ‘name’));
Was even getting desperate as I've spent a lot of time on this and no other answers I found was working so was even trying
echo $_SESSION["name"];
echo $_SESSION[1]["name"];
As well as these previously asked questions but still can't get it to work and no errors are coming up in my error log.
This is the array when I var dump.
ARRAY(1) { ["USER"]=> ARRAY(4) { ["NAME"]=> STRING(4) "TEST" ["EMAIL"]=>
STRING(13) "TEST@TEST.COM" ["STATUS"]=> STRING(1) "A" ["DATA"]=>
STRING(32) "9529C449206201FDBCE28CFA42FD48C8" } }
All I want is to just print the array variable "NAME" which is TEST to the screen. I'm hoping then I can workout how to pull out the user email in order to update my database. As you can see I'm a first time poster on Stack Overflow so apologies if my question is lacking. Thank you.
