Display state of user from Session start when logged in [duplicate] - php

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The 3 different equals
(5 answers)
Closed 2 years ago.
I'm trying to display "active" or "inactive" when a user login with session start with php and mysql.
The problem is that even when I try to login with an account that has an "inactive" state, it keeps showing me that is "active". I already added session_start(); on the start of the page.
Here's a part of my login code:
$sql = "SELECT * FROM `users_tmp` WHERE uidUsers=? OR emailUsers=?;";
$stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt, $sql)){
header("Location: ..index.php?error=sqlerror");
exit();
}
else {
mysqli_stmt_bind_param($stmt, "ss", $mailuid, $password);
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
if ($row = mysqli_fetch_assoc($result)) {
$pwdCheck = $row['pwdUsers'];
if($pwdCheck == false){
header("Location: ../index.php?error=wrongpwd");
exit();
} else if ($pwdCheck == true) {
session_start();
$_SESSION['userId'] = $row['idTmp'];
$_SESSION['idTmp'] = $row['idTmp'];
$_SESSION['stateUser'] = $row['stateUser'];
header("Location: ../user/perfil.php?login=success");
exit();
}
}
else{
header("Location: ../index.php?error=wrongpwds");
exit();
}
And here's what I'm trying to display:
<?php if(isset($_SESSION['idTmp']) ){
$state = $_SESSION['stateUser'];
if($state = "Active"){
echo $state;}
elseif($state = "Inactive"){
echo $state ;}
}
else{echo'nothing';}?>

Answer 1

$state = "Active" is always true, you need to use == operator instead, in the if.