登录时从会话开始显示用户状态

Question

当用户登录会话以 php 和 mysql 开始时，我试图显示“活动”或“非活动”。问题是，即使我尝试使用状态为“非活动”的帐户登录，它也会一直显示“活动”状态。我已经添加了 session_start(); 在页面的开头。这是我的登录代码的一部分：$sql = "SELECT * FROM `users_tmp` WHERE uidUsers=? OR emailUsers=?;";$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'] = $row['stateUser'];            header("Location: ../user/perfil.php?login=success");            exit();        }    }    else{        header("Location: ../index.php?error=wrongpwds");        exit();    }这是我要显示的内容：<?php if(isset($_SESSION['idTmp']) ){  $state = $_SESSION['stateUser'];    if($state = "Active"){    echo $state;}    elseif($state = "Inactive"){echo $state ;}}else{echo'nothing';}?>

Answer 1

$state = "Active"始终为真，您需要==在 if 中改用运算符。