PHP登錄 - 如果聲明執行即使沒有匹配

Question

我是PHP新手，我創建了一個登錄方法。當我提交的數據，即使該值不匹配「如果」正在執行

聲明我的代碼：

<?php 
    // include db configuration 
    include("./config.php"); 

    // converting data to JSON 
    $data = json_decode(file_get_contents("php://input")); 

    //assigning input to variables 
    $user_name = $data->user_name; 
    $password = hash('sha256', $data->password); 
    $tokenUser = hash('sha256', $data->user_name); 

    //checking username and password 
    $getUserInfo = $db->query("SELECT user_name FROM user WHERE user_name='$user_name' AND password='$password'"); 

    //getting user information 
    $getUserInfo = $getUserInfo->fetchAll(); 

    $token; 

    if (count($getUserInfo == 1)) 
    { 
     //creating a token for user authentication 
     $token = $tokenUser . " | " . uniqid() . uniqid() . uniqid(); 

     $q = "UPDATE user SET token=:token WHERE user_name=:user_name AND password=:password"; 

     $query = $db->prepare($q); 

     $execute = $query->execute(array(
      ":token" => $token, 
      ":user_name" => $user_name, 
      ":password" => $password 
    )); 

     $response = array(
      'user_name' => $user_name, 
      'token' => $token, 
      'access' => 'Granted' 
    ); 

     echo json_encode($response); 
    } 
    else 
    { 
     $error = array(
      'status' => 'error', 
      'message' => 'Username or Password is invlid' 
    ); 
     echo json_encode($error); 
    } 

?> 

如何驗證這一點，是代碼正確的，需要一些澄清

Answer 1

這一行：

if (count($getUserInfo == 1)) 

應當如下：

if (count($getUserInfo) == 1)