php oop檢查用戶是否存在？

Question

如果用戶名存在，我想顯示一個錯誤，但不會引發錯誤。

該函數在User.php上，並試圖顯示該函數的錯誤。

我引用了this，但它與OOP方式無關。

user.php的

public function check_user_exists($username) 
{ 
    try{ 
     $stmt = $this->db->prepare("SELECT user_name FROM users WHERE user_name=:username"); 
     $stmt->execute(array(':username'=>$username)); 
     $row = $stmt->fetch(PDO::FETCH_ASSOC); 
     $row['user_name'] == $username; 

    } 

    catch(PDOExeception $e) 
    { 
     echo $e->getMessage(); 
    } 
} 

的index.php

<?php 
session_start(); 

require_once 'User.php'; 
$guest = new User(); 



if($guest->is_logged()) 
{ 
    $guest->redirect('profile'); 
} 


if (isset($_POST['btn_signup'])){ 

    $username = htmlentities($_POST['txt_username']); 
    $unpass = htmlentities($_POST['txt_password']); 
    $password = password_hash($unpass, PASSWORD_BCRYPT, ['cost' => 12]); 
    $unemail = $_POST['txt_email']; 
    $email = filter_var($unemail, FILTER_VALIDATE_EMAIL); 

    $guest = new User(); 


    if($email == ""){ 
     $errors[]= "Enter a Email"; 
    } 

    if($username == ""){ 
     $errors[]= "Enter a Username please"; 

    } 

    if($password == ""){ 
     $errors[]= "Enter a Password"; 
    } 



    if($guest->check_user_exists($username)){ 
     $errors[]= "Username Already Taken"; 
    } 

    if($guest->signup($email,$password,$username)){ 
     $guest->redirect('profile'); 
     die('didnt redirect');  
    } 

    else{ 
     $errors[]= "Invalid Entry"; 
    } 
} 

$title = "Home"; 
require_once 'layouts/header.php'; 


?> 


    <div class="container"> 
     <div class="row"> 
      <div class="col-md-6"> 

      <?php 
      if(isset($errors)) 
      { 
       foreach($errors as $error) 
       { 
        ?> 
        <div class="alert alert-danger"> 
         <i class="glyphicon glyphicon-warning-sign"></i> &nbsp; <?php echo $error; ?> 
        </div> 
        <?php 
       } 
      } 
      else if(isset($_GET['joined'])) 
      { 
       ?> 
       <div class="alert alert-info"> 
         <i class="glyphicon glyphicon-log-in"></i> &nbsp; Successfully registered <a href='index.php'>login</a> here 
       </div> 
       <?php 
      } 
      ?> 

       <h1>Sign Up</h1> 




       <form action ="" method="POST"> 
        <div class="form-group"> 
        <label for="Email">Email address</label> 
        <input type="email" class="form-control" aria-describedby="emailHelp" name="txt_email" placeholder="Enter email"> 
        </div> 

        <div class="form-group"> 
        <label for="Username">Username</label> 
        <input type="text" class="form-control" aria-describedby="emailHelp" name="txt_username" placeholder="Enter Username"> 
        </div> 


        <div class="form-group"> 
        <label for="Password">Password</label> 
        <input type="password" class="form-control" aria-describedby="emailHelp" name="txt_password" placeholder="Enter password"> 
        </div> 


        <button type="submit" name="btn_signup" class="btn btn-primary">Submit</button> 
       </form> 

      </div> 
     </div> 
    </div> 


</body> 
</html> 

Answer 1

public function check_user_exists($username) 
{ 
    try{ 
     $stmt = $this->db->prepare("SELECT user_name FROM users WHERE user_name=:username"); 
     $stmt->execute(array(':username'=>$username)); 
     return $stmt->fetchColumn() > 0; // fetchColumn return the number of rows selected 
    } 

    catch(PDOExeception $e) 
    { 
     echo $e->getMessage(); 
    } 
} 

Answer 2

您的功能實際上並不返回或做任何事情。返回fetch()的結果，如果返回true - 找到結果。如果它返回false，則沒有與用戶名匹配的行。之後您不需要檢查任何內容，因爲fetch()方法只有在找到結果時才爲真。

調整爲，你的函數應該是這樣的

public function check_user_exists($username) { 
    try{ 
     $stmt = $this->db->prepare("SELECT user_name FROM users WHERE user_name=:username"); 
     $stmt->execute(array(':username' => $username)); 
     return $stmt->fetch(PDO::FETCH_ASSOC); 
    } catch(PDOExeception $e) { 
     echo $e->getMessage(); 
    } 
} 

而且，它不是一個好主意，直​​接輸出誤差（在測試/開發環境的很好，但在實際環境中，你應該登錄它（error_log()）來代替。

• http://php.net/manual/en/pdostatement.fetch.php