如何顯示錯誤，因爲用戶名存在或電子郵件存在使用PHP而註冊

Question

嗨朋友試圖顯示錯誤，因爲用戶名已存在或電子郵件存在，而註冊使用PHP。這裏是我的代碼..

if (isset($_POST['username']) && isset($_POST['password'])){ 
    $username = $_POST['username']; 
    $email = $_POST['email']; 
    $password = $_POST['password']; 

    $unique_user = "SELECT * FROM `user` where username='$username' or email='$email'"; 
    $unique_result = mysqli_query($mysql, $unique_user); 
    if(mysqli_num_rows($unique_result)>0) 
     { 
      echo "try again"; 
     } 

現在是重新打印的嘗試，但我怎麼能檢查是否是用戶名或電子郵件已經存在

Answer 1

我建議你只顯示用戶名或電子郵件存在代替的用戶名存在和電子郵件存在。 反正這裏是代碼：

 if (isset($_POST['username']) && isset($_POST['password'])){ 
     $username = $_POST['username']; 
     $email = $_POST['email']; 
     $password = $_POST['password']; 

     $unique_user = "SELECT * FROM `user` where username='$username' or email='$email'"; 
     $unique_result = mysqli_query($mysql, $unique_user); 
     if(mysqli_num_rows($unique_result)>0) 
     { 
      while($res = mysqli_fetch_assoc($unique_result)) 
      { 
       if($res['username'] == $username) 
       { 
       echo "Username Exists"; 
       } 
       if($res['email'] == $email) 
       { 
       echo "Email Exists"; 
       } 
      } 
      echo "Please try again"; 
     } 

Answer 2

$unique_user = "SELECT * FROM `user` where username='$username' or email='$email'"; 
$unique_result = mysqli_query($mysql, $unique_user);     
if(mysqli_num_rows($unique_result)>0) { 
     while ($row = mysqli_fetch_array($unique_result)) { 
       if ($row['username'] === $username) { 
       echo 'Username Exist!'; 
       } else { 
       echo 'Email Exist!'; 
       } 
     } 
}