如何爲唯一電子郵件添加服務器端驗證

Question

我有一個非常簡單的表單，用於獲取電子郵件地址以發送簡報。我想弄清楚如何添加一些服務器端驗證來檢查輸入的電子郵件地址是否已經在我的database中。然後，如果它已經輸入，它不允許表單提交，並且會拋出用戶可以看到的錯誤。

我該如何去檢查這個和顯示錯誤？

<form action="" method="POST" id="newsletter-form"> 
    <input type="email" id="footer-grid1-newsletter-input" placeholder="Your Email Address" pattern=".{3,}" required> 
    <input type="submit" id="footer-grid1-newsletter-submit" name="submit" value='&nbsp'> 
</form> 


$("#newsletter-form").on("submit", function (event) { 
     event.preventDefault(); 

     var newsletter_email = $("#footer-grid1-newsletter-input").val(); 
     var targeted_popup_class = jQuery(this).attr('data-popup-open'); 

     $.ajax({ 
      url: "newsletterSend.php", 
      type: "POST", 
      data: { 
       "newsletter_email": newsletter_email 
      }, 
      success: function (data) { 
      // console.log(data); // data object will return the response when status code is 200 
       if (data == "Error!") { 
        alert("Unable to insert email!"); 
        alert(data); 
       } else { 
        $("#newsletter-form")[0].reset(); 
        $('.newsletter-popup').fadeIn(350).delay(2000).fadeOut(); 
       } 
      }, 
      error: function (xhr, textStatus, errorThrown) { 
       alert(textStatus + " | " + errorThrown); 
       //console.log("error"); //otherwise error if status code is other than 200. 
      } 
     }); 
    }); 

$newsletter_email = $_POST['newsletter_email']; 

try { 

    $con = mysqli_connect("localhost", "", "", ""); 

    if (mysqli_connect_errno()) { 
     printf("Connect failed: %s\n", mysqli_connect_error()); 
     exit(); 
    } 
    $stmt = $con->prepare("INSERT INTO newsletter (email, subscribed) VALUES (?, NOW())"); 
     if (false===$stmt) { 
      die('Newsletter email prepare() failed: ' . htmlspecialchars($con->error)); 
     } 
    $stmt->bind_param('s', $newsletter_email); 
     if (false===$stmt) { 
      die('Newsletter email bind_param() failed: ' . htmlspecialchars($stmt->error)); 
     } 
    $stmt->execute(); 
     if (false===$stmt) { 
      die('Newsletter email execute() failed: ' . htmlspecialchars($stmt->error)); 
     } 

Answer 1

這下面應該可以幫到你。

JS：

$(document).ready(function(){ //newly added 
    $('#footer-grid1-newsletter-submit').click(function() {  
     var emailVal = $('#footer-grid1-newsletter-input').val(); // assuming this is a input text field  
     $.post('checkemail.php', {'email' : emailVal}, function(data) { 
      if(data=='exist'){ 
       $('#alert').html("<p> You are already subscribed to our website</p>"); 
       return false; 
      } 
      else { 
       $('#newsletter-form').submit(); 
      } 
     });  
    }); 

}); 

HTML：

<form action="" method="POST" id="newsletter-form"> 
    <input type="email" id="footer-grid1-newsletter-input" placeholder="Your Email Address" pattern=".{3,}" required> 
    <input type="submit" id="footer-grid1-newsletter-submit" name="submit" value='&nbsp'> 
</form> 

PHP：

$con = mysqli_connect("localhost", "", "", ""); 

if (mysqli_connect_errno()) { 
    printf("Connect failed: %s\n", mysqli_connect_error()); 
    exit(); 
} 


$sql = "SELECT email FROM newsletter WHERE email = '" .$_POST['email'] ."'"; 
$select = mysqli_query($con, $sql); 
$row = mysqli_fetch_assoc($select); 

if (mysqli_num_rows > 0) { 
    echo "exist"; 
} 
else { 
    $stmt = $con->prepare("INSERT INTO newsletter (email, subscribed) VALUES (?, NOW())"); 
}