如果在聯繫表格輸出中發生錯誤，PHP將顯示一個html頁面

Question

我有一個帶有安全問題的簡單表單以防止垃圾郵件。 以問題http://ddry.co.uk/contact_us.html的形式鏈接到我的網頁。

我希望能夠輸出一個HTML頁面，如果用戶輸入一個不正確的答案，而不是純文本。

我有一個重定向到另一個html文件，如果表單成功在我的PHP腳本的底部。看着有人建議使用readfile的其他論壇（「contact-failed.html＃fail」）;顯示html;不過，我並不完全確定將錯誤答案的重定向代碼放在哪裏。我是PHP新手，所以如果有人能夠解釋我做錯了什麼，那會很好。或者，如果somone擁有更好的垃圾郵件過濾器代碼，那麼這也是非常有用的。

反垃圾郵件的html代碼

PHP文件的後。

----- UPDATE --------

我覺得我後是一個if，else語句？ 研究後，我改變了我的代碼，包括一個else語句;然而，由於我缺乏PHP知識，我仍然得到一個空白屏幕，而不是我的錯誤重定向html頁面，它顯示在我的PHP代碼的底部。

問題：如何正確配置if，else語句，如果反垃圾郵件結果錯誤（不等於12），請繼續聯繫failed.html？

在此先感謝

<?php 

    // Email address verification 
    function isEmail($clientEmail) { 
return filter_var($clientEmail, FILTER_VALIDATE_EMAIL);} 

if($_POST) { 

// Enter the email where you want to receive the message 
$myemail = 'info@ddry.co.uk'; 

$name = addslashes(trim($_POST['name'])); 
$clientEmail = addslashes(trim($_POST['email'])); 
$subject = addslashes(trim($_POST['phone'])); 
$phone = addslashes(trim($_POST['phone'])); 
$message = addslashes(trim($_POST['message'])); 
$antispam = addslashes(trim($_POST['antispam'])); 

$array = array('nameMessage' => '','emailMessage' => '', 'phoneMessage' => '', 'messageMessage' => '', 'antispamMessage' => ''); 


if(!isEmail($clientEmail)) { 
    $array['nameMessage'] = 'Empty name'; 
} 
if(!isEmail($clientEmail)) { 
    $array['emailMessage'] = 'Invalid email!'; 
} 
    if($phone == '') { 
    $array['phoneMessage'] = 'Empty phone number!'; 
} 
if($message == '') { 
    $array['messageMessage'] = 'Empty message!'; 
} 
if($antispam != '12') { 
    $array['antispamMessage'] = 'Incorrect Answer!'; 
} 
if(isEmail($clientEmail) && $clientEmail != '' && $message != '' &&  $antispam == '12') { 
    // Send email 
    $to = $myemail; 
$email_subject = "Contact form submission: $name"; 
$email_body = "You have received a new message. ". 
" Here are the details:\n Name: $name \n ". 
"Email: $clientEmail\n Message: \n $message\n Phone: $phone"; 
$headers = "From: $myemail\n"; 
$headers .= "Reply-To: $clientEmail"; 
mail($to,$email_subject,$email_body,$headers); 


echo json_encode($array); 

header('Location: contact-success.html#success'); 
} 

else (isEmail($clientEmail) && $clientEmail != '' && $message != '' &&  $antispam !== '12'){ 
     echo('Location: contact-failed.html#fail');} 
?> 

Answer 1

爲什麼不嘗試這樣簡單的事情？

function isEmail($clientEmail) { 
    return filter_var($clientEmail, FILTER_VALIDATE_EMAIL); 
} 

if($_POST){ 
    // Enter the email where you want to receive the message 
    $myemail = 'info@ddry.co.uk'; 
    $name = addslashes(trim($_POST['name'])); 
    $clientEmail = addslashes(trim($_POST['email'])); 
    $subject = addslashes(trim($_POST['phone'])); 
    $phone = addslashes(trim($_POST['phone'])); 
    $message = addslashes(trim($_POST['message'])); 
    $antispam = addslashes(trim($_POST['antispam'])); 
    if($antispam == '12' && isEmail($clientEmail) && $phone != '' && $message != '' ){ 
     $to = $myemail; 
     $email_subject = "Contact form submission: $name"; 
     $email_body = "You have received a new message. ". 
     " Here are the details:\n Name: $name \n ". 
     "Email: $clientEmail\n Message: \n $message\n Phone: $phone"; 
     $headers = "From: $myemail\n"; 
     $headers .= "Reply-To: $clientEmail"; 
     mail($to,$email_subject,$email_body,$headers); 
     echo json_encode($array); 
     header('Location: contact-success.html#success'); 
    } 
    else { 
     header('Location: contact-failed.html#fail'); 
    } 

Answer 2

我的問題是，爲什麼你需要表現出另一個頁面，而不是在接觸形式的錯誤消息。您正在重定向到另一個成功的頁面，同樣的錯誤也可以應用，您可以重定向到另一個html頁面以顯示不同的版本。