2014-03-02 56 views
0

好了失蹤,所以我創造了我的聯繫形式的PHP文件,這裏是我的PHP代碼(我已刪除了我的電子郵件)PHP數據變量從電子郵件

<?php 
/* Subject and Email Variables */ 


    $emailSubject = 'CV Website!'; 
    $email_to = '-----------'; 

/* Gathering Data Variables */ 


    $nameField = $_POST['name']; 
    $emailField = $_POST['email']; 
    $messageField = $_POST['message']; 

    $body = <<<EOD 
<br><hr><br> 
Name: $name <br> 
Email: $email <br> 
Message: $message <br> 
EOD; 

    $headers = "From: $email\r\n"; 
    $headers .="Content-type: text/html\r\n"; 
    $success = mail($email_to, $emailSubject, $body, $headers); 

這裏是連接形式於:

<form action="process.php" id="contact_form" method="post" class="wow 
fadeInUpBig" data-wow-offset="80" data-wow-duration="2s"> 

    <p> 
     <label>Name</label> 
     <input type="text" name="name" id="name" value="" class="requiredField name" /> 
    </p> 

    <p> 
     <label>Email Address</label>    
     <input type="text" name="email" id="email" value="" class="requiredField email" /> 
    </p> 

     <p> 
     <label>Your Message</label> 
     <textarea name="message" id="message" rows="8" cols="30" class="requiredField message"></textarea> 
     </p> 

     <p> 
     <input name="submitted" id="submitted" value="Submit" class="submit" type="submit" /> 
     </p> 
    </form> 

當我測試的形式,所有我打通了是這樣的消息:

名稱:

電子郵件:

消息:

而且沒有輸入數據,請如果你看到了什麼iv'e做錯了幫助。在此先感謝

+1

有什麼方法你的表單? – Fabio

+1

嘗試開啓錯誤報告 – les

+1

首先,您的PHP變量名稱與您設置爲電子郵件正文的輸入內容不匹配。 – les

回答

0

首先你應該嘗試,因爲你需要這則更改:

$nameField = $_POST['name']; 
$emailField = $_POST['email']; 
$messageField = $_POST['message']; 

$body = <<<EOD 
<br><hr><br> 
Name: $nameField <br> 
Email: $emailField <br> 
Message: $messageField <br> 
EOD; 

$headers = "From: $emailField\r\n"; 
$headers .="Content-type: text/html\r\n"; 
$success = mail($email_to, $emailSubject, $body, $headers); 
+0

謝謝你修復它。 :) – OfficiallyFrex

0

我只是猜測,你需要這個標記爲表單:

<form method="post" ... > 
     ... <!-- Input Fields go here --> 
</form> 
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