php / ajax表单发送邮件并将数据保存到db

Question

我有一个现在可以使用的联系表单，我需要在发送邮件之前将数据存储到db。

FORM：只是一封简化此代码的电子邮件和消息：

<fieldset id="contact_form">
 <div id="result"></div>
 <label for="email"><span>Email : *</span>
   <input type="email" name="email" id="email"  />
 </label>
 <label for="message"><span>Message : *</span>
   <textarea name="message" id="message" ></textarea>
 </label>
 <label>
   <button class="submit_btn" id="submit_btn">Envoyer</button>
 </label>
</fieldset>

继续：使用ajax将数据发送到contact.php

$(document).ready(function() {
$("#submit_btn").click(function() {
var user_email      = $('input[name=email]').val();
var user_message    = $('textarea[name=message]').val();
if(proceed)
{
var post_data = new FormData();
post_data.append( 'userEmail', user_email );
post_data.append( 'userMessage',user_message);
$.ajax({
   url: 'contactus.php',
   data: post_data,
   processData: false,
   contentType: false,
   type: 'POST',
   dataType:'json',
   success: function(data){
   if(data.type == 'error')
     {
     output = '<div class="error">'+data.text+'</div>';
     }else{
     output = '<div class="success">'+data.text+'</div>';
     }
    }
  });
 }
 });
});

CONTACT.php：获取表单数据并发送邮件和显示消息（错误或成功）

include ("connexion.php");
if($_POST)
{
// SAVE DATA
$sql = 'INSERT INTO historique_contact VALUES ("", "tibo", "message")';
mysql_query ($sql) or die ('Erreur SQL !'.$sql.'<br />'.mysql_error());
// SEND DATA
$to     = "mail@gmail.com"; 
$from   = $_POST["userEmail"]; 
$subject        = 'Contact'; 
$user_Email       = filter_var($_POST["userEmail"], FILTER_SANITIZE_EMAIL);
$user_Message     = filter_var($_POST["userMessage"], FILTER_SANITIZE_STRING);
$mail_body       .="Email:".$user_Email;
$mail_body       .="Message:".$user_Message; 
$headers =  'From: '.$from_Email.'' . "\r\n" .
            'Content-Type: text/html; charset="UTF-8"';
            'X-Mailer: PHP/' . phpversion();
$sentMail = mail($to_Email, $subject, $mail_body, $headers);
if(!$sentMail) //output success or failure messages
{
    $output = json_encode(array('type'=>'error', 'text' => 'Error...'));
    die($output);
}else{
    $output = json_encode(array('type'=>'message', 'text' => 'Success'));
    die($output);
}
}