JavaScript 从联系表单获取访问者的先前 URL

javascript to get previous url of visitor from contact form

本文关键字：URL 访问者 获取 联系 表单 JavaScript      更新时间：2023-09-26

我想跟踪我的潜在客户他们访问过我的网站的URL（如fb，g +或mailer）。这一切都是为了了解访问者以获得更好的数字营销。

我的小提琴：http://jsfiddle.net/dSp4g/

请帮助我编写以下代码

提前致谢

.HTML 

<div class="form-group">
<label for="exampleInputEmail">Full name</label>
<input type="text" class="form-control " id="exampleInputName" placeholder="Enter your full name" style="color:#000 !important;">
 </div>
<div class="form-group">
<input type="hidden" class="form-control " id="sourcecode">
</div>                                          
<div class="form-group last">
<button class="mailbtn">Submit</button>
</div>

.JS

$('.mailbtn').live('click',function(){
            name = $('#exampleInputName').val();
            sc = $('#sourcecode').val();
            $(document).ready(function() { 
                if (sourcecode = $.parsequery().s) 
                    {
                        $("#source").val(sourcecode); }
             });

            $.ajax({
            type: "POST",
            async : false,
            url: "mail.php",
                data: { name:name, sourcecode:sc}
            })
            .done(function( msg ) {
            $('.mail_middle').html('');
            $('.mail_middle').html('We will call you to confirm your delivery address.Thank you.');
            return false;
            });

    });

邮件.php

<?php
$to =  array("my_email1","my_email2");
$message .= "<table border='1'>";
$message .= "<tr><td>Name    </td><td>".$_POST['name']."</td></tr>";
$message .= "<tr><td>Source   </td><td>".$_POST['sc']."</td></tr>";
$message .= "</table>";
$headers = "MIME-Version: 1.0'r'n"; 
$headers .= "Content-type: text/html; charset=utf-8'r'n"; 
$headers .=  'from: '.$from .'' . "'r'n" .
            'Reply-To: '.$from.'' . "'r'n" .
            'X-Mailer: PHP/' . phpversion();
foreach($to as $row)
{
   mail($row,$subject,$message,$headers);
}
echo "Mail Sent.";
die;
?>

更改$_POST['sc'] 

$message .= "<tr><td>Source   </td><td>".$_POST['sc']."</td></tr>";

到$_POST['sourcecode'] 

$message .= "<tr><td>Source   </td><td>".$_POST['sourcecode']."</td></tr>";

因为sourcecode是名称（其中sc是值）

data: { name:name, sourcecode:sc}

HTML

添加一个

<input type="hidden" name="the_ref" value="<?= $_SERVER['HTTP_REFERER']; ?>">

.JS

修改为

data: { name:name, sourcecode:sc, the_ref:ref}

.PHP

添加一个

$message .= "<tr><td>Visitor came from </td><td>".$_POST['the_ref']."</td></tr>";

你的jQuery实现失败了，因为你在PHP中你试图读取$_POST['sc']而不是$_POST['sourcecode']

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