在javascript中执行php脚本？

Question

我试图在用户离开我的网站时运行一个快速的PHP脚本,并且还将一个变量从我的javascript传递给php,但我不太确定如何包含php文件并将其传递给var.但它实际上并没有运行PHP脚本.有任何想法吗？

(javascript从外部活动函数中获取用户名,我知道有效,我在一个警报及其那里测试了var.)

我的JavaScript:

<script language="javascript" type="text/javascript">

            var username = null;

            function GetUsername(usernameff)
            {
                username = usernameff;
            }

            window.onbeforeunload = function () 
            {

            if (username != null)
            {

            <?php include("scripts/RemoveUserOnDisconnect.php?username=username");?>
            }

            }


            </script>

我的RemoveUserOnDisconnect.php文件:

<?php
    mysql_connect("mysql.mysql.com", "username", "password");
    mysql_select_db("my_db");
    mysql_query("DELETE FROM my_table WHERE username = '$username'");         
?>

Answer 1

尝试ajax请求.根据您需要的PHP脚本$.post或$.get

jQuery的:

<script language="javascript" type="text/javascript">

    var username = null;

    function GetUsername(usernameff){
        username = usernameff;
    }

    window.onbeforeunload = function(){
        if (username != null){
          $.post('scripts/RemoveUserOnDisconnect.php', {username: username}, function(){
            //successful ajax request
          }).error(function(){
            alert('error... ohh no!');
          });

        }
    }
 </script>

编辑:

$_POST如果您使用上面的代码,您的PHP脚本应该引用该数组.

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

    mysql_connect("mysql.mysql.com", "username", "password");
    mysql_select_db("my_db");
    mysql_query("DELETE FROM my_table WHERE username = '$username'");         
?>