在 javascript 中执行 php 脚本？

发布于 2024-10-20 21:46:18 字数 945 浏览 1 评论 0原文

我试图在用户离开我的网站时运行一个快速的 php 脚本，并将一个变量从我的 javascript 传递到 php，但我不太确定如何包含 php 文件并向其传递一个 var。但它实际上并没有运行 php 脚本。有什么想法吗？

（javascript从外部活动函数获取用户名，我知道我在警报上测试了var并且它在那里。）

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'");         
?>

原文

I'm trying to run a quick php script when users leave my website, and also pass a variable from my javascript to php but i'm not quite sure how to include the php file and pass it a var. But it's not actually running the php script. any ideas?

(the javascript gets the username from an external activity function, and i know that works i tested the var on an alert and its there.)

My 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>

My RemoveUserOnDisconnect.php File:

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

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

日久见人心 2024-10-27 21:46:18

尝试 ajax 请求。根据您的 php 脚本，您将需要 $.post 或 $.get

jQuery：

<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：

您的 php 脚本应该引用 $_POST 数组，如果你使用我上面的代码。

<?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'");         
?>

Try an ajax request. Depending on your php script you will need $.post or $.get

jQuery:

<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>

EDIT:

Your php script should reference the $_POST array if you use my code above.

<?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'");         
?>

回复收藏 0 原文