在javascript执行php脚本
问题说明
当用户离开我的网站时，我正在尝试运行一个快速的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。任何想法？
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?
（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。）
(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.)
我的JavaScript：
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>
我的RemoveUserOnDisconnect.php文件：
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'");
?>
正确答案
尝试ajax请求。根据您的PHP脚本，您需要 $。发布或 $。获取
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>
编辑：
你的php脚本应该如果您使用我上面的代码，请参考 $ _ POST 数组。
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'");
?>
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