PHP get方法AJAX URL读

Question

我有这样的URL

site.com/#request/values-1

site.com/#request/values-2

我如何取这个网址？与PHP你认为这是关于什么？

$request = ? 

Answer 1

编辑︰你需要使用JavaScript来实现。

<script language = "javascript"> 

var query = location.href.split('#'); 

var anchorValueSplit = query[1].split("/"); 

var anchorValue = anchorValueSplit[1]; 

alert(anchorValue); 

</script> 

编辑：另一个解决方案

<script language = "javascript"> 

if(location.href.indexOf("#") > 0) { 

    location.href = location.href.replace("#","?"); 

} 

</script> 
<?php 

$param = array_keys($_GET); 
$param = $param[0]; 

$breakParam = explode("/",$param); 

echo $request = $breakParam[1]; 


?>