php获取当前访问的文件名
方法一:
<?php
$url = $_SERVER[’PHP_SELF’];
$filename= substr( $url , strrpos($url , ‘/’)+1 );
echo $filename;
?>
方法二:
<?php
$url = $_SERVER[’PHP_SELF’];
$arr = explode( ‘/’ , $url );
$filename= $arr[count($arr)-1];
echo $filename;
?>
方法三:
<?php
$url = $_SERVER[’PHP_SELF’];
$filename = end(explode(’/',$url));
echo $filename;
?>
地址栏:$_server['Info_path'];
上传文件:$_file['upfile']['name']
<?php
//获取域名或主机地址
echo $_SERVER['HTTP_HOST']."<br>";
//获取网页地址
echo $_SERVER['PHP_SELF']."<br>";
//获取网址参数
echo $_SERVER["QUERY_STRING"]."<br>";
//来源网页的详细地址
echo $_SERVER['HTTP_REFERER']."<br>";
?>


有没有更好的方法呢?
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