【发布时间】:2009-09-29 09:03:54
【问题描述】:
我正在尝试获取正在运行的脚本的文件名(但不是它正在调用的包含)。
echo basename(__FILE__); # will always output include.php
echo basename($_SERVER['SCRIPT_FILENAME']);
# This will do what I want (echo myscript.php), but I was wondering if there was
# a better way to grab it, as I have had problems with $_SERVER['SCRIPT_FILENAME']
# when running certain scripts from a cron.
有什么建议吗?
<?
#myscript.php
require('include.php');
echo "Hello all";
?>
<?
#include.php
echo basename(__FILE__);
echo basename($_SERVER['SCRIPT_FILENAME']);
?>
谢谢!
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