如何包括另一个PHP文件? [英] how to include another php file?
问题描述
我有一个php文件,我想包含另一个具有css链接标签和javascript源标签的php文件,但是当我尝试包含它们时,它并没有添加到页面中.
I have a php file, and I want to include another php file that have css link tags and javascript source tags, but when I try to include them, it doesn't get added to the page.
我的php页面:
<?php
$root = $_SERVER['SERVER_NAME'] . '/mysite';
$theme = $root . '/includes/php/common.php';
echo $theme;
include($theme);
?>
common.php:
common.php:
<link rel='stylesheet' type='text/css' href='../css/main.css'/>";
有人知道怎么了吗?谢谢
Anyone know whats wrong? Thanks
推荐答案
PHP的 include
是服务器端的,因此您需要使用服务器端的路径.最好使用dirname(__FILE__)
代替$_SERVER['SSCRIPT_NAME']
,但是$_SERVER['SERVER_NAME']
绝对错误.
PHP's include
is server-side, so you need to use the server side path. It is better to use dirname(__FILE__)
instead of $_SERVER['SSCRIPT_NAME']
, but $_SERVER['SERVER_NAME']
is absolutely wrong.
尝试:
include dirname(__FILE__)."/common.php";
或者,如果要包含的文件不在同一目录中,请更改路径.例如,对于父目录,请使用dirname(__FILE__)."/../common.php"
.
Or if the file you want to include is not on the same directory, change the path. For example for a parent directory, use dirname(__FILE__)."/../common.php"
.
请注意,有些人可能建议使用include "./common.php"
或类似名称.此可以起作用,但是当调用include
的脚本实际上已被另一个目录中的另一个脚本包含时,很可能会失败.使用dirname(__FILE__)."/common.php"
可以消除此问题.
Note that some might suggest using include "./common.php"
or similar. This could work, but will most likely fail when the script invoking include
is actually being included by another script in another directory. Using dirname(__FILE__)."/common.php"
will eliminate this problem.
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