PHP：获取页面URL参数减 [英] PHP: Get Page URL minus arguments

问题描述

我如何获取当前页面的URL减去所有的得到的参数（废话= 2及？blah4 = 90 ...）
我知道我能得到完整的URL以$ _ SERVER ['REQUEST_URI']，但我想知道是否有一些更适合我的需要。

How do i get the URL of the current page minus all of the get arguments (?blah=2&blah4=90...) I know i can get the full URL with $_SERVER['REQUEST_URI'] but i was wondering if there was something that more fit my needs.

或者我应该只是做strpos？和SUBSTR砍的争论？ （我想，一个$ _ SERVER变种会更有效 - 如果存在的话）

Or should i just do strpos ? and substr to chop of the arguments? ( i imagine that a $_SERVER var would be more efficient - if one exists)

感谢

推荐答案

$ _ SERVER ['REQUEST_URL']

有时，答案是简单的:)我发现这里的解决方案：的http：// php.net/manual/en/reserved.variables.server.php~~V 通过CTRL + F'ing为查询。

Sometimes answers are easy :) I found the solution here: http://php.net/manual/en/reserved.variables.server.php by CTRL+F'ing for "query".

修改

由于matchu在评论中说，并不是所有的服务器都支持REQUEST_URL。在这种情况下，我会用更优雅的的strtok（$网址，'？'）;

As matchu said in the comments, not all servers support REQUEST_URL. In that case I would use the much less elegant strtok($url, '?');.

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