php 使用 URL 参数获取当前文件名
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Get current filename with URL parameters
提问by Mooseman
Say the user is loading the following page: www.example.com/myDirectory/myPage.php?answer=yes
And I want to use PHP to get myPage.php?answer=yesfrom that. However, basename($_SERVER['PHP_SELF'])only returns myPage.php. How would I do that?
假设用户正在加载以下页面：www.example.com/myDirectory/myPage.php?answer=yes
我想使用 PHP 从中获取myPage.php?answer=yes。但是，basename($_SERVER['PHP_SELF'])只返回myPage.php. 我该怎么做？
回答by Blazemonger
You can append the query string after you getit as an environment variable:
$myurl = basename($_SERVER['PHP_SELF']) . "?" . $_SERVER['QUERY_STRING'];
If you need to consider the possibility of no query string, add a conditional test:
如果需要考虑没有查询字符串的可能性，添加条件测试：
$myurl = strlen($_SERVER['QUERY_STRING']) ? basename($_SERVER['PHP_SELF'])."?".$_SERVER['QUERY_STRING'] : basename($_SERVER['PHP_SELF']);
回答by G?kalp Turan
it is easier?:
它更容易吗？：
basename($_SERVER['REQUEST_URI'])
回答by Shiv Singh
For URLs with trailing parameters the following will place the first part of the current URL (i.e. the part before ?) within the variable $current_url[0] and then print it out.
对于带有尾随参数的 URL，以下内容会将当前 URL 的第一部分（即 ? 之前的部分）放在变量 $current_url[0] 中，然后将其打印出来。
<?
$current_url = explode("?", $_SERVER['REQUEST_URI']);
echo $current_url[0] ;
?>
回答by draxxxeus
$_SERVER['QUERY_STRING']
will give you the paramaters
会给你参数
