Question

我有一个简单的网址：

http://sample.com/?page=1&page=2
http://sample.com/?page=2&page=1

如何使用preg_replace（）删除值副本（页面）

这是代码：

$link = preg_replace("/\b(\w+)\s+\\1\b/i", "$1", $link );

但结果是错误的，我如何修复它，结果如下：

http://sample.com/?page=1&page=2 => http://sample.com/?page=2
http://sample.com/?page=2&page=1 => http://sample.com/?page=1

Answer 1

您可以考虑递归子模式，而不是尝试使用反向引用。

$link = 'http://sample.com/?page=1&page=2';
$link = preg_replace('/\b(\w+=\d+)&((?1))/', '$2', $link);
echo $link; //=> "http://sample.com/?page=2"

使用反向引用我想你可以这样做：

$link = preg_replace('/\b(\w+=)\d+&(\1\d+)/', '$2', $link);

Answer 2

替换：

\?.*?\K([^=]+)(?:=[^&]*)?&(.*)\1=(.*?)(?:&|$)

使用：

\2\1=\3

Demo

\?      (?# match ? literally)
.*?     (?# lazily match unnecessary params)
\K      (?# throw away everything to the left)
([^=]+) (?# capture 1+ non-= characters)
(?:     (?# start non-capturing group)
  =     (?# match = literally)
  [^&]* (?# match 0+ non-& characters)
)?      (?# end optional parameter value)
&       (?# match & literally)
(.*)    (?# capture optional inbetween junk)
\1     (?# a paramater that has already been seen in capture group 1)
=       (?# literally match =)
(.*?)   (?# lazily capture the newest value)
(?:     (?# begin non-capturing group)
  &     (?# literally match &)
 |      (?# OR)
  $     (?# match the end of the URL)
)       (?# end non-capturing group)

如何使用preg_replace（）从URL中删除值副本

2 个答案: