regex - 匹配 URL 的路径，减去文件扩展名

Question

这种情况下最好的正则表达式是什么？

鉴于此网址:

http://php.net/manual/en/function.preg-match.php

我应该如何选择(但不包括)之间的所有内容 http://php.net和 .php :

/manual/en/function.preg-match

这是一个 Nginx配置文件。

Answer 1

像这样:

if (preg_match('/(?<=net).*(?=\.php)/', $subject, $regs)) {
    $result = $regs[0];
}

解释:

"
(?<=      # Assert that the regex below can be matched, with the match ending at this position (positive lookbehind)
   net       # Match the characters “net” literally
)
.         # Match any single character that is not a line break character
   *         # Between zero and unlimited times, as many times as possible, giving back as needed (greedy)
(?=       # Assert that the regex below can be matched, starting at this position (positive lookahead)
   \.        # Match the character “.” literally
   php       # Match the characters “php” literally
)
"