如何使用 preg_match 获取 URL 的特殊部分？

Question

我有一个像这样的url，

https://example.com/folder-name/article-name-xxx-xxx-xxx-xxx-xxx-xxx-5b5964935583202d2beff315.html#id-41

我想要做的是在 url.

中获取 5b5964935583202d2beff315 和 41

我真的很想知道怎么做，我需要帮助。非常感谢您的帮助！

Answer 1

$url = "https://example.com/folder-name/dien-hy-cong-luoc-story-of-yanxi-palace-5b5964935583202d2beff315.html#id-41";

preg_match("/^.+-([^.-]+)\.html#id-(\d+)/", $url, $matches);
print_r($matches);

输出：

Array
(
    [0] => https://example.com/folder-name/dien-hy-cong-luoc-story-of-yanxi-palace-5b5964935583202d2beff315.html#id-41
    [1] => 5b5964935583202d2beff315
    [2] => 41
)

解释：

/               : regex delimiter
  ^             : beginning of line
    .+          : 1 or more any character but newline
    -           : a dash
    ([^.-]+)    : group 1, 1 or more any character that is not a dot or dash
    \.          : a dot
    html#id-    : literally 
    (\d+)       : group 2, 1 or more digits
/

如何使用 preg_match 获取 URL 的特殊部分？

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