perl 正则表达式 - 模式匹配

Question

谁能解释下面正在做什么？

 $name=~m,common/([^/]+)/run.*/([^/]+)/([^/]+)$,;

Answer 1

common , run and / are match themselves. common ， run和/是匹配自己的。
() captures. ()捕获。
[^/]+ matches 1 or more characters that aren't / . [^/]+匹配 1 个或多个不是/字符。
.* matches 0 or more characters that aren't Line Feeds. .*匹配 0 个或多个不是换行符的字符。 ^[1] ^[1]
$ is equivalent to ( \\n?\\z ). $等价于 ( \\n?\\z )。 ^[2] ^[2]
\\n optionally matches a Line Feed. \\n可选地匹配换行符。
\\z matches the end of the string. \\z匹配字符串的结尾。

I think it's trying to match a path of one or both of the following forms:我认为它试图匹配以下一种或两种形式的路径：

Where在哪里

XXX is a sequence of at least one character that doesn't contain / . XXX是至少一个不包含/字符的序列。
YYY is a sequence of any number of characters (incl zero) that doesn't contain / . YYY是不包含/的任意数量字符（包括零）的序列。

It matches more than that, however.然而，它比这更匹配。

It matches uncommon/ XXX /runYYY/ XXX / XXX它匹配uncommon/ XXX /runYYY/ XXX / XXX
It matches common/ XXX /runYYY/XXX/XXX/XXX/XXX/ XXX / XXX它匹配common/ XXX /runYYY/XXX/XXX/XXX/XXX/ XXX / XXX

The parts in bold are captured (available to the caller).粗体部分被捕获（可供调用者使用）。