用 if then else 重写 url 的正则表达式

Question

鉴于这些网址：

1: http://site/page-name-one-123/
2: http://site/page-name-set2/
3: http://site/set20

我写了这个表达式将应用于最后一个 url 段：

(?(?<=set[\d])([\d]+)|([^/]+))

我想要做的是仅当 url 段以 'set' 开头并且紧随其后的数字时才捕获后面跟着 'set' 的每个数字；否则我想使用整个段（不包括斜线）。

当我编写这个正则表达式时，它匹配任何不是“/”的字符。我认为我在测试语句中做错了什么。 有人能指点我吗？

谢谢

更新 感谢Josh 输入，我玩了一会儿，发现这个更适合我的需求：

set-(?P<number>[0-9]+)|(?P<segment>[^/]+)

Answer 1

我希望这个模式可以帮助你，我根据你的要求把它放在一起。您可能想尝试将某些组设置为不捕获，以便仅获得所需的片段。但是，它确实会在开始时单独捕获您的 set URL，而没有 set。

((?<=/{1})(((?<!set)[\w|-]*?)(\d+(?=/?))|((?:set)\d+)))

如果需要，我建议使用RegExr 将其拆开。

Answer 2

试试这个：

((?<=/)set\d+|(?<=/)[^/]+?set\d+)

说明

<!--
Options: ^ and $ match at line breaks

Match the regular expression below and capture its match into backreference number 1 «((?<=/)set\d+|(?<=/)[^/]+?set\d+)»
   Match either the regular expression below (attempting the next alternative only if this one fails) «(?<=/)set\d+»
      Assert that the regex below can be matched, with the match ending at this position (positive lookbehind) «(?<=/)»
         Match the character “/” literally «/»
      Match the characters “set” literally «set»
      Match a single digit 0..9 «\d+»
         Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
   Or match regular expression number 2 below (the entire group fails if this one fails to match) «(?<=/)[^/]+?set\d+»
      Assert that the regex below can be matched, with the match ending at this position (positive lookbehind) «(?<=/)»
         Match the character “/” literally «/»
      Match any character that is NOT a “/” «[^/]+?»
         Between one and unlimited times, as few times as possible, expanding as needed (lazy) «+?»
      Match the characters “set” literally «set»
      Match a single digit 0..9 «\d+»
         Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
-->