提取URL和路径不同部分的正则表达式

Question

考虑一下URL，如

https://stackoverflow.com/v1/summary/1243PQ/details/P1/9981
http://stackoverflow.com/v2/summary/saas?test=123

我需要一个正则表达式来匹配这些URL并将它们转换为

stackoverflow.com:v1:summary:1243PQ:details:P1:9981
stackoverflow.com:v2:summary:saas

我需要使用regex构建一个规则，在这里我可以使用$1、$2等提取路径，而无需使用任何javascript逻辑，因为我需要在分类规则生成器工具中使用它。我尝试过这个URL包含^(([^:/?#]+):)?(//([^/?#]*))?([^?#]*)(\?([^#]*))?(#(.*))?，并提取了返回stackoverflow.com:v1/summary/1243PQ/details/P1/9981的$4:$5

但是，这是不正确的。有人能帮我找出正确的标准吗？

Answer 1

你可以试试这个：

正则表达式

/(?:https?:\/\/([^\/?\s#]+))?\/([^\/?\s#]*)(?:[\?#].*)?/g

替换

$1:$2

(?:                     non-capturing group
    https?:\/\/         "http://" or "https://"
    ([^\/?\s#]+)        capture the domain and put it in group 1
)?                      make this capture optional
\/                      "/"
([^\/?\s#]*)            one segment of the url path, capture it in group 2
(?:[\?#].*)?            an optional non-capturing group for consuming query string or # anchor at the end

检查测试用例

更新

如果不能使用g标志进行替换，那么没有其他更好的方法，只能野蛮地执行所有组合：

您需要为url路径的每个段添加一个\/([^\/?#\s]+)和:$2等：

• https://stackoverflow.com

^https?:\/\/(?:www\.)?([^\/?#\s]+)\/?(?:[#?].*)?$

$1

• https://stackoverflow.com/path1

^https?:\/\/(?:www\.)?([^\/?#\s]+)\/([^\/?#\s]+)\/?(?:[#?].*)?$

$1:$2

• https://stackoverflow.com/path1/path2

^https?:\/\/(?:www\.)?([^\/?#\s]+)\/([^\/?#\s]+)\/([^\/?#\s]+)\/?(?:[#?].*)?$

$1:$2:$3

• https://stackoverflow.com/path1/path2/path3

^https?:\/\/(?:www\.)?([^\/?#\s]+)\/([^\/?#\s]+)\/([^\/?#\s]+)\/([^\/?#\s]+)\/?(?:[#?].*)?$

$1:$2:$3:$4

• https://stackoverflow.com/path1/path2/path3/path4

^https?:\/\/(?:www\.)?([^\/?#\s]+)\/([^\/?#\s]+)\/([^\/?#\s]+)\/([^\/?#\s]+)\/([^\/?#\s]+)\/?(?:[#?].*)?$

$1:$2:$3:$4:$5

• https://stackoverflow.com/path1/path2/path3/path4/path5

^https?:\/\/(?:www\.)?([^\/?#\s]+)\/([^\/?#\s]+)\/([^\/?#\s]+)\/([^\/?#\s]+)\/([^\/?#\s]+)\/([^\/?#\s]+)\/?(?:[#?].*)?$

$1:$2:$3:$4:$5:$6