仅从 URL 获取 URI 段 [英] Get only URI segments from URL

问题描述

我正在尝试使用正则表达式获取 URI 段.

I am trying to get the URI segments using regular expression.

示例 URI:

http://abc.com/hello/hi/bye?humm/ok=hi&ya=yaya/wow/waaah
               ^^^^^ ^^ ^^^                    ^^^ ^^^^^

我正在尝试:

/(?<=\/)[\w-]+(?=(\/|$|\r|\?))/g

但是它不能正常工作.查询字符串没有被排除 (wow/waaah).

But it's not working properly. The query string is not getting excluded (wow/waaah).

因此，当我尝试以下操作时，一切都被排除在外:

So, when I tried the following, everything got excluded:

/(?<!?.+)(?<=\/)[\w-]+(?=(\/|$|\r|\?))/g

这有什么问题吗?

推荐答案

您忘记转义第二个正则表达式中的第二个 ?.它应该是:

You forgot to escape the second ? in the second regex. It should read:

<代码>/(?

注意:您可以使用像这样的字符类来改进正则表达式:

Note: You could improve the regex by using character classes like so:

/(?<!\?.+)(?<=\/)[\w-]+(?=[/\r\n?]|$)/g

对于满足所有不同风格的正则表达式的最低公分母解决方案，您需要一个两步过程:

For a lowest common denominator solution to cater for all the different flavours of regex, you need a two step process:

• 删除尾随的 ? 和所有后续字符(如果存在):
  

  
  ^[^/]+//[^/]+([^?]+)

  保留捕获组 1 中返回的字符串.

• 通过循环提取 URI 段:

  
  /([\w-]+)

  这些段在捕获组 1 中返回.

• Remove the trailing ? and all following chars (if it exists):
  

  
  ^[^/]+//[^/]+([^?]+)

  Keep the string returned in capture group 1.

• Extract the URI segments by looping through:

  
  /([\w-]+)

  The segments are returned in capture group 1.

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