c++ - 获取 URL token 的正则表达式是什么？

Question

假设我有这样的字符串:

bunch of other html<a href="http://domain.com/133742/The_Token_I_Want.zip" more html and stuffbunch of other html<a href="http://domain.com/12345/another_token.zip" more html and stuffbunch of other html<a href="http://domain.com/0981723/YET_ANOTHER_TOKEN.zip" more html and stuff

匹配 The_Token_I_Want、another_token、YET_ANOTHER_TOKEN 的正则表达式是什么？

Answer 1

RFC 2396 的附录 B给出了一个用于将 URI 拆分为其组件的正则表达式，我们可以根据您的情况对其进行调整

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

这在 $6 中留下了 The_Token_I_Want，这是上面的“hashderlined”子表达式。 (请注意，哈希不是模式的一部分。)现场观看:

#! /usr/bin/perl

$_ = "http://domain.com/133742/The_Token_I_Want.zip";    
if (m!^(([^:/?#]+):)?(//([^/?#]*))?([^?#]*/([^.]+)[^?#]*)(\?([^#]*))?(#(.*))?!) {
  print "$6\n";
}
else {
  print "no match\n";
}

输出:

$ ./prog.plThe_Token_I_Want

UPDATE: I see in a comment that you're using boost::regex, so remember to escape the backslash in your C++ program.

#include <boost/foreach.hpp>
#include <boost/regex.hpp>
#include <iostream>
#include <string>

int main()
{
  boost::regex token("^(([^:/?#]+):)?(//([^/?#]*))?([^?#]*"
                     "/([^.]+)"
                   //  ####### I CAN HAZ HASHDERLINE PLZ
                     "[^?#]*)(\\?([^#]*))?(#(.*))?");

  const char * const urls[] = {
    "http://domain.com/133742/The_Token_I_Want.zip",
    "http://domain.com/12345/another_token.zip",
    "http://domain.com/0981723/YET_ANOTHER_TOKEN.zip",
  };

  BOOST_FOREACH(const char *url, urls) {
    std::cout << url << ":\n";

    std::string t;
    boost::cmatch m;
    if (boost::regex_match(url, m, token))
      t = m[6];
    else
      t = "<no match>";

    std::cout << "  - " << m[6] << '\n';
  }

  return 0;
}

输出:

http://domain.com/133742/The_Token_I_Want.zip:  - The_Token_I_Wanthttp://domain.com/12345/another_token.zip:  - another_tokenhttp://domain.com/0981723/YET_ANOTHER_TOKEN.zip:  - YET_ANOTHER_TOKEN