我想在输入中列出所有匹配的组。例如,我想在User-Agent HTTP标头中打印我关心的所有浏览器。一场比赛就足够了。有没有办法摆脱下面的代码中的内部for循环。我查看了boost / regex / sub_match.hpp,我没有任何想法。
#include <string>
#include <iostream>
#include <boost/regex.hpp> // Boost 1.59, no C++14 for me
enum BrowserType
{
FIREFOX = 0 ,
CHROME,
SAFARI,
OPERA,
IE,
EDGE,
OTHER,
};
const boost::regex BROWSERS_REGEX("(Firefox)|(Chrome)|(Safari)|(Opera)|(MSIE)|(Edge)|(Trident)");
int main()
{
// I expect two matches here CHROME and SAFARI
std::string input("Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Safari/537.36 Chrome/62.0.3202.94");
boost::sregex_iterator res(input.begin(), input.end(), BROWSERS_REGEX);
boost::sregex_iterator end;
for(; res != end; ++res)
{
// elude copy here ?
boost::smatch what = *res;
// Can I know the index of the matching group w/o 'for'?
for (int type = 0;type < OTHER;type++)
{
int groupIndex = type+1;
if (what[groupIndex].matched)
std::cout << (BrowserType)type << ",";
}
std::cout << "\n";
}
return 0;
}
重要更新。基于strstr()的代码是短于1K字节的字符串的最快方法
const char *input("Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Safari/537.36 Chrome/62.0.3202.94");
const char *browsers[OTHER+1] = {"Firefox", "Chrome", "Safari", "Opera", "MSIE", "Edge", "Trident"};
int i = 0;
for (const char **browser = &browsers[0];browser <= &browsers[OTHER];browser++, i++)
{
if (strstr(input, *browser))
{
groupindex = i;
}
}
答案 0 :(得分:1)
您可以使用命名捕获。
在你的情况下,我会使用qi :: symbols:
struct browser_type_sym : boost::spirit::qi::symbols<char, BrowserType> {
browser_type_sym() {
this->add
("Firefox", FIREFOX)
("Chrome", CHROME)
("Safari", SAFARI)
("Opera", OPERA)
("MSIE", IE)
("Edge", EDGE)
("Trident", OTHER);
}
} static const browser_type;
您只需将其与BrowserType的任何容器一起使用:
template <typename Types>
bool extract_browser_ids(std::string const& userAgent, Types& into) {
using boost::spirit::repository::qi::seek;
return parse(userAgent.begin(), userAgent.end(), *seek [ browser_type ], into);
}
vector<BrowserType>
正如您将看到的,如果您使用vector<BrowserType
,它将保留订单并重复:
<强> Live On Coliru 强>
int main() {
for (std::string const input : {
"Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Safari/537.36 Chrome/62.0.3202.94",
"Chrome; Opera; Chrome again!"
}) {
std::vector<BrowserType> types;
extract_browser_ids(input, types);
for(auto type : types)
std::cout << type << ",";
std::cout << "\n";
}
}
打印:
2,1,
1,3,1,
set<BrowserType>
使用set<>
时,它会订购和删除重复:
<强> Live On Coliru 强>
std::set<BrowserType> types;
打印:
1,2,
1,3,
只是为了显示小调整:
struct browser_type_sym : boost::spirit::qi::symbols<char, BrowserType> {
browser_type_sym() {
this->add
("firefox", FIREFOX)
("chrome", CHROME)
("safari", SAFARI)
("opera", OPERA)
("msie", IE)
("edge", EDGE)
("trident", OTHER);
}
} static const browser_type;
template <typename Types>
bool extract_browser_ids(std::string const& userAgent, Types& into) {
using boost::spirit::repository::qi::seek;
using boost::spirit::qi::no_case;
return parse(userAgent.begin(), userAgent.end(), *seek [ no_case [ browser_type ] ], into);
}
现在情况并不重要: Live On Coliru
同样,您可以使用Boost Xpressive,它稍微接近regex
方法(尽管internally it still builds a trie of strings from the map)。
语义动作需要更多的手工劳动,使其也不那么通用(如果没有更改,它将无法与std::set
一起使用,这与已经显示的Spirit Qi方法不同。)
然而,为了完整性:
<强> Live On Coliru 强>
std::map<std::string, BrowserType> s_browser_type_map {
{"Firefox", FIREFOX},
{"Chrome", CHROME},
{"Safari", SAFARI},
{"Opera", OPERA},
{"MSIE", IE},
{"Edge", EDGE},
{"Trident", OTHER},
};
#include <boost/xpressive/xpressive.hpp>
#include <boost/xpressive/regex_actions.hpp>
template <typename Types>
void extract_browser_ids(std::string const& userAgent, Types& into) {
using namespace boost::xpressive;
placeholder<Types> _result;
sregex type = (a1 = s_browser_type_map) [ push_back(_result, a1) ];
for (sregex_iterator it(userAgent.begin(), userAgent.end(), type, let(_result=into)),
end; it != end; ++it)
{ } // all side-effects in the semantic action
}
打印与vector<>
Spirit示例相同的输出:
2,1,
1,3,1,