C++ 将第一个 boost::regex 匹配放入一个字符串中
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Put first boost::regex match into a string
提问by Tomá? Zato - Reinstate Monica
Somehow, I've failed to find out, how to put only the first occurrence or regular expression to string. I can create a regex object:
不知何故，我没有找到，如何只将第一次出现或正则表达式放入字符串。我可以创建一个正则表达式对象：
static const boost::regex e("<(From )?([A-Za-z0-9_]+)>(.*?)");
Now, I need to match ([A-Za-z0-9_]+)to std::string, say playername.
现在，我需要匹配([A-Za-z0-9_]+)到std::string，说playername。
std::string chat_input("<Darker> Hello");
std::string playername = e.some_match_method(chat_input, 1); //Get contents of the second (...)
What have I missed?
What should be instead of some_match_methodand what parameters should it take?
我错过了什么？
应该是什么而不是some_match_method它应该采用什么参数？
回答by Loghorn
You can do something like this:
你可以这样做：
static const regex e("<(From )?([A-Za-z0-9_]+)>(.*?)");
string chat_input("<Darker> Hello");
smatch mr;
if (regex_search(begin(chat_input), end(chat_input), mr, e)
string playername = mr[2].str(); //Get contents of the second (...)
Please note that regex is part of C++11, so you don't need boost for it, unless your regular expression is complex (as C++11 and newer still has difficulties processing complex regular expressions).
请注意，regex 是 C++11 的一部分，因此您不需要提升它，除非您的正则表达式很复杂（因为 C++11 和更新版本仍然难以处理复杂的正则表达式）。
回答by Dave S
I think what you're missing is that boost::regexis the regular expression, but it doesn't do the parsing against a given input. You need to actually use it as a parameter to boost::regex_searchor boost::regex_match, which evaluate a string (or iterator pairs) against the regular expression.
我认为您缺少的是boost::regex正则表达式，但它不会对给定的输入进行解析。您需要实际将其用作boost::regex_searchor的参数boost::regex_match，它会根据正则表达式计算字符串（或迭代器对）。
static const boost::regex e("<(From )?([A-Za-z0-9_]+)>(.*?)");
std::string chat_input("<Darker> Hello");
boost::match_results<std::string::const_iterator> results;
if (boost::regex_match(chat_input, results, e))
{
std::string playername = results[2]; //Get contents of the second (...)
}
