Question

我有std::string s=n8Name4Surname。如何在2个字符串中获取姓名和姓氏？ THX

Answer 1

执行此操作的一种方法是使用Boost.Tokenizer。见这个例子：

#include <string>
#include <boost/tokenizer.hpp>
#include <boost/foreach.hpp>
int main()
{
    using namespace std;
    using namespace boost;
    string text="n8Name4Surname.";

    char_separator<char> sep("0123456789");
    tokenizer<char_separator<char> > tokens(text, sep);

    string name, surname;
    int count = 0;
    BOOST_FOREACH(const string& s, tokens)
    {
        if(count == 1)
        {
            name = s;
        }
        if(count == 2)
        {
            surname = s;
        }
        ++count;
    }
}

修改

如果您将结果放在vector中，则代码更少：

#include <string> #include <boost/tokenizer.hpp> #include <boost/foreach.hpp> #include <algorithm> #include <iterator> #include <vector> int main() { using namespace std; using namespace boost; string text="n8Name4Surname."; char_separator<char> sep("0123456789"); tokenizer<char_separator<char> > tokens(text, sep); vector<string> names; tokenizer<char_separator<char> >::iterator iter = tokens.begin(); ++iter; if(iter != tokens.end()) { copy(iter, tokens.end(), back_inserter(names)); } }

Answer 2

您可以使用函数isdigit(mystring.at(position)检测字符串中的数字字符，然后在这些位置之间提取子字符串。

请参阅：

http://www.cplusplus.com/reference/clibrary/cctype/isdigit/

Answer 3

使用数字0-9的Boost tokenizer作为分隔符。然后，扔掉包含“n”的字符串。这太过分了，我意识到......

Answer 4

简单的STL方法：

#include <string>
#include <vector>
#include <iostream>

int main()
{
    std::string s= "n8Name4Surname";

    std::vector<std::string> parts;

    const char digits[] = "0123456789";

    std::string::size_type from=0, to=std::string::npos;

    do
    {
        from = s.find_first_of(digits, from);
        if (std::string::npos != from)
            from = s.find_first_not_of(digits, from);

        if (std::string::npos != from)
        {
            to = s.find_first_of(digits, from);
            if (std::string::npos == to)
                parts.push_back(s.substr(from));
            else
                parts.push_back(s.substr(from, to-from));

            from = to; // could be npos
        } 

    } while (std::string::npos != from);

    for (int i=0; i<parts.size(); i++)
       std::cout << i << ":\t" << parts[i] << std::endl;


    return 0;
}

Answer 5

强制性提升精神样本：

#include <string>
#include <boost/spirit/include/qi.hpp>
#include <iostream>

int main()
{
    std::string s= "n8Name4Surname";

    std::string::const_iterator b(s.begin()), e(s.end());
    std::string ignore, name, surname;

    using namespace boost::spirit::qi;
    rule<std::string::const_iterator, space_type, char()> 
        digit = char_("0123456789"),
        other = (char_ - digit);

    if (phrase_parse(b, e, *other >> +digit >> +other >> +digit >> +other, space, ignore, ignore, name, ignore, surname))
    {
        std::cout << "name = " << name << std::endl;
        std::cout << "surname = " << surname << std::endl;
    }

    return 0;
}

如何拆分包含字符和数字的字符串值

5 个答案: