Question

我想要一个名称 - 值对列表。每个列表都以'。'结尾。和EOL。每个名称 - 值对用'：'分隔。每对由';'分隔在列表中。 E.g。

NAME1: VALUE1; NAME2: VALUE2; NAME3: VALUE3.<EOL>

我遇到的问题是值包含'。'并且最后一个值总是消耗'。'在EOL。我可以使用某种前瞻来确保最后一个'。'在EOL被区别对待之前？

Answer 1

我已经创建了一个样本，大概看起来就像你拥有的那样。调整如下：

value = lexeme [ *(char_ - ';' - ("." >> (eol|eoi))) ];

请注意- ("." >> (eol|eoi)))的含义：排除紧跟行尾或输入结尾的所有.。

测试用例（也在http://liveworkspace.org/code/949b1d711772828606ddc507acf4fb4b上生效）：

const std::string input = 
    "name1: value 1; other name : value #2.\n" 
    "name.sub1: value.with.periods; other.sub2: \"more fun!\"....\n";
bool ok = doParse(input, qi::blank);

输出：

parse success
data: name1 : value 1 ; other name  : value #2 . 
data: name.sub1 : value.with.periods ; other.sub2 : "more fun!"... .

完整代码：

#include <boost/fusion/adapted.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/karma.hpp>
#include <map>
#include <vector>

namespace qi    = boost::spirit::qi;
namespace karma = boost::spirit::karma;
namespace phx   = boost::phoenix;

typedef std::map<std::string, std::string> map_t;
typedef std::vector<map_t> maps_t;

template <typename It, typename Skipper = qi::space_type>
    struct parser : qi::grammar<It, maps_t(), Skipper>
{
    parser() : parser::base_type(start)
    {
        using namespace qi;

        name  = lexeme [ +~char_(':') ];
        value = lexeme [ *(char_ - ';' - ('.' >> (eol|eoi))) ];
        line  = ((name >> ':' >> value) % ';') >> '.';
        start = line % eol;
    }

  private:
    qi::rule<It, std::string(), Skipper> name, value;
    qi::rule<It, map_t(), Skipper> line;
    qi::rule<It, maps_t(), Skipper> start;
};

template <typename C, typename Skipper>
    bool doParse(const C& input, const Skipper& skipper)
{
    auto f(std::begin(input)), l(std::end(input));

    parser<decltype(f), Skipper> p;
    maps_t data;

    try
    {
        bool ok = qi::phrase_parse(f,l,p,skipper,data);
        if (ok)   
        {
            std::cout << "parse success\n";
            for (auto& line : data)
                std::cout << "data: " << karma::format_delimited((karma::string << ':' << karma::string) % ';' << '.', ' ', line) << '\n';
        }
        else    std::cerr << "parse failed: '" << std::string(f,l) << "'\n";

        //if (f!=l) std::cerr << "trailing unparsed: '" << std::string(f,l) << "'\n";
        return ok;
    } catch(const qi::expectation_failure<decltype(f)>& e)
    {
        std::string frag(e.first, e.last);
        std::cerr << e.what() << "'" << frag << "'\n";
    }

    return false;
}

int main()
{
    const std::string input = 
        "name1: value 1; other name : value #2.\n" 
        "name.sub1: value.with.periods; other.sub2: \"more fun!\"....\n";
    bool ok = doParse(input, qi::blank);

    return ok? 0 : 255;
}

用lookahead解决模糊的boost :: spirit :: qi语法

1 个答案: