我想要一个名称-值对的列表。每个列表以“。”结尾。和停产。每个名称/值对之间用“:”分隔。每对之间用“;”分隔在列表中。例如。
NAME1: VALUE1; NAME2: VALUE2; NAME3: VALUE3.<EOL>
我的问题是值包含“。”并且最后一个值始终使用“。”在EOL。我可以使用某种先行方式来确保最后一个'。'在对EOL进行不同对待之前?
最佳答案
我创建了一个样本,大概看起来像您所拥有的。该调整在以下行中:
value = lexeme [ *(char_ - ';' - ("." >> (eol|eoi))) ];
请注意
- ("." >> (eol|eoi)))
的含义:排除行尾或输入结束后紧跟的任何.
。测试用例(也存在于http://liveworkspace.org/code/949b1d711772828606ddc507acf4fb4b上):
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;
}
关于c++ - 提前解决歧义的boost::spirit::qi语法,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/12217515/