您只想确保您也解析“空”字符串的值。
value = +(char_ - ',' - eol) | attr("(unspecified)");
entry = value >> ',' >> value >> ',' >> value >> eol;
查看演示:
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//#define BOOST_SPIRIT_DEBUG
#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/qi.hpp>
namespace qi = boost::spirit::qi;
struct data {
std::string a;
std::string b;
std::string c;
};
BOOST_FUSION_ADAPT_STRUCT(data, (std::string, a)(std::string, b)(std::string, c))
template <typename Iterator, typename skipper = qi::blank_type>
struct google_parser : qi::grammar<Iterator, data(), skipper> {
google_parser() : google_parser::base_type(entry, "contacts") {
using namespace qi;
value = +(char_ - ',' - eol) | attr("(unspecified)");
entry = value >> ',' >> value >> ',' >> value >> eol;
BOOST_SPIRIT_DEBUG_NODES((value)(entry))
}
private:
qi::rule<Iterator, std::string()> value;
qi::rule<Iterator, data(), skipper> entry;
};
int main() {
using It = std::string::const_iterator;
google_parser<It> p;
for (std::string input : {
"something, awful, is\n",
"fine,,just\n",
"like something missing: ,,\n",
})
{
It f = input.begin(), l = input.end();
data parsed;
bool ok = qi::phrase_parse(f,l,p,qi::blank,parsed);
if (ok)
std::cout << "Parsed: '" << parsed.a << "', '" << parsed.b << "', '" << parsed.c << "'\n";
else
std::cout << "Parse failed\n";
if (f!=l)
std::cout << "Remaining unparsed: '" << std::string(f,l) << "'\n";
}
}
打印:
Parsed: 'something', 'awful', 'is'
Parsed: 'fine', '(unspecified)', 'just'
Parsed: 'like something missing: ', '(unspecified)', '(unspecified)'
但是,您有一个更大的问题。 qi::repeat(2) [ value ] 将解析为 2 个字符串的假设是行不通的。
repeat,如operator*、operator+ 和operator% 解析成容器属性。在这种情况下,容器属性(字符串)也将接收来自第二个 value 的输入:
Live On Coliru
Parsed: 'somethingawful', 'is', ''
Parsed: 'fine(unspecified)', 'just', ''
Parsed: 'like something missing: (unspecified)', '(unspecified)', ''
由于这不是您想要的,请重新考虑您的数据类型:
auto_ 方法:
如果你教 Qi 如何提取单个值,你可以使用一个简单的规则,比如
entry = skip(skipper() | ',') [auto_] >> eol;
这样,Spirit 本身将为给定的 Fusion 序列生成正确数量的值提取!
这是一个快速而肮脏的方法:
CAVEAT 像这样直接专门针对std::string 可能不是最好的主意(它可能并不总是合适的,并且可能与其他解析器交互不良)。但是,默认情况下,create_parser<std::string> 没有定义(因为,它会做什么?)所以我抓住了这个演示的机会:
namespace boost { namespace spirit { namespace traits {
template <> struct create_parser<std::string> {
typedef proto::result_of::deep_copy<
BOOST_TYPEOF(
qi::lexeme [+(qi::char_ - ',' - qi::eol)] | qi::attr("(unspecified)")
)
>::type type;
static type call() {
return proto::deep_copy(
qi::lexeme [+(qi::char_ - ',' - qi::eol)] | qi::attr("(unspecified)")
);
}
};
}}}
再次查看演示输出:
Live On Coliru
Parsed: 'something', 'awful', 'is'
Parsed: 'fine', 'just', '(unspecified)'
Parsed: 'like something missing: ', '(unspecified)', '(unspecified)'
注意 有一些高级魔法可以让船长“恰到好处”地工作(参见skip()[] 和lexeme[])。一些一般性的解释可以在这里找到:Boost spirit skipper issues
更新
容器方法
这很微妙。实际上是两个。所以这里有一个演示:
Live On Coliru
//#define BOOST_SPIRIT_DEBUG
#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/qi.hpp>
namespace qi = boost::spirit::qi;
struct data {
std::vector<std::string> parts;
};
BOOST_FUSION_ADAPT_STRUCT(data, (std::vector<std::string>, parts))
template <typename Iterator, typename skipper = qi::blank_type>
struct google_parser : qi::grammar<Iterator, data(), skipper> {
google_parser() : google_parser::base_type(entry, "contacts") {
using namespace qi;
qi::as<std::vector<std::string> > strings;
value = +(char_ - ',' - eol) | attr("(unspecified)");
entry = strings [ repeat(2) [ value >> ',' ] >> value ] >> eol;
BOOST_SPIRIT_DEBUG_NODES((value)(entry))
}
private:
qi::rule<Iterator, std::string()> value;
qi::rule<Iterator, data(), skipper> entry;
};
int main() {
using It = std::string::const_iterator;
google_parser<It> p;
for (std::string input : {
"something, awful, is\n",
"fine,,just\n",
"like something missing: ,,\n",
})
{
It f = input.begin(), l = input.end();
data parsed;
bool ok = qi::phrase_parse(f,l,p,qi::blank,parsed);
if (ok) {
std::cout << "Parsed: ";
for (auto& part : parsed.parts)
std::cout << "'" << part << "' ";
std::cout << "\n";
}
else
std::cout << "Parse failed\n";
if (f!=l)
std::cout << "Remaining unparsed: '" << std::string(f,l) << "'\n";
}
}
微妙之处在于: