Question

我有以下行 / 90pv-RKSJ-UCS2C usecmap

std::string const line = "/90pv-RKSJ-UCS2C usecmap";

auto first = line.begin(), last = line.end();

std::string label, token;
bool ok = qi::phrase_parse(
        first, last, 
        qi::lexeme [ "/" >> +~qi::char_(" ") ] >> ' ' >>  qi::lexeme[+~qi::char_(' ')] , qi::space, label, token);


if (ok)
    std::cout << "Parse success: label='" << label << "', token='" << token << "'\n";
else
    std::cout << "Parse failed\n";

if (first!=last)
    std::cout << "Remaining unparsed input: '" << std::string(first, last) << "'\n";

我想在标签变量中 90pv-RKSJ-UCS2C标签和 usecmap 。

我提取90pv-RKSJ-UCS2C值但不提取usecmap

Answer 1

使用space船长，您无法匹配' '（它会被跳过！）。另见：Boost spirit skipper issues

所以，要么不使用船长，要么允许船长吃掉它：

bool ok = qi::phrase_parse(
        first, last, 
        qi::lexeme [ "/" >> +qi::graph ] >> qi::lexeme[+qi::graph], qi::blank, label, token);

注意：

我使用qi::graph代替~qi::char_(" ")制定
我使用了blank_type，因为你说

我有以下行

这意味着不应该跳过行尾

演示

<强> Live On Coliru

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

namespace qi = boost::spirit::qi;

int main()
{
    std::string const line = "/90pv-rksj-ucs2c usecmap";

    auto first = line.begin(), last = line.end();

    std::string label, token;
    bool ok = qi::phrase_parse(
            first, last, 
            qi::lexeme [ "/" >> +qi::graph ] >> qi::lexeme[+qi::graph], qi::blank, label, token);

    if (ok)
        std::cout << "parse success: label='" << label << "', token='" << token << "'\n";
    else
        std::cout << "parse failed\n";

    if (first!=last)
        std::cout << "remaining unparsed input: '" << std::string(first, last) << "'\n";
}

打印：

parse success: label='90pv-rksj-ucs2c', token='usecmap'

Answer 2

如果您使用的是C ++ 11，我建议使用正则表达式。

#include <iostream>
#include <regex>
using namespace std;
int main() {
    regex re("^/([^\\s]*)\\s([^\\s]*)"); // 1st () captures
                                         // 90pv-RKSJ-UCS2C and 2nd () captures usecmap
    smatch sm;
    string s="/90pv-RKSJ-UCS2C usecmap";
    regex_match(s,sm,re);
    for(int i=0;i<sm.size();i++) {
        cout<<sm[i]<<endl;
    }
    string label=sm[1],token=sm[2];
    system("pause");
}

使用精神解析器从字符串中提取值

2 个答案:

演示