如何使用boost :: spirit解析csv

Question

我有这个csv线

std::string s = R"(1997,Ford,E350,"ac, abs, moon","some "rusty" parts",3000.00)";

我可以用boost::tokenizer以下方法解析它:

typedef boost::tokenizer< boost::escaped_list_separator<char> , std::string::const_iterator, std::string> Tokenizer;
boost::escaped_list_separator<char> seps('\\', ',', '\"');
Tokenizer tok(s, seps);
for (auto i : tok)
{
    std::cout << i << std::endl;
}

它是正确的,除了令牌"生锈"应该有双引号被剥离.

这是我尝试使用boost :: spirit

boost::spirit::classic::rule<> list_csv_item = !(boost::spirit::classic::confix_p('\"', *boost::spirit::classic::c_escape_ch_p, '\"') | boost::spirit::classic::longest_d[boost::spirit::classic::real_p | boost::spirit::classic::int_p]);
std::vector<std::string> vec_item;
std::vector<std::string>  vec_list;
boost::spirit::classic::rule<> list_csv = boost::spirit::classic::list_p(list_csv_item[boost::spirit::classic::push_back_a(vec_item)],',')[boost::spirit::classic::push_back_a(vec_list)];
boost::spirit::classic::parse_info<> result = parse(s.c_str(), list_csv);
if (result.hit)
{
  for (auto i : vec_item)
  {
    cout << i << endl;
   }
}

问题:

1. 不起作用,仅打印第一个令牌

2. 为什么要提升::精神::经典？找不到使用Spirit V2的例子

3. 设置是残酷的...但我可以忍受这个

**我真的很想使用,boost::spirit因为它往往很快

预期产量:

1997
Ford
E350
ac, abs, moon
some "rusty" parts

3000.00

Answer 1

有关解析(可选)引用分隔字段(包括不同引号字符(',"))的背景信息,请参阅此处:

• 使用boost :: spirit解析引用的字符串

对于一个非常非常完整的示例,完全支持部分引用的值和a

splitInto(input, output, ' ');

采用'任意'输出容器和分隔符表达式的方法,请参见此处:

• 如何使我的分割只在一个实线上工作,并能够跳过字符串的引用部分？

解决您的具体问题,假设无论是引用或不带引号的字段(没有引号部分里面的字段值),使用精神V2:

让我们采用可能有效的最简单的"抽象数据类型":

using Column  = std::string;
using Columns = std::vector<Column>;
using CsvLine = Columns;
using CsvFile = std::vector<CsvLine>;

重复的双引号转义了双引号语义(正如我在评论中指出的那样),你应该可以使用类似的东西:

static const char colsep = ',';

start  = -line % eol;
line   = column % colsep;
column = quoted | *~char_(colsep);
quoted = '"' >> *("\"\"" | ~char_('"')) >> '"';

打印以下完整的测试程序

[1997][Ford][E350][ac, abs, moon][rusty][3001.00]

(注意BOOST_SPIRIT_DEBUG定义以便于调试).在科利鲁看到它

完整的演示

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

namespace qi = boost::spirit::qi;

using Column  = std::string;
using Columns = std::vector<Column>;
using CsvLine = Columns;
using CsvFile = std::vector<CsvLine>;

template <typename It>
struct CsvGrammar : qi::grammar<It, CsvFile(), qi::blank_type>
{
    CsvGrammar() : CsvGrammar::base_type(start)
    {
        using namespace qi;

        static const char colsep = ',';

        start  = -line % eol;
        line   = column % colsep;
        column = quoted | *~char_(colsep);
        quoted = '"' >> *("\"\"" | ~char_('"')) >> '"';

        BOOST_SPIRIT_DEBUG_NODES((start)(line)(column)(quoted));
    }
  private:
    qi::rule<It, CsvFile(), qi::blank_type> start;
    qi::rule<It, CsvLine(), qi::blank_type> line;
    qi::rule<It, Column(),  qi::blank_type> column;
    qi::rule<It, std::string()> quoted;
};

int main()
{
    const std::string s = R"(1997,Ford,E350,"ac, abs, moon","""rusty""",3001.00)";

    auto f(begin(s)), l(end(s));
    CsvGrammar<std::string::const_iterator> p;

    CsvFile parsed;
    bool ok = qi::phrase_parse(f,l,p,qi::blank,parsed);

    if (ok)
    {
        for(auto& line : parsed) {
            for(auto& col : line)
                std::cout << '[' << col << ']';
            std::cout << std::endl;
        }
    } else
    {
        std::cout << "Parse failed\n";
    }

    if (f!=l)
        std::cout << "Remaining unparsed: '" << std::string(f,l) << "'\n";
}

Answer 2

Sehe的帖子看起来比我的更干净,但我把它放在一起有点,所以这里它是反正的:

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

namespace qi = boost::spirit::qi;

int main() {
    const std::string s = R"(1997,Ford,E350,"ac, abs, moon",""rusty"",3000.00)";

    // Tokenizer
    typedef boost::tokenizer< boost::escaped_list_separator<char> , std::string::const_iterator, std::string> Tokenizer;
    boost::escaped_list_separator<char> seps('\\', ',', '\"');
    Tokenizer tok(s, seps);
    for (auto i : tok)
        std::cout << i << "\n";
    std::cout << "\n";

    // Boost Spirit Qi
    qi::rule<std::string::const_iterator, std::string()> quoted_string = '"' >> *(qi::char_ - '"') >> '"';
    qi::rule<std::string::const_iterator, std::string()> valid_characters = qi::char_ - '"' - ',';
    qi::rule<std::string::const_iterator, std::string()> item = *(quoted_string | valid_characters );
    qi::rule<std::string::const_iterator, std::vector<std::string>()> csv_parser = item % ',';

    std::string::const_iterator s_begin = s.begin();
    std::string::const_iterator s_end = s.end();
    std::vector<std::string> result;

    bool r = boost::spirit::qi::parse(s_begin, s_end, csv_parser, result);
    assert(r == true);
    assert(s_begin == s_end);

    for (auto i : result)
        std::cout << i << std::endl;
    std::cout << "\n";
}   

这输出:

1997
Ford
E350
ac, abs, moon
rusty
3000.00

1997
Ford
E350
ac, abs, moon
rusty
3000.00

值得注意的事情:这不会实现完整的CSV解析器.您还需要查看转义字符或实现所需的任何其他内容.

另外:如果您正在查看文档,那么您知道,在Qi中,'a'相当于boost::spirit::qi::lit('a')和"abc"等同于boost::spirit::qi::lit("abc").

关于双引号:因此,正如Sehe在上面的评论中指出的那样,并不直接清楚""输入文本中围绕a的规则意味着什么.如果您希望""不在引用字符串中的所有实例都转换为a ",那么类似下面的内容将起作用.

qi::rule<std::string::const_iterator, std::string()> double_quote_char = "\"\"" >> qi::attr('"');
qi::rule<std::string::const_iterator, std::string()> item = *(double_quote_char | quoted_string | valid_characters );