试图用Boost-Spirit解析SQL like语句 [英] Attempting to parse SQL like statement with Boost-Spirit

问题描述

我是boost ::精神的新手。我写了程序来解析一个SQL语句，比如select * from table where条件。它编译失败。报告大量模板错误。那么，有人会帮助我吗？

I'm a newbie in boost::spirit. I wrote the program to parse a SQL statement, like "select * from table where conditions". It compile failed. A large of template errors reported. So, would somebody help me?

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

namespace qi = boost::spirit::qi;
namespace ascii = boost::spirit::ascii;

struct db_select {
    void exec() {}

    std::string filed;
    std::string table;
    std::string condition;
};

std::ostream& operator<<(std::ostream& os, const db_select& se) {
    return os << "filed: " << se.filed << " table: " << se.table << " condition: " << se.condition;
}

template <class Iterator>
struct selecter : qi::grammar<Iterator, db_select (), ascii::space_type> {
    selecter() : selecter::base_type(se) {
            se %= "select" >> +qi::char_ << "from" << +qi::char_ << "where" << +qi::char_;
    } 

    qi::rule<Iterator, db_select (), ascii::space_type> se;
};

int main(int argc, char* argv[]) {
    if (argc < 2)
            return -1;

    std::string str(argv[1]);
    const char* first = str.c_str();
    const char* last = &str[str.size()];
    selecter<const char*> se;
    db_select rst;

    bool r = qi::phrase_parse(first, last, se, ascii::space, rst);
    if (!r || first != last) {
            std::cout << "parse failed, at: " << std::string(first, last) << std::endl;
            return -1;
    } else
            std::cout << "success, " << rst << std::endl;

    return 0;
}

推荐答案

^{编辑最后在计算机后面修订回答：}

^{Edit Finally behind a computer, revised answer:}

有三件事要注意：

解析器表达式包含错误（<< ，而不是 > ）。这占很大的编译错误。注意编译器错误中出现 ******* ：

The parser expression contained errors (<< instead of >> as intended). This accounts for a lot of compile errors. Note the appearance of ******* in the compiler error:

/.../qi/nonterminal/rule.hpp|176 col 13| error: no matching function for call to ‘assertion_failed(mpl_::failed************ 

这是为了引导您到源代码中的相应注释：

which is designed to lead you to the corresponding comment in the source code:

// Report invalid expression error as early as possible.
// If you got an error_invalid_expression error message here,
// then the expression (expr) is not a valid spirit qi expression.
BOOST_SPIRIT_ASSERT_MATCH(qi::domain, Expr);

轻松修复一个，两个去！

Easily fixed. One down, two to go!

为了将你的数据类型指定为规则属性，你需要使它融合最方便的方法：

In order to assign to your datatype as an rule attribute, you need to make it fusion compatible. The most convenient way:

 BOOST_FUSION_ADAPT_STRUCT(db_select,
    (std::string,filed)(std::string,table)(std::string,condition));

编译，但解析失败了：

此外，您还需要采取措施，避免使用+ qi :: char_表达式搜索您的查询关键字。

Further more you will need to take measures to avoid 'eating' your query keywords with the +qi::char_ expressions.

作为基础，考虑写它例如

As a basis, consider writing it something like

   lexeme [  (!lit("where")  >> +qi::graph) % +qi::space ]

• code> lexeme 禁止包含表达式的队长


• 必须不匹配
  
  • lexeme inhibits the skipper for the enclosed expression
  • ! Asserts that the specified keyword must not match

  最后，查看 qi :: no_case

  Lastly, look at the docs for qi::no_case to do case-insensitive matching.

  #include <iostream>
  #include <string>
  #include <boost/fusion/adapted.hpp>
  #include <boost/spirit/include/qi.hpp>
  
  namespace qi = boost::spirit::qi;
  
  struct db_select {
      void exec() {}
  
      std::string field;
      std::string table;
      std::string condition;
  };
  
  BOOST_FUSION_ADAPT_STRUCT(db_select,(std::string,field)(std::string,table)(std::string,condition));
  
  std::ostream& operator<<(std::ostream& os, const db_select& se) {
      return os << "field: " << se.field << " table: " << se.table << " condition: " << se.condition;
  }
  
  template <class Iterator>
  struct selecter : qi::grammar<Iterator, db_select (), qi::space_type> {
      selecter() : selecter::base_type(se) {
          using namespace qi;
          se %= "select"
              >> lexeme [ (!lit("from")  >> +graph) % +space ] >> "from"
              >> lexeme [ (!lit("where") >> +graph) % +space ] >> "where" 
              >> +qi::char_;
      } 
  
      qi::rule<Iterator, db_select (), qi::space_type> se;
  };
  
  int main(int argc, char* argv[]) {
      if (argc < 2)
              return -1;
  
      std::string str(argv[1]);
      const char* first = str.c_str();
      const char* last = &str[str.size()];
      selecter<const char*> se;
      db_select rst;
  
      bool r = qi::phrase_parse(first, last, se, qi::space, rst);
      if (!r || first != last) {
              std::cout << "parse failed, at: " << std::string(first, last) << std::endl;
              return -1;
      } else
              std::cout << "success, " << rst << std::endl;
  
      return 0;
  }
  

  
  
  

  测试：

  
  
  

  Test:

  g++ test.cpp -o test
  ./test "select aap, noot, mies from table where field = 'value'"
  

  输出：

  success, field: aap,noot,mies table: table condition: field='value'
  

  这篇关于试图用Boost-Spirit解析SQL like语句的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！