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无法使用boost :: spirit :: qi解析条件的SQL类型

[英]Unable to parse SQL type where condition using boost::spirit::qi

I may be asking a very trivial question but am not getting blocks out of my brain to crack it. 我可能问的是一个非常琐碎的问题,但是我并没有从大脑中脱颖而出。 Trying to parse a SQL like where clause as shown below using boost::spirit::qi to generate a vector of pairs 尝试使用boost :: spirit :: qi解析如下所示的SQL where子句,以生成向量对

std::string input = "book.author_id = '1234' and book.isbn = 'xy99' and book.type = 'abc' and book.lang = 'Eng'"

I have gone through the following threads but still unable to do it :-( Thread5 Thread4 Thread3 Thread2 Thread1 我已经通过下面的线程,但仍无法做到这一点:-(了Thread5 Thread4 Thread3 线程2 线程1

[Thread1][6]
[Thread2][7]
[Thread3][8]
[Thread4][9]
[Thread5][10]

I genuinely request, kindly help me understand how to achieve this ... may be I had not completely given my 100% but please be kind .... 我真诚地要求,请帮助我理解如何实现此目标...可能是我没有完全给我100%的费用,但请保持友好....

Here is the full code (some part commented which I wish to do), as a first step I was just checking if I can get all tokens in a Vector and then parse each Vector element to generate another vector of std::pair 这是完整的代码(我想做一部分注释),作为第一步,我只是检查是否可以在Vector中获取所有标记,然后解析每个Vector元素以生成另一个std :: pair向量。

#include <boost/fusion/adapted.hpp>
#include <boost/spirit/include/qi.hpp>
#include <map>
#include <vector>

namespace qi    = boost::spirit::qi;
namespace phx   = boost::phoenix;

typedef std::string str_t;
typedef std::pair<str_t, str_t> pair_t;
typedef std::vector<pair_t> pairs_t;

typedef std::vector<str_t> strings_t;
//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, pairs_t(), Skipper>
    struct parser : qi::grammar<It, strings_t(), Skipper>
{
    parser() : parser::base_type(start)
    {
        using namespace qi;

        cond    = lexeme [ *(char_) ];
        conds   =  *(char_) >> cond % (lit("and"));

        //conds =  *(char_ - lit("and")) >>(cond % lit("and"));
        /*cond  = lexeme [ *(char_ - lit("and")) ];
        cond    = key >> "=" >> value;
        key     = *(char_ - "=");
        value   = ('\'' >> *(~char_('\'')) >> '\'');
        kv_pair = key >> value;*/
        start   = conds;
        //cond  = key >> "=" >> value;
        //key       = *(char_ - "=");
        //value = ('\'' >> *(~char_('\'')) >> '\'');
  //      kv_pair   = key >> value;
  //      start = kv_pair;
    }

  private:
    qi::rule<It, str_t(), Skipper> cond;
    qi::rule<It, strings_t(), Skipper> conds;
    //qi::rule<It, std::string(), Skipper> key, value;//, cond;
    //qi::rule<It, pair_t(), Skipper> kv_pair;
    //qi::rule<It, pairs_t(), Skipper> start;
    qi::rule<It, strings_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;
    strings_t data;

    try
    {
        bool ok = qi::phrase_parse(f,l,p,skipper,data);
        if (ok)   
        {
            std::cout << "parse success\n";
            std::cout << "No Of Key-Value Pairs=  "<<data.size()<<"\n";
        }
        else    std::cerr << "parse failed: '" << 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()
{
    std::cout<<"Pair Test \n";
    const std::string input = "book.author_id = '1234' and book.isbn = 'xy99' and book.type = 'abc' and book.lang = 'Eng'";
    bool ok = doParse(input, qi::space);
    std::cout<< input <<"\n";
    return ok? 0 : 255;
}

OUTPUT: OUTPUT:

Pair Test
parse success
No Of Key-Value Pairs=  2
book.author_id = '1234' and book.isbn = 'xy99' and book.type = 'abc' and book.lang = 'Eng'

Which I expect 4 ... since there are 4 conditions !! 我期望4 ...因为有4个条件!

Thanks in Advance Regards, Vivek 在此先感谢Vivek

some example to work out- live on coliru 一些可以在大肠杆菌上工作的例子

I'm sorry to break it to you, but your grammar is far more broken than you imagined. 很抱歉把它给您打破,但是您的语法比您想像的要复杂得多。

    conds   =  *(char_) // ...

Right here, you're basically just parsing all the input into a single string, with whitespace skipped. 在这里,您基本上只是将所有输入解析为单个字符串,而忽略了空格。 In fact, adding 实际上,添加

    for (auto& el : data)
        std::cout << "'" << el << "'\n";

after parsing prints: 解析打印后:

Pair Test 
parse success
No Of Key-Value Pairs=  2
'book.author_id='1234'andbook.isbn='xy99'andbook.type='abc'andbook.lang='Eng''
''

As you can see, the first element is the string that *char_ parsed, and you get an empty element for free due to the fact that both conds and cond match on empty input. 正如你所看到的,第一个元素是字符串*char_解析,你会得到一个空元素免费由于这两个condscond对空输入匹配。

I would strongly suggest you to start simple . 我强烈建议您从简单开始。 And I mean, much simpler. 我的意思是,要简单得多。

Slowly build your grammar up from the ground. 从头慢慢建立语法。 Spirit is a very good tool to tackle with test-driven development (except for the compile times, but hey, you get more time to think!). Spirit是解决测试驱动开发的一个非常好的工具(编译时间除外,但是,您有更多的时间思考!)。

Here's something that I just made up, starting thinking from the very first building block, the indent ifier, and working my way up to the higher-level elements: 这是我刚刚完成的工作,从第一个构建块, indent符开始思考,然后逐步发展到更高层次的元素:

// lexemes (no skipper)
ident     = +char_("a-zA-Z.");
op        = no_case [ lit("=") | "<>" | "LIKE" | "IS" ];
nulllit   = no_case [ "NULL" ];
and_      = no_case [ "AND" ];
stringlit = "'" >> *~char_("'") >> "'";

// other productions
field     = ident;
value     = stringlit | nulllit;
condition = field >> op >> value;

conjunction = condition % and_;
start       = conjunction;

These are close to the simplest thing that I suppose could parse your grammar (with a few creative notes left and right, where they don't seem too intrusive). 这些接近于我认为可以解析您的语法的最简单的东西(左右有一些创造性的注释,在这些注释中似乎不太令人讨厌)。

UPDATE So this is where I got in 20 minutes: 更新所以这是我在20分钟内到达的地方:

I always start out with mapping the types that I want the rules to expose: 我总是从映射我希望规则公开的类型开始:

namespace ast
{
    enum op { op_equal, op_inequal, op_like, op_is };

    struct null { };

    typedef boost::variant<null, std::string> value;

    struct condition
    {
        std::string _field;
        op _op;
        value _value;
    };

    typedef std::vector<condition> conditions;
}

Only condition cannot be "naturally" used in a Spirit grammar without adaptation: 如果没有适应,则condition不能在Spirit语法中“自然地”使用:

BOOST_FUSION_ADAPT_STRUCT(ast::condition, (std::string,_field)(ast::op,_op)(ast::value,_value))

Now comes the grammar itself: 现在是语法本身:

    // lexemes (no skipper)
    ident       = +char_("a-zA-Z._");
    op_token.add
        ("=",    ast::op_equal)
        ("<>",   ast::op_inequal)
        ("like", ast::op_like)
        ("is",   ast::op_is);
    op          = no_case [ op_token ];
    nulllit     = no_case [ "NULL" >> attr(ast::null()) ];
    and_        = no_case [ "AND" ];
    stringlit   = "'" >> *~char_("'") >> "'";

    //// other productions
    field       = ident;
    value       = stringlit | nulllit;
    condition   = field >> op >> value;

    whereclause = condition % and_;
    start       = whereclause;

You can see minor deviations from my original sketch, that's interesting: 您可以看到与原始草图略有偏差,这很有趣:

  • added _ to identifier chars 在标识符char中添加了_
  • move op_token into a symbol matcher (because that's easier to map the enum values) op_token移至符号匹配器中(因为这样更容易映射枚举值)

See it all Live And Working On Coliru , output: 实时查看全部内容, 在Coliru上工作 ,输出:

Pair Test 
parse success
No Of Key-Value Pairs=  4
( [book.author_id] = 1234 )
( [book.isbn] LIKE xy99 )
( [book.type] = abc )
( [book.lang] IS NULL )

book.author_id = '1234' and book.isbn liKE 'xy99' and book.type = 'abc' and book.lang IS null

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