如何使用Boost Spirit提取修剪的文本? [英] How to extract trimmed text using Boost Spirit?

问题描述

使用增强精神，我想提取一个字符串，然后在括号中添加一些数据.相关字符串与左括号之间用空格隔开.不幸的是，字符串本身可能包含空格.我正在寻找一种简洁的解决方案，该解决方案可以返回不带尾部空格的字符串.

Using boost spirit, I'd like to extract a string that is followed by some data in parentheses. The relevant string is separated by a space from the opening parenthesis. Unfortunately, the string itself may contain spaces. I'm looking for a concise solution that returns the string without a trailing space.

以下代码说明了该问题:

The following code illustrates the problem:

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

namespace qi = boost::spirit::qi;
using std::string;
using std::cout;
using std::endl;

void
test_input(const string &input)
{
    string::const_iterator b = input.begin();
    string::const_iterator e = input.end();
    string parsed;
    bool const r = qi::parse(b, e,
        *(qi::char_ - qi::char_("(")) >> qi::lit("(Spirit)"),
            parsed
    );
    if(r) {
        cout << "PASSED:" << endl;
    } else {
        cout << "FAILED:" << endl;
    }
    cout << "  Parsed: \"" << parsed << "\"" << endl;
    cout << "  Rest: \"" << string(b, e) << "\"" << endl;
}

int main()
{
    test_input("Fine (Spirit)");
    test_input("Hello, World (Spirit)");

    return 0;
}

其输出是:

PASSED:
  Parsed: "Fine "
  Rest: ""
PASSED:
  Parsed: "Hello, World "
  Rest: ""

通过这种简单的语法，提取的字符串始终后面跟一个空格(我想消除该空格).

With this simple grammar, the extracted string is always followed by a space (that I 'd like to eliminate).

该解决方案应在Spirit内运行，因为这只是较大语法的一部分. (因此，解析后修剪提取的字符串可能很笨拙.)

The solution should work within Spirit since this is only part of a larger grammar. (Thus, it would probably be clumsy to trim the extracted strings after parsing.)

谢谢.

推荐答案

就像注释中所说的那样，在单个空格的情况下，您可以对其进行硬编码.如果您需要更加灵活或宽容:

Like the comment said, in the case of a single space, you can just hard code it. If you need to be more flexible or tolerant:

我会使用

I'd use a skipper with raw to "cheat" the skipper for your purposes:

bool const r = qi::phrase_parse(b, e,
    qi::raw [ *(qi::char_ - qi::char_("(")) ] >> qi::lit("(Spirit)"),
    qi::space,
    parsed
);

这行得通，并且打印出来

This works, and prints

PASSED:
  Parsed: "Fine"
  Rest: ""
PASSED:
  Parsed: "Hello, World"
  Rest: ""

查看 在Coliru上直播

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完整程序供参考:

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

namespace qi = boost::spirit::qi;
using std::string;
using std::cout;
using std::endl;

void
test_input(const string &input)
{
    string::const_iterator b = input.begin();
    string::const_iterator e = input.end();
    string parsed;
    bool const r = qi::phrase_parse(b, e,
        qi::raw [ *(qi::char_ - qi::char_("(")) ] >> qi::lit("(Spirit)"),
        qi::space,
        parsed
    );
    if(r) {
        cout << "PASSED:" << endl;
    } else {
        cout << "FAILED:" << endl;
    }
    cout << "  Parsed: \"" << parsed << "\"" << endl;
    cout << "  Rest: \"" << string(b, e) << "\"" << endl;
}

int main()
{
    test_input("Fine (Spirit)");
    test_input("Hello, World (Spirit)");

    return 0;
}

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