regex - Boost regex用于解析字符串中的键/值对

Question

我尝试解析其键值对的以下字符串:

#include <iostream>
#include <string>
#include <map>

#include <boost/regex.hpp>

int main()
{
    std::string deliveryReceipt = "id:pgl01130529155035239084 sub:001 dlvrd:001 submit date:1305291550 done date:1305291550 stat:DELIVRD err:0";

    std::map<std::string, std::string> results;
    boost::regex re("(?:([^:]+):([^,]+)(?:,|$))+"); // key - value pair

    boost::sregex_iterator it(deliveryReceipt.begin(), deliveryReceipt.end(), re), end;
    for ( ; it != end; ++it){
      results[(*it)[1]] = (*it)[2];
    }    

    std::map<std::string, std::string>::iterator resultsIter = results.begin();
    while (resultsIter != results.end())
    {
        std::cout << "key:" << resultsIter->first << " value:" << resultsIter->second << std::endl;
        resultsIter++;
    }
}

我得到以下输出:

键:id值:pgl01130529155035239084子:001 dlvrd:001提交日期:1305291550完成日期:1305291550统计:DELIVRD错误:0

如何修复此正则表达式以正确解析键/值对？

Answer 1

我会选择这样的东西(更新了)
"\\s*(?<!\\S)([^:]+)\\s*:(\\S+)(?!\\S)"
https://regex101.com/r/Sufx5m/1

讲解

 \s*              # Optional whitespace trim
 (?<! \S)         # Whitespace boundary delimiter
                  #   (also matches at beginning of string)
 ( [^:]+ )        # (1), Key - not any ':' colon chars
 \s*              # Optional whitespace trim
 :                # Colon 
 ( \S+ )          # (2), Value - not whitespace chars
 (?! \S )         # Whitespace boundary delimiter.
                  #   (also matches at end of string)