如何将十六进制表示从URL(%)转换为std::string(中文)?

How to convert hex representation from URL (%) to std::string (chinese text)?

简介

我有一些输入需要转换为正确的汉字,但我想我在最后一个数字到字符串的转换上遇到了困难。我已经使用 this hex to text converter online tool 检查过 e6b9af 对应于文本 .

MWE

这是我为说明问题而制作的一个最小示例。输入是 "%e6%b9%af"(从其他地方的 URL 获得)。

#include <iostream>
#include <string>

std::string attempt(std::string path)
{
  std::size_t i = path.find("%");
  while (i != std::string::npos)
  {
    std::string sub = path.substr(i, 9);
    sub.erase(i + 6, 1);
    sub.erase(i + 3, 1);
    sub.erase(i, 1);
    std::size_t s = std::stoul(sub, nullptr, 16);
    path.replace(i, 9, std::to_string(s));
    i = path.find("%");
  }
  return path;
}

int main()
{
  std::string input = "%E6%B9%AF";
  std::string goal = "湯";

  // convert input to goal
  input = attempt(input);
  
  std::cout << goal << " and " << input << (input == goal ? " are the same" : " are not the same") << std::endl;

  return 0;
}

输出

湯 and 15120815 are not the same

预期输出

湯 and 湯 are the same

附加问题

外文的字符都是3字节还是只有中文?由于我的尝试假定块为 3 个字节,这是一个好的假设吗?

根据您的建议并更改 this other post 中的示例。这是我想出来的。

#include <iostream>
#include <string>
#include <sstream>

std::string decode_url(const std::string& path)
{
  std::stringstream decoded;
  for (std::size_t i = 0; i < path.size(); i++)
  {
    if (path[i] != '%')
    {
      if (path[i] == '+')
        decoded << ' ';
      else
        decoded << path[i];
    }
    else
    {
      unsigned int j;
      sscanf(path.substr(i + 1, 2).c_str(), "%x", &j);
      decoded << static_cast<char>(j);
      i += 2;
    }
  }
  return decoded.str();
}

int main()
{
  std::string input = "%E6%B9%AF";
  std::string goal = "湯";

  // convert input to goal
  input = decode_url(input);

  std::cout << goal << " and " << input << (input == goal ? " are the same" : " are not the same") << std::endl;

  return 0;
}

输出

湯 and 湯 are the same