Question

我正在搜索一种技术来转换由包含以下内容的服务器发送的字符串（JSON）： ...."Test \u00e9\u00e9\u00e9".....类似于：“Testééé” 我找到了一个解决方案：boost::replace_all(listFolder, "\\u00e9", "é"); 我正在使用这个提升功能与其他字母àùèê等....这很痛苦！

我想知道是否有一个函数自动执行这种转换。

否则，我想告诉你一些事情，如果我使用这个函数，服务器将正确处理我发送给它的字符串并包含带重音的字母：

std::string fromLocale(std::string localeStr)
{
    boost::locale::generator g;
    g.locale_cache_enabled(true);
    std::locale loc = g(boost::locale::util::get_system_locale());
    return boost::locale::conv::to_utf<char>(localeStr,loc);
}

不幸的是，该代码的反转不能处理服务器发送的字符串。

std::string toLocale(std::string utf8Str)
{
    boost::locale::generator g;
    g.locale_cache_enabled(true);
    std::locale loc = g(boost::locale::util::get_system_locale());
    return boost::locale::conv::from_utf<char>(utf8Str,loc);
}

Answer 1

JSON specification允许Unicode字符的"\uXXXX"个序列（amonst其他\X转义序列）。如果您没有使用处理解码此类序列的现有JSON解析器，则必须手动解码它们，例如：

// JSON uses Unicode, but is commonly encoded as UTF-8. However, Unicode
// characters that are encoded in "\uXXXX" format are expressed as UTF-16
// codeunit values, using surrogate pairs for codepoint values U+10000 and
// higher. This example uses C++11's std::u16string to handle UTF-16 parsing.
// If you are not using C++11 or later, you can replace it with std::wstring
// on platforms where wchar_t is 16bit, for instance.  If you want to handle
// the JSON using std::string/UTF-8 instead, you will have to tweak this
// parsing accordingly...

std::u16string str = ...; // JSON quoted-string value, eg: "Test \u00e9\u00e9\u00e9"...
std::u16string::size_type idx = 0;
do
{
    idx = str.find(u'\\', idx);
    if (idx == std::u16string::npos) break;

    std::u16string replaceStr;
    std::u16string::size_type len = 2;

    char16_t ch = str.at(idx+1);
    switch (ch)
    {
        case u'\"':
        case u'\\':
        case u'/':
            replaceStr = ch;
            break;

        case u'b':
            replaceStr = u'\b';
            break;

        case u'f':
            replaceStr = u'\f';
            break;

        case u'n':
            replaceStr = u'\n';
            break;

        case u'r':
            replaceStr = u'\r';
            break;

        case u't':
            replaceStr = u'\t';
            break;

        case u'u':
        {
            std::u16string hexStr = str.substr(idx+2, 4);
            len += hexStr.size();

            std::basic_istringstream<char16_t> iss(hexStr);
            unsigned short value;
            iss >> std::hex >> value;
            if (!iss)
            {
                // illegal value, do something
            }

            replaceStr = (char_t) value;
            break;
        }

        default:
            // illegal sequence, do something
            break;
    }

    str.replace(idx, len, replaceStr);
    idx += replaceStr.size();
}
while (true);

Answer 2

我发现的解决方案是使用RapidJson。

使用ASCII / UTF8中的重音转换Unicode字母

2 个答案: