如何将 time-uuid(存储在 boost uuid 中)转换为时间戳/自纪元以来的时间?

How do I convert time-uuid (stored in boost uuid) to a timestamp/time since epoch?

本文关键字:时间戳 纪元 时间 转换 time-uuid 存储 uuid boost      更新时间:2023-10-16

根据规范,从 UUID 时间戳转换为秒似乎很容易,也基于 Cassandra 基于其结构定义的 C++ 驱动程序源代码。

但是,当我尝试这样做时,我总是得到错误的值。我做错了什么,我无法弄清楚它是什么。

为此,我使用了从此处和此处提供的示例 UUID 值。

所要做的就是从 UUID 原始数据中获取第一个uint64_t,屏蔽其前四个 MSb,减去差值并除以一个数字。

这是我的最小完整示例:

#include <boost/date_time.hpp>
#include <boost/uuid/uuid.hpp>
#include <boost/uuid/uuid_generators.hpp>
#include <boost/uuid/uuid_io.hpp>
#include <cstdint>
#include <iostream>
uint64_t TimestampFromUUID(const boost::uuids::uuid& uuid) {
static constexpr const int UUID_SIZE = 16;
static_assert(sizeof(uuid) == UUID_SIZE, "Invalid size of uuid");
static constexpr const int MS_FROM_100NS_FACTOR = 10000;
static constexpr const uint64_t OFFSET_FROM_15_10_1582_TO_EPOCH = 122192928000000000;
struct two64s {
uint64_t n1;
uint64_t n2;
} contents;
std::memcpy(&contents, uuid.data, UUID_SIZE);
//    contents.n1 = __builtin_bswap64(contents.n1);
uint64_t timestamp = contents.n1 & UINT64_C(0x0FFFFFFFFFFFFFFF);
return (timestamp - OFFSET_FROM_15_10_1582_TO_EPOCH) / MS_FROM_100NS_FACTOR;
}
int main() {
std::cout << "Time now: " << (boost::posix_time::second_clock::universal_time() - boost::posix_time::ptime(boost::gregorian::date(1970, 1, 1))).total_milliseconds() << std::endl;
auto gen = boost::uuids::string_generator();
std::cout << "UUID: " << gen("49cbda60-961b-11e8-9854-134d5b3f9cf8") << std::endl;
std::cout << "Time from UUID: " << TimestampFromUUID(gen("49cbda60-961b-11e8-9854-134d5b3f9cf8")) << std::endl;
std::cout << "UUID: " << gen("58e0a7d7-eebc-11d8-9669-0800200c9a66") << std::endl;
std::cout << "Time from UUID: " << TimestampFromUUID(gen("58e0a7d7-eebc-11d8-9669-0800200c9a66")) << std::endl;
return 0;
}

该程序的输出是:

Time now: 1571735685000
UUID: 49cbda60-961b-11e8-9854-134d5b3f9cf8
Time from UUID: 45908323159150
UUID: 58e0a7d7-eebc-11d8-9669-0800200c9a66
Time from UUID: 45926063291384

你可以在这里玩这个源代码。

为什么我的结果甚至没有接近当前时间戳?我做错了什么?

我认为通过将 UUID 处理为字符串并使用字符串操作提取时间戳信息,然后将其转换为数值会更容易理解。诀窍是时间戳信息存储在 UUID 中的方式。从规范:

UUID 字符串表示形式的正式定义是 由以下ABNF [7]提供:

UUID                   = time-low "-" time-mid "-"
time-high-and-version "-"
clock-seq-and-reserved
clock-seq-low "-" node
time-low               = 4hexOctet
time-mid               = 2hexOctet
time-high-and-version  = 2hexOctet
clock-seq-and-reserved = hexOctet
clock-seq-low          = hexOctet
node                   = 6hexOctet
hexOctet               = hexDigit hexDigit
hexDigit =
"0" / "1" / "2" / "3" / "4" / "5" / "6" / "7" / "8" / "9" /
"a" / "b" / "c" / "d" / "e" / "f" /
"A" / "B" / "C" / "D" / "E" / "F"

以下是 UUID 的字符串表示形式示例 作为骨灰盒:

urn:uuid:f81d4fae-7dec-11d0-a765-00a0c91e6bf6

即 UUID 的第一部分(在 '-' 之前(是时间低音版本,第二部分是时间中段,第三部分是时间高版本,第一个字符是 UUID 版本。因此,我们需要拆分 UUID 并重新组合这些时间戳部分以创建完整的时间戳字符串,如下所示: {time-high minus-version}{time-mid}{time-low}

这是修改后的代码段。我以这个漂亮的javascript示例作为参考:https://stackoverflow.com/a/26915856/3694234

#include <boost/date_time.hpp>
#include <boost/uuid/uuid.hpp>
#include <boost/uuid/uuid_generators.hpp>
#include <boost/uuid/uuid_io.hpp>
#include <boost/algorithm/string.hpp>
#include <string>
#include <vector>
#include <cstdint>
#include <iostream>
uint64_t TimestampFromUUID(const boost::uuids::uuid& uuid) {
static constexpr const int UUID_SIZE = 16;
static_assert(sizeof(uuid) == UUID_SIZE, "Invalid size of uuid");
static constexpr const int MS_FROM_100NS_FACTOR = 10000;
static constexpr const uint64_t OFFSET_FROM_15_10_1582_TO_EPOCH = 122192928000000000;
/* convert uuid to string for manipulation */
std::string uuid_str = boost::uuids::to_string(uuid);
/* store uuid parts in a vector */
std::vector<std::string> uuid_parts;
/* split uuid with '-' as delimiter */
boost::split(uuid_parts, uuid_str, [](char c){return c == '-';});
/* first part of uuid is time-low
second part is time-mid
third part is time high with most significant 4 bits as uuid version
*/
std::string uuid_timestamp = uuid_parts[2].substr(1) + uuid_parts[1] + uuid_parts[0];
std::cout << std::endl << "UUID Timestamp : " << uuid_timestamp << std::endl;
uint64_t timestamp = std::stoul(uuid_timestamp, nullptr, 16);
return (timestamp - OFFSET_FROM_15_10_1582_TO_EPOCH) / MS_FROM_100NS_FACTOR;
}
int main() {
std::cout << "Time now: " << (boost::posix_time::second_clock::universal_time() - boost::posix_time::ptime(boost::gregorian::date(1970, 1, 1))).total_milliseconds() << std::endl;
auto gen = boost::uuids::string_generator();
std::cout << "UUID: " << gen("49cbda60-961b-11e8-9854-134d5b3f9cf8") << std::endl;
std::cout << "Time from UUID: " << TimestampFromUUID(gen("49cbda60-961b-11e8-9854-134d5b3f9cf8")) << std::endl;
std::cout << "UUID: " << gen("58e0a7d7-eebc-11d8-9669-0800200c9a66") << std::endl;
std::cout << "Time from UUID: " << TimestampFromUUID(gen("58e0a7d7-eebc-11d8-9669-0800200c9a66")) << std::endl;
return 0;
}

输出

Time now: 1571838175000
UUID: 49cbda60-961b-11e8-9854-134d5b3f9cf8
Time from UUID: 
UUID Timestamp : 1e8961b49cbda60
1533190458118
UUID: 58e0a7d7-eebc-11d8-9669-0800200c9a66
Time from UUID: 
UUID Timestamp : 1d8eebc58e0a7d7
1092575371981

恕我直言,你一直做得不好。阅读您提供的文档,我尝试从UUID重新生成时间戳。这是我的代码:

uint64_t TimestampFromUUID(const boost::uuids::uuid& uuid) {
static constexpr const int UUID_SIZE = 16;
static_assert(sizeof(uuid) == UUID_SIZE, "Invalid size of uuid");
static constexpr const int MS_FROM_100NS_FACTOR = 10000;
static constexpr const uint64_t OFFSET_FROM_15_10_1582_TO_EPOCH = 122192928000000000;
uint64_t timestamp = uuid.data[3] + (uuid.data[2] << 8) + (uuid.data[1] << 16) + (uuid.data[0] << 24);
timestamp += ((uint64_t)uuid.data[4] << 40) + ((uint64_t)uuid.data[5] << 32);
timestamp += ((uint64_t)uuid.data[7] << 48) + ((uint64_t)(uuid.data[6] & 0x0F) << 56);
return (timestamp - OFFSET_FROM_15_10_1582_TO_EPOCH) / MS_FROM_100NS_FACTOR;
}