在 windows 将时间字符串转换为纪元时间

Question

我有一个获取文件的界面，文件名是内部数据有效时的时间戳。我正在按顺序处理文件并将它们提供给另一个 API，它自 1970 年 1 月 1 日起以毫秒为单位。

所以，我以 YYYYMMDD_HHmmSS.sss 格式的字符串形式获取文件名，并且必须将其转换为时间。我假设处理和收集发生在同一时区等，只需将字符转换为数字即可。

uint64_t foo::convertFileNameToTimestamp(const std::string& filename) const
{
    const uint64_t yyyy = atoi(filename.substr(0,4).c_str());
    const uint64_t MM = atoi(filename.substr(4,2).c_str());
    const uint64_t DD = atoi(filename.substr(6,2).c_str());
    const uint64_t HH = atoi(filename.substr(9,2).c_str());
    const uint64_t MI = atoi(filename.substr(11,2).c_str());
    const uint64_t SS = atoi(filename.substr(13,2).c_str());
    const uint64_t sss = atoi(filename.substr(16,3).c_str());

    // number of milliseconds in a day
    const uint64_t DAY = 24 * 60 * 60 * 1000;
    int MD = 0;
    if (MM == 2) MD = 31; // currently feb, so add all of january's days
    else if (MM == 3) MD = 31+28; // ...
    else if (MM == 4) MD = 31+28+31;
    else if (MM == 5) MD = 31+28+31+30;
    else if (MM == 6) MD = 31+28+31+30+31;
    else if (MM == 7) MD = 31+28+31+30+31+30;
    else if (MM == 8) MD = 31+28+31+30+31+30+31;
    else if (MM == 9) MD = 31+28+31+30+31+30+31+31;
    else if (MM == 10) MD = 31+28+31+30+31+30+31+31+30;
    else if (MM == 11) MD = 31+28+31+30+31+30+31+31+30+31;
    else if (MM == 12) MD = 31+28+31+30+31+30+31+31+30+31+30;

    // year 2000 wasn't a leap year
    uint64_t YLD = ((yyyy-1970) / 4);
    if (yyyy > 2000) --YLD;

    uint64_t temp = sss;
    temp += SS * 1000;
    temp += MI * 60 * 1000;
    temp += HH * 60 * 60 * 1000;
    temp += (DD-1) * DAY;
    temp += (MD) * DAY;
    temp += (yyyy-1970) * 365 * DAY + YLD*DAY;

    return temp;
}

显然，在这里重新发明轮子。似乎应该有某种功能。另外..我如何计算闰秒？处理闰日已经够烦人的了。时间戳都是从 2015 年及以后开始的，并且永远都是，但我不认为我可以盲目地添加 26 秒。最终我们将有 27 个或备份到 25 个。在之前的函数中，我已经验证了字符串格式正确、文件存在以及所有爵士乐。我正在运行 windows 8.1 使用 VS 2010 编译 64 位。

我看过 Convert Epoch Time string to Time，它建议 ctime()，但它似乎不处理构造函数中的毫秒数，甚至不处理任何 get方法，它不接受通用格式的字符串输入。我假设我必须调用一些时间类 CTOR，它将接受文件名字符串，然后在其上调用一些访问器以获取自 1970 年以来的毫秒时间，包括闰秒等。

我没有使用 boost，也没有 access/permission 来使用它。

Answer 1

您可以使用这段代码（您不必担心闰年和所有相关问题）。 @Edit1：修改代码以考虑闰秒；还将其重组为 class:

Foo.h:

#ifndef __FOO__H__
#define __FOO__H__

#include <string>
#include <Windows.h>
#include <stdint.h>

class CFoo {
private:
    const static int kLeapSecsDim = 26;
    static uint64_t msecsBetweenEpochs;
    static SYSTEMTIME leapSecs[kLeapSecsDim];

    ULARGE_INTEGER leapSecsUi[kLeapSecsDim];

    int CFoo::getLeapSeconds(ULARGE_INTEGER ui) const;
public:
    CFoo();
    ~CFoo() {};

    uint64_t toEpoch(const std::string& filename) const;
};

#endif  //__FOO__H__

Foo.cpp:

#include "Foo.h"

uint64_t CFoo::msecsBetweenEpochs = 11644473600000; /* Milliseconds between 1.1.1601 and 1.1.1970 */
SYSTEMTIME CFoo::leapSecs[CFoo::kLeapSecsDim] = 
                            {{1972, 06, 0, 30, 23, 59, 59, 999},
                             {1972, 12, 0, 31, 23, 59, 59, 999},
                             {1973, 12, 0, 31, 23, 59, 59, 999},
                             {1974, 12, 0, 31, 23, 59, 59, 999},
                             {1975, 12, 0, 31, 23, 59, 59, 999},
                             {1976, 12, 0, 31, 23, 59, 59, 999},
                             {1977, 12, 0, 31, 23, 59, 59, 999},
                             {1978, 12, 0, 31, 23, 59, 59, 999},
                             {1979, 12, 0, 31, 23, 59, 59, 999},
                             {1981, 06, 0, 30, 23, 59, 59, 999},
                             {1982, 06, 0, 30, 23, 59, 59, 999},
                             {1983, 06, 0, 30, 23, 59, 59, 999},
                             {1985, 06, 0, 30, 23, 59, 59, 999},
                             {1987, 12, 0, 31, 23, 59, 59, 999},
                             {1989, 12, 0, 31, 23, 59, 59, 999},
                             {1990, 12, 0, 31, 23, 59, 59, 999},
                             {1992, 06, 0, 30, 23, 59, 59, 999},
                             {1993, 06, 0, 30, 23, 59, 59, 999},
                             {1994, 06, 0, 30, 23, 59, 59, 999},
                             {1995, 12, 0, 31, 23, 59, 59, 999},
                             {1997, 06, 0, 30, 23, 59, 59, 999},
                             {1998, 12, 0, 31, 23, 59, 59, 999},
                             {2005, 12, 0, 31, 23, 59, 59, 999},
                             {2008, 12, 0, 31, 23, 59, 59, 999},
                             {2012, 06, 0, 30, 23, 59, 59, 999},
                             {2015, 06, 0, 30, 23, 59, 59, 999},
                            };


int CFoo::getLeapSeconds(ULARGE_INTEGER ui) const {
    int ret = 0;
    for (int i = 0; i < kLeapSecsDim; i++) {
        if (ui.QuadPart <= this->leapSecsUi[i].QuadPart)
            break;
        ret++;
    }
    return ret;
}

CFoo::CFoo() {
    FILETIME ft;
    BOOL res;
    for (int i = 0; i < this->kLeapSecsDim; i++) {
        res = SystemTimeToFileTime(&(this->leapSecs[i]), &ft);
        if (res == FALSE)
            throw std::exception("SystemTimeToFileTime error", GetLastError());
        this->leapSecsUi[i].LowPart = ft.dwLowDateTime;
        this->leapSecsUi[i].HighPart = ft.dwHighDateTime;
    }
}

uint64_t CFoo::toEpoch(const std::string& filename) const {
    SYSTEMTIME st;
    FILETIME ft;
    ULARGE_INTEGER ui;
    st.wYear = atoi(filename.substr(0, 4).c_str());
    st.wMonth = atoi(filename.substr(4, 2).c_str());
    st.wDay = atoi(filename.substr(6, 2).c_str());
    st.wHour = atoi(filename.substr(9, 2).c_str());
    st.wMinute = atoi(filename.substr(11, 2).c_str());
    st.wSecond = atoi(filename.substr(13, 2).c_str());
    st.wMilliseconds = atoi(filename.substr(16, 3).c_str());
    BOOL result = SystemTimeToFileTime(&st, &ft);
    if (result == FALSE)
        throw std::exception("SystemTimeToFileTime error", GetLastError());
    ui.HighPart = ft.dwHighDateTime;
    ui.LowPart = ft.dwLowDateTime;
    //printf("%016I64X - %I64u\n", ui.QuadPart, ui.QuadPart);
    //printf("%016I64X - %I64u\n", ui.QuadPart/10000, ui.QuadPart/10000);
    return (ui.QuadPart / 10000) - this->msecsBetweenEpochs + this->getLeapSeconds(ui) * 1000;
}

备注:

无效dates/timesSystemTimeToFileTime会失败
常量CFoo::msecsBetweenEpochs我觉得可以在Google上找到；我从 Python(2.7.10) 的 posixmodule.c 拿来的（实际上有秒数；我只必须乘以 1000)
您的实施产生的结果不是很准确（我使用 http://www.epochconverter.com 作为参考）。
根据SystemTimeToFileTime，时间戳为UTC。

Answer 2

这是一个适用于任何支持 C++11 或 C++14 的平台的答案。它建立在 C++11 中引入的 std::chrono 库之上。它还使用免费、开源、跨平台的库来简化算法（通常被认为是 lawyer-friendly）的 MIT 许可证。

如果你不需要考虑闰秒，你可以使用this date library，它看起来像这样：

#include <string>
#include "date.h"

using time_stamp = std::chrono::time_point<std::chrono::system_clock,
                                           std::chrono::milliseconds>;

time_stamp
convertFileNameToTimestamp(const std::string& filename)
{
    using namespace std::chrono;
    using namespace date;
    const uint64_t yyyy = atoi(filename.substr(0,4).c_str());
    const uint64_t MM = atoi(filename.substr(4,2).c_str());
    const uint64_t DD = atoi(filename.substr(6,2).c_str());
    const uint64_t HH = atoi(filename.substr(9,2).c_str());
    const uint64_t MI = atoi(filename.substr(11,2).c_str());
    const uint64_t SS = atoi(filename.substr(13,2).c_str());
    const uint64_t sss = atoi(filename.substr(16,3).c_str());
    return sys_days{year(yyyy)/MM/DD}
         + hours{HH} + minutes{MI} + seconds{SS} + milliseconds{sss};
}

从文件名中解析出数字后，创建一个 type-safe std::chrono::time_point 非常简单，它仅保存自 1970-01-01 以来的整数毫秒数（作为 int64_t).

如果要考虑闰秒，需要this higher-level library which is a complete parser of the IANA timezone database. You will also need to keep an updated copy of the IANA timezone database downloaded for my timezone/leap-second library解析。但是一旦设置好，您的转换器的源代码与上面的非常相似，而且几乎一样简单：

#include <string>
#include "tz.h"

using time_stamp_ls = std::chrono::time_point<date::utc_clock,
                                              std::chrono::milliseconds>;

time_stamp_ls
convertFileNameToTimestamp_ls(const std::string& filename)
{
    using namespace std::chrono;
    using namespace date;
    const uint64_t yyyy = atoi(filename.substr(0,4).c_str());
    const uint64_t MM = atoi(filename.substr(4,2).c_str());
    const uint64_t DD = atoi(filename.substr(6,2).c_str());
    const uint64_t HH = atoi(filename.substr(9,2).c_str());
    const uint64_t MI = atoi(filename.substr(11,2).c_str());
    const uint64_t SS = atoi(filename.substr(13,2).c_str());
    const uint64_t sss = atoi(filename.substr(16,3).c_str());
    return utc_clock::sys_to_utc(sys_days{year(yyyy)/MM/DD}
         + hours{HH} + minutes{MI} + seconds{SS} + milliseconds{sss});
}

这两个功能都可以通过以下 HelloWorld 来实现：

#include <iostream>

int
main()
{
    std::string filename = "20150830_002120.123";
    std::cout << convertFileNameToTimestamp   (filename).time_since_epoch().count() << '\n';
    std::cout << convertFileNameToTimestamp_ls(filename).time_since_epoch().count() << '\n';
}

输出：

1440894080123
1440894106123

请注意，这些时间戳恰好相隔 26,000 毫秒。

更新

"tz.h" header 现在包含一个 parse 函数，这使得编写这些函数更更容易：

date::sys_time<std::chrono::milliseconds>
convertFileNameToTimestamp(const std::string& filename)
{
    using namespace std::chrono;
    using namespace date;
    std::istringstream in{filename};
    sys_time<milliseconds> tp;
    parse(in, "%Y%m%d_%H%M%S", tp);
    return tp;
}

date::utc_time<std::chrono::milliseconds>
convertFileNameToTimestamp_ls(const std::string& filename)
{
    return date::to_utc_time(convertFileNameToTimestamp(filename));
}

在 windows 将时间字符串转换为纪元时间

Converting time string into epoch time on windows

c++

windows

time