将日期时间转换回Windows 64位文件时间

Question

我想创建NT格式的网络时间戳。

我已经能够用这个函数将它们转换为可读时间:

NetworkStamp = "xc0x65x31x50xdex09xd2x01"
def GetTime(NetworkStamp):
    Ftime = int(struct.unpack('<Q',NetworkStamp)[0])
    Epoch = divmod(Ftime - 116444736000000000, 10000000)
    Actualtime = datetime.datetime.fromtimestamp(Epoch[0])
    return Actualtime, Actualtime.strftime('%Y-%m-%d %H:%M:%S')
print GetTime(NetworkStamp)

输出:

(datetime.datetime(2016, 9, 8, 11, 35, 57), '2016-09-08 11:35:57')

现在我想做相反的事情，将'2016/09/08 11:35:57'秒转换为以下格式:

 "xc0x65x31x50xdex09xd2x01"

Answer 1

      将窗口的FILETIME值转换为datetime.datetime值的代码并不像它可能的那样准确-它截断了可能存在的任何小数秒(因为它忽略了divmod()结果的其余部分)。这在代码创建的可读字符串中是不明显的，因为它只显示整个秒。
      即使包含小数秒，也不能完全按照您的需要进行操作，因为Windows FILETIME结构的值间隔为100纳秒(纳秒)。1微秒)，但Python的datetime只支持精确到整个微秒。因此，最好的方法就是接近原始值，因为即使进行最精确的转换，也会丢失信息。

以下是Python 2和Python 3的代码，使用问题中的NetworkStamp测试值来演示这一点:

import datetime
import struct
import time
WINDOWS_TICKS = int(1/10**-7)  # 10,000,000 (100 nanoseconds or .1 microseconds)
WINDOWS_EPOCH = datetime.datetime.strptime('1601-01-01 00:00:00',
                                           '%Y-%m-%d %H:%M:%S')
POSIX_EPOCH = datetime.datetime.strptime('1970-01-01 00:00:00',
                                         '%Y-%m-%d %H:%M:%S')
EPOCH_DIFF = (POSIX_EPOCH - WINDOWS_EPOCH).total_seconds()  # 11644473600.0
WINDOWS_TICKS_TO_POSIX_EPOCH = EPOCH_DIFF * WINDOWS_TICKS  # 116444736000000000.0
def get_time(filetime):
    """Convert windows filetime winticks to python datetime.datetime."""
    winticks = struct.unpack('<Q', filetime)[0]
    microsecs = (winticks - WINDOWS_TICKS_TO_POSIX_EPOCH) / WINDOWS_TICKS
    return datetime.datetime.fromtimestamp(microsecs)
def convert_back(timestamp_string):
    """Convert a timestamp in Y=M=D H:M:S.f format into a windows filetime."""
    dt = datetime.datetime.strptime(timestamp_string, '%Y-%m-%d %H:%M:%S.%f')
    posix_secs = int(time.mktime(dt.timetuple()))
    winticks = (posix_secs + int(EPOCH_DIFF)) * WINDOWS_TICKS
    return winticks
def int_to_bytes(n, minlen=0):  # helper function
    """ int/long to bytes (little-endian byte order).
        Note: built-in int.to_bytes() method could be used in Python 3.
    """
    nbits = n.bit_length() + (1 if n < 0 else 0)  # plus one for any sign bit
    nbytes = (nbits+7) // 8  # number of whole bytes
    ba = bytearray()
    for _ in range(nbytes):
        ba.append(n & 0xff)
        n >>= 8
    if minlen > 0 and len(ba) < minlen:  # zero pad?
        ba.extend([0] * (minlen-len(ba)))
    return ba  # with low bytes first
def hexbytes(s):  # formatting helper function
    """Convert string to string of hex character values."""
    ba = bytearray(s)
    return ''.join('\x{:02x}'.format(b) for b in ba)
win_timestamp = b'xc0x65x31x50xdex09xd2x01'
print('win timestamp: b"{}"'.format(hexbytes(win_timestamp)))
dtime = get_time(win_timestamp)
readable = dtime.strftime('%Y-%m-%d %H:%M:%S.%f')  # includes fractional secs
print('datetime repr: "{}"'.format(readable))
winticks = convert_back(readable)  # includes fractional secs
new_timestamp = int_to_bytes(winticks)
print('new timestamp: b"{}"'.format(hexbytes(new_timestamp)))

输出:

win timestamp: b"xc0x65x31x50xdex09xd2x01"
datetime repr: "2016-09-08 07:35:57.996998"
new timestamp: b"x80x44x99x4fxdex09xd2x01"

Answer 2

如果您了解如何在一个方向上执行转换，那么反向执行转换基本上就是以相反顺序使用每个方法的逆。只需查看您正在使用的模块/类的文档:

1. strftime有对应的strptime
2. fromtimestamp由timestamp匹配(如果你使用3.3之前的Python, timestamp不存在，但你可以在函数外定义FILETIME_epoch = datetime.datetime(1601, 1, 1) - datetime.timedelta(seconds=time.altzone if time.daylight else time.timezone)来预先计算一个datetime，它代表你的时区的FILETIME epoch，然后使用int((mydatetime - FILETIME_epoch).total_seconds())直接获得FILETIME epoch以来的int秒，而无需手动调整FILETIME和Unix epoch之间的差异)
3. divmod(你并不真正需要，因为你只使用商，而不是余数，你可以只做Epoch = (Ftime - 116444736000000000) // 10000000，以后避免索引)是微不足道的可逆的(只是乘和加，如果你使用我的技巧直接从#2转换为FILETIME epoch秒，则不需要添加)
4. struct.unpack匹配struct.pack

我没有提供确切的代码，因为你真的应该自己学习使用这些东西(必要时阅读文档);我猜你的前向代码是在不理解它在做什么的情况下编写的，因为如果你理解了它，反向代码应该是显而易见的;

Answer 3

基本转换:

from datetime import datetime
from calendar import timegm
EPOCH_AS_FILETIME = 116444736000000000  # January 1, 1970 as MS file time
HUNDREDS_OF_NANOSECONDS = 10000000
def dt_to_filetime(dt):
    return EPOCH_AS_FILETIME + (timegm(dt.timetuple()) * HUNDREDS_OF_NANOSECONDS)

答案是基于这个要点我发现:https://gist.github.com/Mostafa-Hamdy-Elgiar/9714475f1b3bc224ea063af81566d873

Gist增加了对时区和其他方向的转换的支持。