C++ 在 Unix/Linux 中将 Windows 文件时间转换为秒
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Convert Windows Filetime to second in Unix/Linux
提问by ARH
I have a trace file that each transaction time represented in Windows filetime format. These time numbers are something like this:
我有一个跟踪文件，每个事务时间都以 Windows 文件时间格式表示。这些时间数字是这样的：
- 128166372003061629
- 128166372016382155
- 128166372026382245
- 128166372003061629
- 128166372016382155
- 128166372026382245
Would you please let me know if there are any C/C++ library in Unix/Linux to extract actual time (specially second) from these numbers ? May I write my own extraction function ?
您能否告诉我 Unix/Linux 中是否有任何 C/C++ 库可以从这些数字中提取实际时间（特别是秒）？我可以编写自己的提取函数吗？
回答by Eugene
it's quite simple: the windows epoch starts 1601-01-01T00:00:00Z. It's 11644473600 seconds before the UNIX/Linux epoch (1970-01-01T00:00:00Z). The Windows ticks are in 100 nanoseconds. Thus, a function to get seconds from the UNIX epoch will be as follows:
这很简单：Windows 纪元从 1601-01-01T00:00:00Z 开始。它比 UNIX/Linux 时代 (1970-01-01T00:00:00Z) 早 11644473600 秒。Windows 滴答以 100 纳秒为单位。因此，从 UNIX 时代获取秒数的函数如下：
#define WINDOWS_TICK 10000000
#define SEC_TO_UNIX_EPOCH 11644473600LL
unsigned WindowsTickToUnixSeconds(long long windowsTicks)
{
return (unsigned)(windowsTicks / WINDOWS_TICK - SEC_TO_UNIX_EPOCH);
}
回答by ashaw
FILETIME type is is the number 100 ns increments since January 1 1601.
FILETIME 类型是自 1601 年 1 月 1 日以来 100 ns 增量的数字。
To convert this into a unix time_t you can use the following.
要将其转换为 unix time_t，您可以使用以下命令。
#define TICKS_PER_SECOND 10000000
#define EPOCH_DIFFERENCE 11644473600LL
time_t convertWindowsTimeToUnixTime(long long int input){
long long int temp;
temp = input / TICKS_PER_SECOND; //convert from 100ns intervals to seconds;
temp = temp - EPOCH_DIFFERENCE; //subtract number of seconds between epochs
return (time_t) temp;
}
you may then use the ctime functions to manipulate it.
然后您可以使用 ctime 函数来操作它。
回答by Howard Hinnant
New answer for old question.
旧问题的新答案。
Using C++11's <chrono>plus this free, open-source library:
使用 C++11<chrono>加上这个免费的开源库：
https://github.com/HowardHinnant/date
https://github.com/HowardHinnant/date
One can very easily convert these timestamps to std::chrono::system_clock::time_point, and also convert these timestamps to human-readable format in the Gregorian calendar:
人们可以很容易地将这些时间戳转换为std::chrono::system_clock::time_point，也可以将这些时间戳转换为公历中人类可读的格式：
#include "date.h"
#include <iostream>
std::chrono::system_clock::time_point
from_windows_filetime(long long t)
{
using namespace std::chrono;
using namespace date;
using wfs = duration<long long, std::ratio<1, 10'000'000>>;
return system_clock::time_point{floor<system_clock::duration>(wfs{t} -
(sys_days{1970_y/jan/1} - sys_days{1601_y/jan/1}))};
}
int
main()
{
using namespace date;
std::cout << from_windows_filetime(128166372003061629) << '\n';
std::cout << from_windows_filetime(128166372016382155) << '\n';
std::cout << from_windows_filetime(128166372026382245) << '\n';
}
For me this outputs:
对我来说，这个输出：
2007-02-22 17:00:00.306162
2007-02-22 17:00:01.638215
2007-02-22 17:00:02.638224
On Windows, you can actually skip the floor, and get that last decimal digit of precision:
在 Windows 上，您实际上可以跳过floor, 并获得最后一个十进制数字的精度：
return system_clock::time_point{wfs{t} -
(sys_days{1970_y/jan/1} - sys_days{1601_y/jan/1})};
2007-02-22 17:00:00.3061629
2007-02-22 17:00:01.6382155
2007-02-22 17:00:02.6382245
With optimizations on, the sub-expression (sys_days{1970_y/jan/1} - sys_days{1601_y/jan/1})will translate at compile time to days{134774}which will further compile-time-convert to whatever units the full-expression requires (seconds, 100-nanoseconds, whatever). Bottom line: This is both very readable and very efficient.
优化后，子表达式(sys_days{1970_y/jan/1} - sys_days{1601_y/jan/1})将在编译时days{134774}转换为将进一步编译时转换为完整表达式所需的任何单位（秒、100 纳秒等）。底线：这既非常易读又非常有效。
回答by Stan Sieler
(I discovered I can't enter readable code in a comment, so...)
（我发现我无法在评论中输入可读代码，所以...）
Note that Windows can represent times outside the range of POSIX epoch times, and thus a conversion routine should return an "out-of-range" indication as appropriate. The simplest method is:
请注意，Windows 可以表示超出 POSIX 纪元时间范围的时间，因此转换例程应适当返回“超出范围”指示。最简单的方法是：
... (as above)
long long secs;
time_t t;
secs = (windowsTicks / WINDOWS_TICK - SEC_TO_UNIX_EPOCH);
t = (time_t) secs;
if (secs != (long long) t) // checks for truncation/overflow/underflow
return (time_t) -1; // value not representable as a POSIX time
return t;
回答by Johan K?hler
The solution that divides and adds will not work correctly with daylight savings.
使用夏令时，分割和添加的解决方案将无法正常工作。
Here is a snippet that works, but it is for windows.
这是一个有效的片段，但它适用于 Windows。
time_t FileTime_to_POSIX(FILETIME ft)
{
FILETIME localFileTime;
FileTimeToLocalFileTime(&ft,&localFileTime);
SYSTEMTIME sysTime;
FileTimeToSystemTime(&localFileTime,&sysTime);
struct tm tmtime = {0};
tmtime.tm_year = sysTime.wYear - 1900;
tmtime.tm_mon = sysTime.wMonth - 1;
tmtime.tm_mday = sysTime.wDay;
tmtime.tm_hour = sysTime.wHour;
tmtime.tm_min = sysTime.wMinute;
tmtime.tm_sec = sysTime.wSecond;
tmtime.tm_wday = 0;
tmtime.tm_yday = 0;
tmtime.tm_isdst = -1;
time_t ret = mktime(&tmtime);
return ret;
}
回答by Johan K?hler
Assuming you are asking about the FILETIMEStructure, then FileTimeToSystemTimedoes what you want, you can get the seconds from the SYSTEMTIMEstructure it produces.
假设您询问FILETIME结构，然后FileTimeToSystemTime执行您想要的操作，您可以从它生成的SYSTEMTIME结构中获取秒数。
回答by Bluebaron
Here's essentially the same solution except this one encodes negative numbers from Ldap properly and lops off the last 7 digits before conversion.
这里基本上是相同的解决方案，除了这个解决方案正确编码来自 Ldap 的负数并在转换前去掉最后 7 位数字。
public static int LdapValueAsUnixTimestamp(SearchResult searchResult, string fieldName)
{
var strValue = LdapValue(searchResult, fieldName);
if (strValue == "0") return 0;
if (strValue == "9223372036854775807") return -1;
return (int)(long.Parse(strValue.Substring(0, strValue.Length - 7)) - 11644473600);
}
回答by Bluebaron
Also here's a pure C#ian way to do it.
这里还有一个纯粹的 C#ian 方式来做到这一点。
(Int32)(DateTime.FromFileTimeUtc(129477880901875000).Subtract(new DateTime(1970, 1, 1))).TotalSeconds;
Here's the result of both methods in my immediate window:
这是我的即时窗口中两种方法的结果：
(Int32)(DateTime.FromFileTimeUtc(long.Parse(strValue)).Subtract(new DateTime(1970, 1, 1))).TotalSeconds;
1303314490
(int)(long.Parse(strValue.Substring(0, strValue.Length - 7)) - 11644473600)
1303314490
DateTime.FromFileTimeUtc(long.Parse(strValue))
{2011-04-20 3:48:10 PM}
Date: {2011-04-20 12:00:00 AM}
Day: 20
DayOfWeek: Wednesday
DayOfYear: 110
Hour: 15
InternalKind: 4611686018427387904
InternalTicks: 634389112901875000
Kind: Utc
Millisecond: 187
Minute: 48
Month: 4
Second: 10
Ticks: 634389112901875000
TimeOfDay: {System.TimeSpan}
Year: 2011
dateData: 5246075131329262904
回答by lexxmt
If somebody need convert it in MySQL
如果有人需要在 MySQL 中转换它
SELECT timestamp,
FROM_UNIXTIME(ROUND((((timestamp) / CAST(10000000 AS UNSIGNED INTEGER)))
- CAST(11644473600 AS UNSIGNED INTEGER),0))
AS Converted FROM events LIMIT 100
