我有一个界面,我获取文件,文件名是内部数据有效时的时间戳。我按顺序处理文件并将它们提供给另一个API,自1970年1月1日起,该时间为毫秒。
因此,我将文件名作为字符串以YYYYMMDD_HHmmSS.sss的格式获取,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,只需将字符转换为数字。
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?处理闰日很烦人。时间戳都来自2015年及以后,而且总是如此,但我不认为我可以盲目地增加26秒。最终我们将有27或最多25个。在以前的函数中,我已经验证了字符串是正确的格式和存在的文件以及所有爵士乐。我使用VS 2010运行Windows 8.1编译64位。
我查看了提出 ctime()的Convert Epoch Time string to Time,但它似乎没有处理构造函数中的毫秒,甚至任何get方法并且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得自1970年以来的毫秒时间,包括闰秒等。
我没有使用提升功能,也没有权限使用它。
答案 0 :(得分:0)
你可以使用这段代码(你不必担心闰年和所有相关的东西)。 @ Edit1:修改代码以考虑闰秒;还将其重组为一个类:
<强> Foo.h
强>:
#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__
<强> Foo.cpp
强>:
#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;
}
备注强>:
对于无效的日期/时间SystemTimeToFileTime
将失败
常数CFoo::msecsBetweenEpochs
我认为它可以在Google上找到;我从 Python(2.7.10)&#39; s posixmodule.c 中取出它(实际上有秒数;我只需将它乘以1000)
您的实施会产生不太准确的结果(我使用http://www.epochconverter.com作为参考)。
根据SystemTimeToFileTime,时间戳位于 UTC 。
答案 1 :(得分:0)
这个答案适用于任何支持C ++ 11或C ++ 14的平台。它建立在C ++ 11中引入的std::chrono
库之上。它还使用免费的开源跨平台库来简化算术(MIT许可证通常被认为是律师友好型)。
如果您不需要考虑闰秒,可以使用this date library,它将如下所示:
#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};
}
解析文件名后的数字后,创建一个类型安全std::chrono::time_point
非常简单,它自1970-01-01以来只保留一个整数毫秒(作为int64_t
)
如果您想考虑闰秒,则需要this higher-level library,这是IANA timezone database的完整解析器。您还需要保留为IANA timezone database下载的my timezone/leap-second library的更新副本以进行解析。但是一旦设置好,转换器的源代码与上面的代码非常相似,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});
}
这两个函数都可以使用以下HelloWorld来执行:
#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。
<强>更新强>
"tz.h"
标题现在包含parse
函数,这使得编写这些函数更多更容易:
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));
}