[英]Convert `time_t` to decimal year in C++
如何将time_t
结构转换为十进制年份?
例如,对于2015-07-18 00:00:00
的日期,我想获得2015.625
。
在我的评论中寻求关于你如何来到.625
更多信息,我将假设你实际上意味着.55
因为2015年7月18日是一年的第199天。
您需要首先使用get_time
从字符串中获取时间到std::tm
结构。 然后,在mktime
帮助下,我们应该能够获得一年中的这一天。 接下来,我们可以执行快速计算以查看年份是否为闰年,然后执行除法以获得我们的比率:
#include <ctime>
#include <iostream>
#include <sstream>
#include <locale>
#include <iomanip>
#include <string.h>
int main()
{
std::tm theTime = {};
// initialize timeToConvert with the Year-Month-Day Hours:Minutes:Seconds string you want
std::string timeToConvert = "2015-07-18 00:00:00";
std::istringstream timeStream(timeToConvert);
// need to use your locale (en-US)
timeStream.imbue(std::locale("en_US.UTF-8"));
timeStream >> std::get_time(&theTime, "%Y-%m-%d %H:%M:%S");
if (timeStream.fail())
{
std::cerr << "Parse failed\n";
exit(0);
}
// call mktime to fill out other files in theTime
std::mktime(&theTime);
// get years since 1900
int year = theTime.tm_year + 1900;
/* determine if year is leap year:
If the year is evenly divisible by 4, go to step 2. ...
If the year is evenly divisible by 100, go to step 3. ...
If the year is evenly divisible by 400, go to step 4. ...
The year is a leap year (it has 366 days).
The year is not a leap year (it has 365 days).
*/
bool isLeapYear = year % 4 == 0 &&
year % 100 == 0 &&
year % 400 == 0;
// get number of days since January 1st
int days = theTime.tm_yday+1; // Let January 1st be the 1st day of year, not 0th
// get number of days in this year (either 365 or 366 if leap year)
int daysInYear = isLeapYear ? 366 : 365;
double yearAsFloat = static_cast<double>(year) + static_cast<double>(days)/static_cast<double>(daysInYear);
std::cout << timeToConvert << " is " << yearAsFloat << std::endl;
}
输出:
2015-07-18 00:00:00是2015.55
您可以使用例如std::gmtime
或std::localtime
将其转换为细分时间 。
细分时间结构包含年份, tm_yday
成员是自1月1日以来的日期。您可以使用此tm_yday
成员计算小数点后的部分。
如果您想要更高的分辨率,请使用小时,分钟和秒。
使用strftime
struct stat info;
char buff[20];
struct tm * timeinfo;
stat(workingFile, &info);
timeinfo = localtime (&(info.st_mtime));
strftime(buff, 20, "%b %d %H:%M", timeinfo);
printf("%s",buff);
格式:
%b - The abbreviated month name according to the current locale.
%d - The day of the month as a decimal number (range 01 to 31).
%H - The hour as a decimal number using a 24-hour clock (range 00 to 23).
%M - The minute as a decimal number (range 00 to 59).
使用以下步骤:
std::gmtime
将月份与月份分开。 X
=日期和月份的天数(请记住,年份可能是闰年 ); 例如,对于2月3日, X
总是31 + 3 = 34;
3月3日,闰年为31 + 28 + 3 = 62
或31 + 29 + 3 = 63
; Y
=一年中的天数 (X * 100.) / Y
并显示三位小数
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