Question

我希望获得这些值，以便提前10分钟制作一个日期时间。然后以这种格式返回新的日期时间：yyyy / mm / dd / HH / MM

int yyyy = 2010;
int month = 11;
int day = 18;
int hour = 12;
int minute = 10;

Datetime?

Answer 1

查看Boost :: Date_Time。

http://www.boost.org/doc/libs/1_44_0/doc/html/date_time.html

编辑：以下是http://www.boost.org/doc/libs/1_36_0/doc/html/date_time/examples.html的示例。

 /* Some simple examples of constructing and calculating with times
   * Output:
   * 2002-Feb-01 00:00:00 - 2002-Feb-01 05:04:02.001000000 = -5:04:02.001000000
   */

  #include "boost/date_time/posix_time/posix_time.hpp"
  #include <iostream>

  int
  main() 
  {
    using namespace boost::posix_time;
    using namespace boost::gregorian;

    date d(2002,Feb,1); //an arbitrary date
    //construct a time by adding up some durations durations
    ptime t1(d, hours(5)+minutes(4)+seconds(2)+millisec(1));
    //construct a new time by subtracting some times
    ptime t2 = t1 - hours(5)- minutes(4)- seconds(2)- millisec(1);
    //construct a duration by taking the difference between times
    time_duration td = t2 - t1;

    std::cout << to_simple_string(t2) << " - " 
              << to_simple_string(t1) << " = "
              << to_simple_string(td) << std::endl;

    return 0;
  }

Answer 2

#include <boost/date_time.hpp>
#include <iostream>
#include <sstream>
#include <locale>

int main(int argc, char* argv[])
{
    int yyyy = 2010;
    int month = 11;
    int day = 18;
    int hour = 12;
    int minute = 10;

    boost::gregorian::date date(yyyy, month, day);
    boost::posix_time::ptime time(date,
        boost::posix_time::hours(hour) +
        boost::posix_time::minutes(minute)
    );

    boost::posix_time::time_facet* facet = new boost::posix_time::time_facet();
    facet->format("%Y/%m/%d/%H/%M");

    std::ostringstream oss;
    oss.imbue(std::locale(oss.getloc(), facet));
    oss << time;

    std::cout << oss.str() << std::endl;
}

Answer 3

以下是可能有用的代码段。查看MSDN以获取有关所用函数的更多详细信息。不要忘记用零填充结构。

struct tm now;
memset(&now, 0, sizeof(struct tm));

now.tm_year = 2010 - 1900; // years start from 1900
now.tm_mon = 11 - 1; // Months start from 0
now.tm_mday = 18;
now.tm_hour = 12;
now.tm_min = 10; // you have day here, but I guess it's minutes

time_t afterTime = mktime(&now) + 10 * 60; // time_t is time in seconds

struct tm *after = localtime(&afterTime);

编辑：我犹豫是否写了将日期时间写入字符串的函数，因为它是C而不是C ++，但有时人们只需要一个解决方案而不介意它来自哪个库。所以：

char output[17]; // 4+1+2+1+2+1+2+1+2 format  +1 zero terminator
if (strftime(output, sizeof(output), "%Y/%m/%d/%H/%M", after) == 0)
   handle_error();

C ++：提前10分钟的日期时间实例

3 个答案: