C++：将 20160120 形式的日期转换为纪元以来的天数

Question

假设我们有一个表示日期的整数，格式为 20160120。我想将其转换为自纪元 (1970-01-01) 以来的天数，以便它对应于 POSIX 样式的日期。因此，例如 int 20160718 将变为 int 17000。

我的尝试如下，我首先将 int 转换为字符串，然后转换为自纪元以来的秒数，然后转换为自纪元以来的天数。这样做的一个问题是我必须将日期移动 3 小时才能获得正确的日期（请参阅std::to_string(d*100 + 3)）行。我怀疑这与时区有关？我在UTC - 1。我不知道如何处理。

所以我想知道是否有一种不太复杂的方法，如果没有，我该如何解决时区问题。

#include <iostream>
#include <chrono>
#include <time.h>
#include <iomanip>
#include <sstream>
#include <string>

typedef std::chrono::duration<int, std::ratio<60 * 60 * 24>> days_type;
typedef std::chrono::system_clock sysclock;

int convert(const unsigned d){
    std::tm t = {};
    std::istringstream ss(std::to_string(d*100 + 3));   
    ss >> std::get_time(&t, "%Y%m%d%H");

    time_t t_ =  mktime(&t);

    /*
     Now convert seconds since epoch to days since epoch via chrono...
     */

    sysclock::time_point tp = sysclock::from_time_t(t_);

    std::chrono::time_point<sysclock, days_type> tp_day = 
    std::chrono::time_point_cast<days_type>(tp);

    return tp_day.time_since_epoch().count();
}


int main(){
    int d = 20160718;
    std::cout << d << ": " << convert(d) << std::endl; //17000
    return 0;
}

Answer 1

使用date library：

#include "date.h"
#include <iostream>

int convert(unsigned d) {
  date::year_month_day ymd{date::year(d / 100 / 100),
                           date::month(d / 100 % 100), date::day(d % 100)};
  return date::sys_days{ymd}.time_since_epoch().count();
}

int main() {
  int d = 20160718;
  std::cout << d << ": " << convert(d) << '\n'; // 17000
}

See it run

当然，convert() API 会通过更强的类型来改进，例如直接返回一个date::sys_days 来表示时间点。

Answer 2

time_t 可能不是您系统上 POSIX 纪元开始时的 0。

这是一个显示不同结果的程序：

#include "date.h" // uses this library: https://github.com/HowardHinnant/date
#include <iostream>
#include <chrono>
#include <ctime>
#include <iomanip>
#include <string>

time_t convert_local(int d)
{
    std::tm t{};
    auto ts{std::to_string(d)};
    ts.insert(8, " ");
    ts.insert(6, " ");
    ts.insert(4, " "); // this is a workaround for a VS2015 bug: http://stackoverflow.com/q/21172767/1460794
    std::istringstream ss(ts);
    ss >> std::get_time(&t, "%Y %m %d");
    if(!ss) std::cout << "parse fail.\n";

    time_t t_{mktime(&t)};
    return t_;
}

time_t convert_UTC(int d)
{
    std::tm base_t{};
    base_t.tm_mday = 1;
    base_t.tm_mon = 0;
    base_t.tm_year = 70;
    base_t.tm_wday = 4;
    auto base_tt{std::mktime(&base_t)};

    std::tm t{};
    auto ts{std::to_string(d)};
    ts.insert(8, " ");
    ts.insert(6, " ");
    ts.insert(4, " "); // this is a workaround for a VS2015 bug: http://stackoverflow.com/q/21172767/1460794
    std::istringstream ss(ts);
    ss >> std::get_time(&t, "%Y %m %d");
    if(!ss) std::cout << "parse fail.\n";

    time_t t_{mktime(&t)};
    return t_ - base_tt;
}

typedef std::chrono::duration<int, std::ratio<60 * 60 * 24>> days_type;
typedef std::chrono::system_clock sysclock;

int days_since_epoch(time_t t)
{
    sysclock::time_point tp = sysclock::from_time_t(t);
    std::chrono::time_point<sysclock, days_type> tp_day =
        std::chrono::time_point_cast<days_type>(tp);
    return tp_day.time_since_epoch().count();
}

int main()
{
    std::tm t{};

    t.tm_mday = 1;
    t.tm_mon = 0;
    t.tm_year = 70;
    t.tm_wday = 4;

    auto tt{std::mktime(&t)};
    auto tp{std::chrono::system_clock::from_time_t(tt)};

    time_t t0{0};
    auto tp0{std::chrono::system_clock::from_time_t(t0)};

    using namespace date;
    std::cout << tp << " for Jan 1 1970 00:00 UTC in my timezone (-6)\n";
    std::cout << std::chrono::system_clock::now() << " is the time now in my timezone (-6)\n";
    std::cout << tp0 << " is what we get from time_t{0} in my timezone (-6)\n";

    int d = 20160718;
    std::cout << "\n\n";
    std::cout << d << " converted to time_t is " << convert_local(d) << " but it considers local timezone on my system.\n";
    std::cout << "Now, using this time_t, converted to days since epoch: " << days_since_epoch(convert_local(d));

    std::cout << "\n\n";
    std::cout << d << " converted to time_t, adjusted for local timezone is " << convert_UTC(d) << ".\n";
    std::cout << "Now, using this time_t, converted to days since epoch: " << days_since_epoch(convert_UTC(d));

    return 0;
}

在我的系统上它产生：

1970-01-01 06:00:00.0000000 for Jan 1 1970 00:00 UTC in my timezone (-6)
2016-11-05 14:42:19.1299886 is the time now in my timezone (-6)
1970-01-01 00:00:00.0000000 is what we get from time_t{0} in my timezone (-6)


20160718 converted to time_t is 1468821600 but it considers local timezone on my system.
Now, using this time_t, converted to days since epoch: 17000

20160718 converted to time_t, adjusted for local timezone is 1468800000.
Now, using this time_t, converted to days since epoch: 17000

在另一台服务器 (live demo) 上生成：

1970-01-01 00:00:00.000000000 for Jan 1 1970 00:00 UTC in my timezone
2016-11-05 14:37:15.661541264 is the time now in my timezone
1970-01-01 00:00:00.000000000 is what we get from time_t{0} in my timezone


20160718 converted to time_t is 1468800000 but it considers local timezone on my system.
Now, using this time_t, converted to days since epoch: 17000

20160718 converted to time_t, adjusted for local timezone is 1468800000.
Now, using this time_t, converted to days since epoch: 17000

---

最后，最好为每个时间点明确地获取您想要使用的两个时间点（通过指定日、月和年）。

这样您就可以使用system_clock，这并不重要，因为我们只希望以天为单位计算持续时间：

#include <iostream>
#include <chrono>
#include <ctime>
#include <iomanip>
#include <string>
#include <sstream>

auto get_epoch_tp()
{
    std::tm base_t{};
    base_t.tm_mday = 1;
    base_t.tm_mon = 0;
    base_t.tm_year = 70;
    base_t.tm_wday = 4;
    auto base_tt{std::mktime(&base_t)};
    auto tp{std::chrono::system_clock::from_time_t(base_tt)};
    return tp;
}

int get_days_since_epoch(int d)
{
    std::tm t{};
    auto ts{std::to_string(d)};
    ts.insert(8, " ");
    ts.insert(6, " ");
    ts.insert(4, " "); // this is a workaround for a VS2015 bug: http://stackoverflow.com/q/21172767/1460794
    std::istringstream ss(ts);
    ss >> std::get_time(&t, "%Y %m %d");
    if(!ss) std::cout << "parse fail.\n";
    time_t tt{mktime(&t)};

    auto tp{std::chrono::system_clock::from_time_t(tt)};

    using days_type = std::chrono::duration<int, std::ratio<60 * 60 * 24>>;
    auto duration{tp - get_epoch_tp()};
    auto days{std::chrono::duration_cast<days_type>(duration)};
    return days.count();
}

int main()
{
    int d = 20160718;
    std::cout << d << ": " << get_days_since_epoch(d) << std::endl; //17000
}

demo

Answer 3

您可以在转换为时间后进行简单的数学运算。

#include <iostream>
#include <time.h>
#include <iomanip>
#include <sstream>

int convert(const unsigned d){
    std::tm t = {};
    std::istringstream ss(std::to_string(d));  
    ss >> std::get_time(&t, "%Y%m%d");

    time_t t_ = mktime(&t);
    return (int)t_/(60*60*24);
}


int main(){
    int d = 20160718;
    std::cout << d << ": " << convert(d) << std::endl; //17000
    return 0;
}