请教一个问题：两个时间相减，怎么得到多天，多少时，多少天分，多少秒
如题：例如我有两个时间：
Timer1：“2020-12-03 23:59:59”，
Timer2：“2013-04-08 08:00:00”

Timer1 － Timer2 得到的时间，怎么获知是多天，多少时，多少天分，多少秒？
[解决办法]

AnsiString sStr2 = "2013-04-08 08:00:00";
AnsiString sStr2 = "2020-12-03 23:59:59";
int nDay = StrToDateTime(sStr2) -  StrToDateTime(sStr); //天
int nHours = double(StrToDateTime(sStr2) -  StrToDateTime(sStr)) * 24; //小时
int nMin = double(StrToDateTime(sStr2) -  StrToDateTime(sStr)) * 24 * 60; //分钟
int nSec = double(StrToDateTime(sStr2) -  StrToDateTime(sStr)) * 24 * 60 * 60;//秒

[解决办法]
#include <DateUtils.hpp>

用下面这些现成的函数

int __fastcall YearsBetween(const System::TDateTime ANow, const System::TDateTime AThen);
 int __fastcall MonthsBetween(const System::TDateTime ANow, const System::TDateTime AThen);
 int __fastcall WeeksBetween(const System::TDateTime ANow, const System::TDateTime AThen);
 int __fastcall DaysBetween(const System::TDateTime ANow, const System::TDateTime AThen);
 __int64 __fastcall HoursBetween(const System::TDateTime ANow, const System::TDateTime AThen);
 __int64 __fastcall MinutesBetween(const System::TDateTime ANow, const System::TDateTime AThen);
 __int64 __fastcall SecondsBetween(const System::TDateTime ANow, const System::TDateTime AThen);
 __int64 __fastcall MilliSecondsBetween(const System::TDateTime ANow, const System::TDateTime AThen);