如何在不诉诸TimeToStr的情况下从TTime中删除秒（const datetime：TDateTime; const formatsettings：TFormatSettings）

Question

如何在不使用TimeToStr（const datetime：TDateTime; const formatsettings：TFormatSettings）来获取没有秒的TTime值的额外开销的情况下从TTime变量中删除秒数？

即这是我需要的 - ＆gt; HH：MM：00

是否有某种数学运算（如ANDing或ORing值与某些东西）可以执行？

Answer 1

uses
  ..., DateUtils;

var
  t: TTime;
begin
  t := ...;
  t := RecodeSecond(t, 0);
end;

Answer 2

var
  t: TTime;
  iHour, iMin, iSec, iMSec: Word;

DecodeTime(t, iHour, iMin, iSec, iMSec);
t := EncodeTime(iHour, iMin, 0, 0);

Answer 3

var
t : TTime;

t := Trunc(t * MinsPerDay) / MinsPerDay;

编辑：

这是截断秒数的更准确的功能。 它会在截断前将时间四舍五入到最接近的毫秒。

uses
  SysUtils;

const
  FMSecsPerDay: Double = MSecsPerDay;
  FMSecsPerMinute: Double = SecsPerMin * MSecsPerSec;  
  FMinsPerDay: Double = HoursPerDay * MinsPerHour;


function TruncateSeconds(aTime: TDateTime): TDateTime;
begin
  { - Round to closest millisecond before truncating the seconds }
  Result := Trunc(Round(aTime * FMSecsPerDay) / FMSecsPerMinute) / FMinsPerDay;
  //              -- Time in Milliseconds --
  //        ------------- Time in minutes -----------------------
end;

Answer 4

LongTimeFormat="hh:mm";
Label1->Caption=TimeToStr(Time());