求Delphi中日期时间相减的方法
如题。
s1, s2: string; //日期时间是字符串型

  s1 := '2013-06-28 15:44:50';
  s2 := '2013-06-28 16:47:51';
//我如何得到格式为 '00天01小时03分01秒' 的结果？

[解决办法]
DateUtils单元中的SecondsBetween

返回的是int64，表示的是秒数，然后你自己算就好了
[解决办法]
先将字符串转换成日期格式，然后用下面的代码处理：http://bbs.csdn.net/topics/390395202

unit Unit1;

interface

uses
  Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms,
  Dialogs, DateUtils, ComCtrls, StdCtrls;

type
  TForm1 = class(TForm)
    Button1: TButton;
    Memo1: TMemo;
    DateTimePicker1: TDateTimePicker;
    DateTimePicker2: TDateTimePicker;
    DateTimePicker3: TDateTimePicker;
    procedure Button1Click(Sender: TObject);
  private
    { Private declarations }
    function TimeBetween(A, B: TDateTime): string;
  public
    { Public declarations }
  end;

var
  Form1: TForm1;

implementation

{$R *.dfm}

procedure TForm1.Button1Click(Sender: TObject);
begin
  Memo1.Lines.Add(TimeBetween(DateTimePicker1.Date, DateTimePicker2.Date));
end;

function TForm1.TimeBetween(A, B: TDateTime): string;
var
  Y, M, Day: Integer;
  C, D: TDateTime;
  sA, sB: string;
begin
  if Trunc(A) > Trunc(B) then
  begin
    Result := 'Error';
    exit;
  end;

  if Trunc(A) = Trunc(B) then
  begin
    Result := '0Y0M0D';
    exit; 
 
  end;

  Y := YearsBetween(B, A);

  M := MonthsBetween(B, A);
  M := M - Y * MonthsPerYear;
  Day := DaysBetween(B, A) - Trunc(Y * ApproxDaysPerYear) - Trunc(M * ApproxDaysPerMonth);
  Result := IntToStr(Y) +'Y'+ IntToStr(M) +'M'+ IntToStr(Day) +'D';
end;

end.

[解决办法]

const
  s1 = '2013-06-28 15:44:50';
  s2 = '2013-06-28 16:47:51';
var t,tmp:TDatetime;
    d,h,n,s:integer;
    str:string;
begin
  t:=abs(strtodatetime(s2)-strtodatetime(s1));
  d:=round(t);
  h:=round((t-d)*24);
  tmp:=d+h/24;
  n:=round((t-tmp)*1440);
  tmp:=tmp+n/1440;
  s:=round((t-tmp)*86400);
  str:=inttostr(d)+'天'+inttostr(h)+'小时'+inttostr(n)+'分'+inttostr(s)+'秒';
  if s1>s2 then str:='负了'+str;
  showmessage('时间：'+#13+'    '+s2
              +#13+'减去'+#13+'    '+s1
              +#13+#13+'结果为：'+str);

[解决办法]
转成日期，相减，再按整数对应天，1/24对应1小时，1/1440对应1分钟，1/86400对应1秒钟来组成字符串。
[解决办法]
完美无错运行

uses
  SysUtils, DataUtils;

var
  S1, S2: string;
  T1, T2: TDateTime;
  D, H, M, S: Integer;
  Value: Int64;
begin
  DateSeparator := '-';
  S1 := '2013-06-28 15:44:50';
  S2 := '2013-06-28 16:47:51';
  T1 := StrToDateTime(S1);
  T2 := StrToDateTime(S2);

  Value := SecondsBetween(T1, T2);
  D := Value div SecsPerDay;
  H := Value mod SecsPerDay div SecsPerHour;
  M := Value mod SecsPerDay mod SecsPerHour div SecsPerMin;
  S := Value mod SecsPerDay mod SecsPerHour mod SecsPerMin;
  Caption := Format('%.2d天%.2d小时%.2d分%.2d秒', [D, H, M, S]); 
 
end;