Solved

number of days in a month

Posted on 2001-06-02
13
212 Views
Last Modified: 2010-04-06
how can i get the number of days in a month?
ex.
   On DECEMBER have 31 days
      November have only 30 days
      February (have leap year) on 28 or ..
0
Comment
Question by:ryan_sabarre
13 Comments
 

Author Comment

by:ryan_sabarre
ID: 6148596
some sourcecode will do
0
 
LVL 4

Expert Comment

by:fva
ID: 6148635
if month<12 then
days:=EncodeDate(year,month+1,1)-EncodeDate(year,month,1)
else
days:=EncodeDate(year+1,1,1)-EncodeDate(year,12,1);

F.
0
 

Author Comment

by:ryan_sabarre
ID: 6148689
I think you code is lacking and cant run with my program
0
 

Author Comment

by:ryan_sabarre
ID: 6148691
can you complete it
0
 
LVL 4

Expert Comment

by:fva
ID: 6148738

function DaysInMonth(year:integer;month:integer):integer;
begin
     if (month<1) or (month>12) then
        raise Exception.Create('Invalid month in DaysInMonth call')
     else if month<12 then
        Result:=EncodeDate(year,month+1,1)-EncodeDate(year,month,1)
     else
        Result:=EncodeDate(year+1,1,1)-EncodeDate(year,12,1);
end;

F.
0
 
LVL 4

Accepted Solution

by:
fva earned 50 total points
ID: 6148739
Oh, sorry; my mistake. It should be:

function DaysInMonth(year:integer;month:integer):integer;
begin
    if (month<1) or (month>12) then
       raise Exception.Create('Invalid month in DaysInMonth call')
    else if month<12 then
       Result:=Trunc(EncodeDate(year,month+1,1)-EncodeDate(year,month,1))
    else
       Result:=Trunc(EncodeDate(year+1,1,1)-EncodeDate(year,12,1));
end;

0
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LVL 4

Expert Comment

by:DMN
ID: 6148769
procedure DaysPerMonth(Month,Year:integer):integer;
const
  dmp:array[1..12] of integer=(
  31,28,31,30,31,30,31,31,30,31,30,31
  );
begin
  Result := dmp[Month];
  if Month=2 then
    if Year mod 100 = 0 then
      if Year mod 400 = 0 then
        Inc(Result)
      else
    else
      if Year mod 4 = 0 then
        Inc(Result);
end;
0
 
LVL 1

Expert Comment

by:martin_g
ID: 6148863
Hi,
This is no original, I copied from the JEDI source found in JCLDateTime.pas which has just about any kind of date/ time functions imaginable!

DaysInMonths: array [1..12] of Integer = (31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);

function DaysInMonth(const DateTime: TDateTime): Integer;
var
  M: Integer;
begin
  M := MonthOfDate(DateTime);
  Result := DaysInMonths[M];
  if M = 2 then
    if IsLeapYear(DateTime) then
      Result := 29;
end;

{Don't forget to add overload to the 2 IsLeapYear functions below in your declaration section
 function IsLeapYear(const Year: Integer): Boolean; overload;
function IsLeapYear(const DateTime: TDateTime): Boolean; overload;
Is Leap year:  The original Gregorian rule for all who want to learn it
 Result := (Year mod 4 = 0) and ((Year mod 100 <> 0) or (Year mod 400 = 0)); }


//-----------------------------------------------------------------------------

function IsLeapYear(const DateTime: TDateTime): Boolean;
begin
  Result := IsLeapYear(YearOfDate(DateTime));
end;

//------------------------------------------------------------------------------

function IsLeapYear(const Year: Integer): Boolean;
begin
  Result := SysUtils.IsLeapYear(Year);
end;
0
 
LVL 1

Expert Comment

by:bakry99
ID: 6148975
try this :


Function GetMonthDays( Month, Year   :Integer)   :integer;
Begin
  If Month = 2 Then
  Begin
    if Year Mod 4=0 Then
      Result:= 29
    Else
      Result:= 28
  End
  Else
    If Month<8 Then
      Result:= 30 + Month Mod 2
    Else
      Result:=  31 - Month Mod 2;
End;

Regards Bakry
0
 
LVL 6

Expert Comment

by:edey
ID: 6149022
Interesting sugestion Bakry - just one thing.  Year that end with 00 (1900, 2000, 2100 etc) will be leap years _only_ if they are evenly divisible by 400 (1600, 2000, 2400 etc)


GL
Mike
0
 

Author Comment

by:ryan_sabarre
ID: 6149050
Hey fva i just come up with other way i just
change something about your source.

because some errors occured about the
Incompatible types ..integer and Extented ERROR


I comeup with this

function DaysInMonth(year:integer;month:integer):Integer;
var
 nTemp:TdateTime;
begin
    if (month<1) or (month>12) then
       raise Exception.Create('Invalid month in DaysInMonth call')
    else if month<12 then
       nTemp:=EncodeDate(year,month+1,1)-EncodeDate(year,month,1)
    else
       nTemp:=EncodeDate(year+1,1,1)-EncodeDate(year,12,1);
   Result:=strtoint(formatdatetime('dd',nTemp)) + 1;
end;
0
 

Author Comment

by:ryan_sabarre
ID: 6149054
Thanks for the Idea
0
 
LVL 4

Expert Comment

by:fva
ID: 6151519
Don't mention it, but why C?

F.
0

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