I need help with a problem with dates

[Anon]

first off not girl date time dates.

I need to write a program that computes the number of days between two dates.
I have this which I "borrowed" from someone, I am lost otherwise Help a redman out PLEASE.

int ckdate(int mm,int dd,int yyyy);
int leapYear(int year);
void printMsg(int value);
void showDate(int type, int day, int mth, int year);
void getDate(int &mm, int &dd, int &yyyy); //pass by reference Alias

#include <iostream.h>
#include <conio.h>
#include <ctype.h>
#include <stdio.h>
void main()
{
int mm1,dd1,yyyy1,mm2,dd2,yyyy2,mm3,dd3,yyyy3,ret;
char cont, fill;
do
{
getDate(mm1, dd1, yyyy1);
ret=ckdate(mm1,dd1,yyyy1);
printMsg(ret);
showDate(1,dd1,mm1,yyyy1);

	  cout << "\n\n Enter date #2 = mm/dd/yyyy  "; cin >>mm2 >> fill >> dd2 >> fill >> yyyy2;

	  ret=ckdate(mm2,dd2,yyyy2);
	  printMsg(ret);
	  showDate(2,dd2,mm2,yyyy2);

	  cout << "\n\n Enter date #3 = mm/dd/yyyy  "; cin >>mm3 >> fill >> dd3 >> fill >> yyyy3;

	  ret=ckdate(mm3,dd3,yyyy3);
	  printMsg(ret);
	  showDate(3,dd3,mm3,yyyy3);


	  cout << "\n\n Continue y/n ";
	  cin >> cont;
	}
	while (toupper(cont) != 'N');
	cout << "\n\nProgram completed\n";
}

int ckdate(int mm,int dd,int yyyy)
{
if( yyyy < 1000 || yyyy > 9999 )
return 1;

switch(mm)
{	  
	case 1:	case 3:	case 5:	case 7:	case 8:
    case 10: case 12:
		if(dd > 31 || dd < 1)
			return 3;
		break;
	case 4:	case 6:	case 9:	case 11:
    	if(dd > 30 || dd < 1)
			return 4;
		break;
	case 2:
		if( leapYear(yyyy) == 1 )
		{
			if(dd < 1 || dd > 29)
				return 5;
		}
		else
		{
			if(dd < 1 || dd > 28)
				return 6;
		}
		break;
	default:
		return 2;
}

return 0;

}

int leapYear(int yyyy)
{
if(yyyy%400 == 0) //leap century
return 1;

int yy = yyyy%100; // find yy-99 from yyyy-1999
if (yy%4 == 0 && yy != 0)
return 1;

return 0;
}

void printMsg(int value)
{

switch(value)
{
	 case 0:
		cout << "\n Good DATE";
		break;
	 case 1:
		cout << "\n Bad Year";
		break;
	 case 2:
		cout << "\n Bad Month";
		break;
	 case 3:
		cout << "\n Bad DAY not 1-31";
		break;
	 case 4:
		cout << "\n Bad Day not 1-30";
		break;
	 case 5:
		cout << "\n Bad Day not 1-29";
		break;
	 case 6:
		cout << "\n Bad Day not 1-28";
		break;
	 default:
		cout << "\n Bad RET CODE";
}

}

void showDate(int type, int day, int mth, int year)
{
cout << "\n ";
switch(type)
{
case 1:
cout << mth << "/" << day << "/" << year;
break;
case 2:
cout << day << "/" << mth << "/" << year;
break;
case 3:
cout << year << "/" << mth << "/" << day;
break;
}

}

void getDate(int &mm, int &dd, int &yyyy)
{

char fill;
cout << "\n\n Enter date = mm/dd/yyyy  "; cin >>mm >> fill >> dd >> fill >> yyyy;

}

[cheald]

Much easier way to do this: use unix timestamps.

Then, do this:

function getDiff($d1, $m1, $y1, $d2, $m2, $y2) {

$date1 = mktime(0,0,0,$m1,$d1,$y1);
$date2 = mktime(0,0,0,$m2,$d2,$y2);

$diff = abs($date1 - $date2);
$days = $diff / (60 60 24);
return floor($days);
}

There you go. Lots less code 🙂