Validate number of days in a given month

Answer 1

If the month isn't February, get the number from the array. Otherwise, check if the year is leap to return 29, or return 28. Is there a problem with that?

Answer 2

I'm mostly agreeing w/ Moayad. I'd use a table lookup, with an if check on February and the year.

pseudocode:

Last_Day = Last_Day_Of_Month[Month];
Last_Day += (Month == February && Leap_Year(Year)) ? 1 : 0;

Note that Leap_Year() can't be implemented simply as (Year % 4 == 0), because the rules for leap years are way more complex than that. Here's an algorithm cribbed from Wikipedia

bool Leap_Year (int year) {
   return ((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0);
}

Answer 3

function caldays(m,y)
{

        if(m==01||m==03||m==05||m==07||m==08||m==10||m==12)
    {
     var dmax = 31;   
     return dmax;         

    }
    else if (m==04||m==06||m==09||m==11)
    {

            var dmax = 30;  
     return dmax;    

    }
    else
    {

     if((y%400==0) || (y%400==0 && y%100!=0))
     {

      dmax = 29;   
      return dmax;

     }
                else 
                {
                    dmax = 28;     
                }
     return dmax;

    }

}

source: http://www.dotnetspider.com/resources/20979-Javascript-code-get-number-days-perticuler-month-year.aspx

Answer 4

I agree with Moayad and TED. Stick with the lookup table unless the month is February. If you need an algorithm for checking leap years, wikipedia has two:

if year modulo 400 is 0 then leap
 else if year modulo 100 is 0 then no_leap
 else if year modulo 4 is 0 then leap
 else no_leap

A more direct algorithm (terms may be grouped either way):

function isLeapYear (year):
 if ((year modulo 4 is 0) and (year modulo 100 is not 0)) or (year modulo 400 is 0)
  then true
 else false

Answer 5

I've been doing this using the Date object (assuming it's compiled, and hence blindingly fast compared to scripting).

The trick is that if you enter a too high number for the date part, the Date object wraps over into the next month. So:

var year = 2009;
var month = 1;
var date = 29;

var presumedDate = new Date(year, month, date);

if (presumedDate.getDate() != date)
    WScript.Echo("Invalid date");
else
    WScript.Echo("Valid date");

This will echo "Invalid date" because presumedDate is actually March 1st.

This leaves all the trouble of leap years etc to the Date object, where I don't have to worry about it.

Neat trick, eh? Dirty, but that's scripting for you...

Answer 6

function daysInMonth(m, y) { // m is 0 indexed: 0-11
    switch (m) {
        case 1 :
            return (y % 4 == 0 && y % 100) || y % 400 == 0 ? 29 : 28;
        case 8 : case 3 : case 5 : case 10 :
            return 30;
        default :
            return 31
    }
}

function isValid(d, m, y) {
    return m >= 0 && m < 12 && d > 0 && d <= daysInMonth(m, y);
}

Answer 7

This will not perform as well as the accepted answer. I threw this in here because I think it is the simplest code. Most people would not need to optimize this function.

function validateDaysInMonth(year, month, day)
{
    if (day < 1 || day > 31 || (new Date(year, month, day)).getMonth() != month)
        throw new Error("Frack!");
}

It takes advantage of the fact that the javascript Date constructor will perform date arithmetic on dates that are out of range, e.g., if you do:

var year = 2001; //not a leap year!
var month = 1 //February
var day = 29; //not a valid date for this year
new Date(year, month, day);

the object will return Mar 1st, 2001 as the date.

Answer 8

Assuming the JS Date object standard where months are numbered from 0, and you have your daysInMonth array:

var days = daysInMonth[month] + ((month === 1) && (year % 4 === 0) && ((year % 100 !== 0) || (year % 400 === 0)));

will give you the number of days in the month, with 28 increased to 29 iff the month is February and the year is a leap year.

Answer 9

all this logic is already built in to the javascript engine... Why recode it ? Unless you are doing this just as an exercise, you can use the javascript Date object:

Like this:

function daysInMonth(aDate) {
      return new Date(aDate.getYear(), aDate.getMonth()+1, 0).getDate();      
   }

Answer 10

You can use DateTime to solve this:

new DateTime('20090901')->format('t'); // gives the days of the month