And again about calculate the number of days between two dates using JavaScript

Question

Seems that solution from How to calculate the number of days between two dates using JavaScript? work fine exept for case : 30.4.2016 - 1.5.2016 it calculate 2 days ( in real 1 ) 29.2.2016 - 1.3.2016 it calculate 3 days ( in real 1 )

May be someone already found solution for these cases ?

This function from link
function count_days(){ // Expecting resuls: positive when date1 > date2  ; negative  when date1 < date2  
            var $obj = "dd.mm.YYYY";
            var $obj2 = "dd.mm.YYYY";
            if($obj2.value){
                $start=$obj.value.split(".");
                var date1 = new Date($start[2],$start[1],$start[0]); // Converted date to ("YYYY,mm,dd")
                var $stop=$obj2.value.split(".");
                var date2 = new Date($stop[2],$stop[1],$stop[0]); // Converted date to ("YYYY,mm,dd")
                var oneDay = 24*60*60*1000; // hours*minutes*seconds*milliseconds
                var diffDays = (date2.getTime() - date1.getTime())/(oneDay);
                if(diffDays >= 0){
                    return Math.abs(diffDays)+1;
                } else {
                    return diffDays-1;
                }
            }
return false; // No date2 nothing to compare  
}
// Expecting resuls in full days (24h):  
//(5.5.2016 - 5.5.2016) =  1 day (days equal)  
//(29.2.2016 - 1.3.2016) =  1 day  
//(30.4.2016 - 1.5.2016) =  1 day  
//(4.5.2016 - 5.5.2016) =  2 days  
//(29.2.2016 - null) = False   
//(1.3.2016 - 29.2.2016) = -1 day  
//(1.5.2016 - 30.4.2016) = -1 day  
//(5.5.2016 - 4.5.2016)  = -2 day  

Answer 1

Note that new Date()'s month value starts from 0 (0 - January, 1 - February and etc...).

var 
  a = new Date(2016, 01, 29), // Feb 29 2016 00:00:00
  b = new Date(2016, 02, 1); // Mar 01 2016 00:00:00

alert((+b - +a) / 1000 / 60 / 60 / 24); // 1 day

An expression (+b - +a) - returns you the difference between two dates in milliseconds (1 second = 1000 millisecond).

As result you can convert:

• milliseconds to seconds: (+b - +a) / 1000 = 86400 seconds

• seconds to minutes: (+b - +a) / 1000 / 60 = 1440 minutes

• minutes to hours: (+b - +a) / 1000 / 60 / 60 = 24 hours

• hours to days: (+b - +a) / 1000 / 60 / 60 / 24 = 1 day

+a and +b - give you integer value representing the number of milliseconds since 1 January 1970 00:00:00 for each dates.

Read more about new Date()

The JavaScript date is based on a time value that is milliseconds since midnight 01 January, 1970 UTC. A day holds 86,400,000 milliseconds. The JavaScript Date object range is -100,000,000 days to 100,000,000 days relative to 01 January, 1970 UTC.

Answer 2

This code works for every case:

      function fmtDate( when ) {
    function D2( val ) {
      return ( val < 10 ) ? '0' + val : '' + val;
    }
    return D2( when.getMonth() + 1 ) + '/' + D2( when.getDate() ) + '/' + when.getFullYear();
  }

  function workdays( d1, d2 ) {
    var result = 0;
    var d0 = new Date();
    var negative = 1;
    if ( ( typeof( d1 ) == typeof( d2 ) ) && ( typeof( d1 ) == typeof( d0 ) ) ) {
      if ( d2 < d1 ) {  // Exchange/swap the dates
        d0 = d2;
        d2 = d1;
        d1 = d0;
        negative = -1;
      }
      for ( d = d1; d < d2; d.setDate( d.getDate() + 1 ) ) {
       dow = d.getDay();     // Day Of Week: 0 (Sun) .. 6 (Sat)
       when = fmtDate( d );  // Formatted date
        if ( ( dow >= 0 ) && ( dow <= 6 )) {

            result++;
        }
      }
    } else {
      alert( 'workdays() - parameter error - date objects required.' );
    }
    return result
  }

and if you remove "=" from this line:

if ( ( dow >= 0 ) && ( dow <= 6 ))

You can use this for working out how many workdays (excluding Sat and Sun)

Answer 3

Finaly good working code

function count_days(){ 
            var $obj = "dd.mm.YYYY";
            var $obj2 = "dd.mm.YYYY";
            if($obj2.value){
                $start=$obj.value.split(".");
                var date1 = new Date($start[2],$start[1]-1,$start[0]); // Converted date to ("YYYY,mm,dd")
                var $stop=$obj2.value.split(".");
                var date2 = new Date($stop[2],$stop[1]-1,$stop[0]); // Converted date to ("YYYY,mm,dd")
                var oneDay = 24*60*60*1000; // hours*minutes*seconds*milliseconds
                var diffDays = (date2.getTime() - date1.getTime())/(oneDay);
                if(diffDays >= 0){
                    return Math.abs(diffDays)+1; // positive as date1 > date2    
                } else {
                    return diffDays-1;  //negative  as date1 < date2
                }
            }
return false; // No date2 nothing to compare  
}