再次关于使用 JavaScript 计算两个日期之间的天数

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  
}