我有 2 个输入,其中选择了日期 #startdate 和 #enddate。我正在使用 datepicker,目前禁用了周末和节假日,这很好用!
然后我有一个#days 输入,它计算两个日期之间的差异,然后这个日期差异有一个周末计数。
我想创建一个假期计数,这样我也可以从天数差中减去这个。
所以我最终会得到
$('#days').val((Math.abs(($d2new-$d1new)/86400000) - weekend_count - holidaycount) + 1);
我当前的代码如下(这是从其他 stackoverflow 问题中提取的,并且运行良好:)底部是我正在努力解决的 var holidaycount。任何帮助将不胜感激。
//holidays
var natDays = [
[1, 1, 'uk'],
[1, 2, 'uk'],
[1, 3, 'uk'],
[1, 4, 'uk'],
[12, 24, 'uk'],
[12, 25, 'uk'],
[12, 26, 'uk'],
[12, 27, 'uk'],
[12, 28, 'uk'],
[12, 29, 'uk'],
[12, 30, 'uk'],
[12, 31, 'uk']
];
function noWeekendsOrHolidays(date) {
var noWeekend = $.datepicker.noWeekends(date);
if (noWeekend[0]) {
return nationalDays(date);
} else {
return noWeekend;
}
}
function nationalDays(date) {
for (i = 0; i < natDays.length; i++) {
if (date.getMonth() == natDays[i][0] - 1 && date.getDate() == natDays[i][1]) {
return [false, natDays[i][2] + '_day'];
}
}
return [true, ''];
}
// do initialization here
$("#startdate").datepicker({
dateFormat: 'dd-mm-yy',
changeMonth: true,
changeYear: true,
firstDay: 1,
yearRange: '0:+100',
beforeShowDay: noWeekendsOrHolidays,
onSelect: function( selectedDate ) {
$("#enddate").datepicker("option","minDate",selectedDate );
$("#enddate2").datepicker("option","minDate",selectedDate );
},
minDate: '+1d',
maxDate: '+' + DAY_DIFFERENCE + 'd'
});
// do initialization here
$("#enddate").datepicker({
dateFormat: 'dd-mm-yy',
changeMonth: true,
changeYear: true,
firstDay: 1,
yearRange: '0:+100',
beforeShowDay: noWeekendsOrHolidays,
maxDate: '+' + DAY_DIFFERENCE + 'd'
});
// do initialization here
$("#enddate2").datepicker({
dateFormat: 'dd-mm-yy',
changeMonth: true,
changeYear: true,
firstDay: 1,
yearRange: '0:+100',
beforeShowDay: noWeekendsOrHolidays,
onSelect: function( selectedDate ) {
$d1 = $('#startdate').val();
$d2 = $('#enddate2').val();
$myDateParts1 = $d1.split("-");
$myDateParts2 = $d2.split("-");
$d1flip = new Date($myDateParts1[2], ($myDateParts1[1]-1), $myDateParts1[0]);
$d2flip = new Date($myDateParts2[2], ($myDateParts2[1]-1), $myDateParts2[0]);
$newdate1 = $d1flip.format("ddd mmm dd hh:MM:ss yyyy");
$newdate2 = $d2flip.format("ddd mmm dd hh:MM:ss yyyy");
// For Opera and older winXP IE n such
$d1new = Date.parse($newdate1);
$d2new = Date.parse($newdate2);
var weekend_count = 0;
for (i = $d1new.valueOf(); i <= $d2new.valueOf(); i+= 86400000){
var temp = new Date(i);
if (temp.getDay() == 0 || temp.getDay() == 6) {
weekend_count++;
}
}
var holidaycount = 0;
for (i = $d1new.valueOf(); i <= $d2new.valueOf(); i+= 86400000){
var temp = new Date(i);
if (**date in var natDays & is in selected difference range**) {
holidaycount++;
}
}
console.log(weekend_count);
console.log(holidaycount);
$('#days').val((Math.abs(($d2new-$d1new)/86400000) - weekend_count - holidaycount) + 1);
}
});
}, 'html');
return false;
});
}
编辑
从答案中,我插入了函数并尝试了这个.... console.log(holidaycount); 返回 0 但我选择的日期应该是 10
var holidaycount = 0;
for (i = $d1new.valueOf(); i <= $d2new.valueOf(); i+= 86400000){
var temp = new Date(i);
if (isHoliday(temp)) {
holidaycount++;
}
}