我如何使用javascript返回两个日期之间的假期天数，不包括假期？

Question

我有一个代码返回使用 jquery datapicker 选择的两个日期之间的天数。

我想为假期添加一个函数，它将排除数组中的所有假期； var 假期 [25-12-2016,26-12-2016,1-1-2017];

请在下面找到代码：

            <script>
            <!--Calculate Leave days excluding weekends
            function calcBusinessDays(start, end) {
                // This makes no effort to account for holidays
                // Counts end day, does not count start day

                // make copies we can normalize without changing passed in objects    
                var start = new Date(start);
                var end = new Date(end);

                // initial total
                var totalBusinessDays = 0;

                // normalize both start and end to beginning of the day
                start.setHours(0,0,0,0);
                end.setHours(0,0,0,0);

                var current = new Date(start);
                current.setDate(current.getDate() + 1);
                var day;
                // loop through each day, checking
                while (current <= end) {
                    day = current.getDay();
                    if (day >= 1 && day <= 5) {
                        ++totalBusinessDays;
                    }
                    current.setDate(current.getDate() + 1);
                }
                return totalBusinessDays;
            }

                $(function() {
                $( "#start_date" ).datepicker({ minDate:0, showOn: 'button', buttonImageOnly: true, buttonImage: 'images/calendar.png', beforeShowDay: $.datepicker.noWeekends });
                    $( "#end_date" ).datepicker({ minDate:0, showOn: 'button', buttonImageOnly: true, buttonImage: 'images/calendar.png',beforeShowDay: $.datepicker.noWeekends,
                    onSelect: function (dateStr) {
                    var max = $(this).datepicker('getDate'); // Get selected date
                    $('#datepicker').datepicker('option', 'maxDate', max || '+1Y+12M'); // Set other max, default to +18 months
                    var start = $("#start_date").datepicker("getDate");
                    var end = $("#end_date").datepicker("getDate");
                    var days = (end - start) / (1000 * 60 * 60 * 24);
                    var diff = calcBusinessDays(start,end);
                    $("#leave_days").val(diff);

                } });

              });



            </script>

  <input name="start_date" type="text"   id="start_date"  />  

 <input name="end_date" type="text"   id="end_date" /> 

 <input  name="leave_days" type="text"  id="leave_days" size="32" class="form-control"/>  

Answer 1

就像在 cmets 中所说的，您必须定义假期数组。

对于这个例子，我定义了两个日期：2016-11-23 和 2016-12-02

您可以使用数据库或在脚本中手动执行，以便随着时间的推移维护相关日期。
这部分这里就不解释了，但是在脚本中，我使用了the default MySQL date format，也就是YYYY-MM-DD。
从数据库中获取假期日期应该很容易。

一个附加函数用于将current日期转换成这个MySQL日期格式，以便比较它。 然后，在while 循环中，我们检查日期是否为假日，如果是，则设置一个布尔标志，用于在条件中向计数器添加“工作日”或不添加“工作日”。

var holiday_array=["2016-11-23", "2016-12-02"];   // YYYY-MM-DD (Default MySQL date format)

function dateToMySQL (x){
    var MySQL_day = x.getDate();
    if(MySQL_day<10){
        MySQL_day = "0"+MySQL_day;      // Leading zero on day...
    }
    var MySQL_month = x.getMonth()+1;   // Months are zero-based.
    if(MySQL_month<10){
        MySQL_month = "0"+MySQL_month;  // Leading zero on month...
    }
    var MySQL_year = x.getYear()+1900;  // Years are 1900 based.
    var MySQL_date = MySQL_year+"-"+MySQL_month+"-"+MySQL_day;
    return MySQL_date;
}

function calcBusinessDays(start, end) {
    // This makes no effort to account for holidays
    // Counts end day, does not count start day

    // make copies we can normalize without changing passed in objects    
    var start = new Date(start);
    var end = new Date(end);

    // initial total
    var totalBusinessDays = 0;

    // normalize both start and end to beginning of the day
    start.setHours(0,0,0,0);
    end.setHours(0,0,0,0);

    // Prepare loop's variables
    var current = new Date(start);
    current.setDate(current.getDate() + 1);
    var day;
    var holidayFound=false;

    // loop through each day, checking
    while (current <= end) {
        //console.log("current: "+current);

        // Check if current is in the holiday array
        var MySQLdate = dateToMySQL(current);
        console.log("MySQL date: "+MySQLdate);

        if($.inArray(MySQLdate,holiday_array)!=-1){
            console.log("                  ^----------- Holiday!!!");
            holidayFound=true;     // "flag"
        }

        // If current is monday to friday and NOT a holiday
        day = current.getDay();
        if (day >= 1 && day <= 5 && !holidayFound) {
            ++totalBusinessDays;
        }

        // For next iteration
        current.setDate(current.getDate() + 1);
        holidayFound=false;
    }
    return totalBusinessDays;
}

$(function() {
    $( "#start_date" ).datepicker({
        minDate:0,
        showOn: 'button',
        buttonImageOnly: true,
        buttonImage: 'http://www.nscale.net/forums/images/misc/Tab-Calendar.png',    //'images/calendar.png',
        beforeShowDay: $.datepicker.noWeekends
    });
    $( "#end_date" ).datepicker({
        minDate:0,
        showOn: 'button',
        buttonImageOnly: true,
        buttonImage: 'http://www.nscale.net/forums/images/misc/Tab-Calendar.png',    //'images/calendar.png',
        beforeShowDay: $.datepicker.noWeekends,
        onSelect: function (dateStr) {
            var max = $(this).datepicker('getDate'); // Get selected date
            $('#datepicker').datepicker('option', 'maxDate', max || '+1Y+12M'); // Set other max, default to +18 months
            var start = $("#start_date").datepicker("getDate");
            var end = $("#end_date").datepicker("getDate");
            var days = (end - start) / (1000 * 60 * 60 * 24);
            var diff = calcBusinessDays(start,end);
            $("#leave_days").val(diff);

        }
    });
});

见CodePen（检查控制台；））

Answer 2

你可以使用这个逻辑：

把周末当作假期

function workingDaysBetweenDates(startDate, endDate) {
    var millisecondsPerDay = 86400 * 1000; 
    startDate.setHours(0,0,0,1);  
    endDate.setHours(23,59,59,999);  
    var diff = endDate - startDate;     
    var days = Math.ceil(diff / millisecondsPerDay);
    
    // Subtract two weekend days for every week in between
    var weeks = Math.floor(days / 7);
    days = days - (weeks * 2);

    // Handle special cases
    var startDay = startDate.getDay();
    var endDay = endDate.getDay();
    
    // Remove weekend not previously removed.   
    if (startDay - endDay > 1)         
        days = days - 2;      
    
    // Remove start day if span starts on Sunday but ends before Saturday
    if (startDay === 0 && endDay != 6)
        days = days - 1 ; 
            
    // Remove end day if span ends on Saturday but starts after Sunday
    if (endDay === 6 && startDay !== 0)
        days = days - 1  ;
    
    return days;
}

var a = new Date(2015, 10, 16); 
var b = new Date(2016, 01, 20);
var t = workingDaysBetweenDates(a,b);
alert(t);

希望这会有所帮助！