jQuery显示明天日期无周末或每年的假期 [英] jQuery display Tomorrows date No weekend or Holidays by year

问题描述

请有人帮我更新此信息，以便也可以从数组中检测年份.我要尝试的目标是使标签不包括假日或周末.因此，当列出假日或周末时，它将显示下一个可用的日期.例如，如果今天(星期四)10/15/15，标签将显示"DUE 10/16/15 @ 5:00".但是，如果明天(星期五)10/16/15，标签将显示"DUE 10/19/15 @ 5:00".同样的做法也适用于假期，这将在下一个工作日而不是周末.

Will someone please help me update this to be able to detect the year from the array as well. The goal I am trying to do is have the label not include the holiday days or weekends. So when a holiday is listed or its the weekend it will display the next available day. For example if today (thursday) 10/15/15, the label will show "DUE 10/16/15 @ 5:00". But if tomorrow (friday) 10/16/15, the label will show "DUE 10/19/15 @ 5:00". And the same would apply for holidays it would be the next available day and not on the weekend.

这就是我所拥有的: http://jsfiddle.net/byyeh83t/1/其中在将年份考虑在内之前，它似乎一直有效，因为该阵列仅按月份和日期运行.因此，我现在还需要添加一些内容来检查年份.

This is what I have: http://jsfiddle.net/byyeh83t/1/ which appears to work until you take the year into consideration because the array is only going by the month and day. So I now need to add something to check the year as well.

请有人帮我调整一下以检查正确的年份. http://jsfiddle.net/byyeh83t/3/

Will someone please help me adjust this to check for the correct year as well. http://jsfiddle.net/byyeh83t/3/

如果1/18/15是星期日，则输出应为1/20/15，因为1/19/15是MLK假期，但返回1/19/15

If 1/18/15 is the date which is a sunday the output should be 1/20/15 since 1/19/15 is MLK holiday but instead it returns 1/19/15

$(document).ready(function() {

var natDays = [
    [2014, 1, 1, 'New Year'], 
    [2014, 1, 20, 'Martin Luther King'], 
    [2014, 2, 17, 'Washingtons Birthday'],       
    [2014, 5, 26, 'Memorial Day'], 
    [2014, 7, 4, 'Independence Day'], 
    [2014, 9, 1, 'Labour Day'], 
    [2014, 10, 13, 'Columbus Day'], 
    [2014, 11, 11, 'Veterans Day'], 
    [2014, 11, 27, 'Thanksgiving Day'], 
    [2014, 11, 28, 'Thanksgiving Day'],
    [2014, 12, 25, 'Christmas'],  
    [2014, 12, 26, 'Christmas'], 
    [2015, 1, 1, 'New Year'], 
    [2015, 1, 19, 'Martin Luther King'], 
    [2015, 2, 16, 'Washingtons Birthday'],       
    [2015, 5, 25, 'Memorial Day'], 
    [2015, 7, 3, 'Independence Day'], 
    [2015, 9, 7, 'Labour Day'], 
    [2015, 10, 12, 'Columbus Day'], 
    [2015, 11, 11, 'Veterans Day'], 
    [2015, 11, 26, 'Thanksgiving Day'], 
    [2015, 11, 27, 'Thanksgiving Day'],
    [2015, 12, 24, 'Christmas'],  
    [2015, 12, 25, 'Christmas']  
    ];


// dateMin is the minimum delivery date
var dateMin = new Date("1/18/2015");
dateMin.setDate(dateMin.getDate() + (dateMin.getHours() >= 14 ? 1 : 0));


function AddBusinessDays(curdate, weekDaysToAdd) {
    var date = new Date(curdate.getTime());
    while (weekDaysToAdd > 0) {
        date.setDate(date.getDate() + 1);
        //check if current day is business day
        if (noWeekendsOrHolidays(date)[0]) {
            weekDaysToAdd--;
        }
    }
    return date;
}


function noWeekendsOrHolidays(date) {
    var noWeekend = $.datepicker.noWeekends(date);
    return (noWeekend[0] ? nationalDays(date) : 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, ''];
}


function setDeliveryDate(date) {
    $('#delivery-date').text($.datepicker.formatDate('mm/dd/yy', date));
}


setDeliveryDate(AddBusinessDays(dateMin, 1));

});

任何帮助将不胜感激.

推荐答案

您应该:

• 将natDays[i][0](以前是月份)更改为natDays[i][1]
• 将natDays[i][1](以前是一天)更改为natDays[i][2]
• 将natDays[i][2](以前是描述)更改为natDays[i][3]
• 将此条件添加到nationalDays函数中的if:date.getFullYear() == natDays[i][0]

• Change natDays[i][0] (which used to be the month) to natDays[i][1]
• Change natDays[i][1] (which used to be the day) to natDays[i][2]
• Change natDays[i][2] (which used to be the description) to natDays[i][3]
• Add this condition to the if in nationalDays function: date.getFullYear() == natDays[i][0]

赞:

if (date.getFullYear() == natDays[i][0] && date.getMonth() == natDays[i][1] - 1 && date.getDate() == natDays[i][2]) {
    return [false, natDays[i][3] + '_day'];
}

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