jQuery UI datepicker-禁用公共假期，周末，上午10点后的第二天，并且只允许将星期二，星期三和星期四作为可选日期

Question

我正在使用来自JQuery UI Datepicker下一天第二天中午12点之后的代码来禁用周末，公共假期和第二天（如果在上午10点之后选择），但是我只能选择星期二，星期三和星期四。

// dates
var dateMin = new Date();
var weekDays = AddWeekDays(1);
dateMin.setDate(dateMin.getDate() + weekDays);
var natDays = [
  [1, 1, 'uk'],
  [12, 25, 'uk'],
  [12, 26, 'uk']
];
function noWeekendsOrHolidays(date) {
    var noWeekend = $j.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, ''];
}
function AddWeekDays(weekDaysToAdd) {
    var mydate = new Date();
    if (mydate.getHours()>=10) 
    var daysToAdd = 1;
    else var daysToAdd = 0;
    var day = mydate.getDay()
    weekDaysToAdd = weekDaysToAdd - (5 - day)
    if ((5 - day) < weekDaysToAdd || weekDaysToAdd == 1) {
        daysToAdd = (5 - day) + 2 + daysToAdd
    } else { // (5-day) >= weekDaysToAdd
        daysToAdd = (5 - day) + daysToAdd
    }
    while (weekDaysToAdd != 0) {
        var week = weekDaysToAdd - 5
        if (week > 0) {
            daysToAdd = 7 + daysToAdd
            weekDaysToAdd = weekDaysToAdd - 5
        } else { // week < 0
            daysToAdd = (5 + week) + daysToAdd
            weekDaysToAdd = weekDaysToAdd - (5 + week)
        }
    }

    return daysToAdd;
}   

$j('.input-text.addon.addon-custom').datepicker({
    beforeShowDay: noWeekendsOrHolidays,
    minDate : dateMin,
    defaultDate: +1,
    firstDay: 1,
    changeFirstDay: true,
    dateFormat: "DD, dd MM yy"
});

任何帮助将不胜感激。

在这里提琴： http : //jsfiddle.net/prydonian/4k4gga6j/

Answer 1

在您的datepicker函数上，像这样添加“ beforeShowDay ”选项。

jQuery('#datepicker').datepicker({
    minDate: dateMin,
    defaultDate: +1,
    firstDay: 1,
    changeFirstDay: true,
    dateFormat: "DD, dd MM yy",
    beforeShowDay: function(day){
         if (day.getDay()<2 || day.getDay()>4){
            return [false, ""];
        }
        return noWeekendsOrHolidays(day);
    }
});

这是小提琴更新： http : //jsfiddle.net/4k4gga6j/3/

Answer 2

如果我理解正确，那么您需要排除星期二，星期三和星期四以外的其他日子吗？ 如果答案是肯定的，则应将以下代码添加到现有的noWeekendsOrHolidays方法中：

($.inArray(date.getDay(), [2, 3, 4]) != -1)

这是更新的jsFiddle http://jsfiddle.net/4k4gga6j/4/