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

如果在上午10点之后选择，我将使用中的代码禁用周末、公共假日和第二天，但我一直无法确定如何只允许选择星期二、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"
});

任何帮助都将不胜感激

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在这里拨弄：

在datepicker函数中，添加如下选项

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);
    }
});

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这是最新的小提琴：如果我没听错，你需要排除星期二、星期三和星期四以外的任何一天吗？如果答案为“是”，则应将以下代码添加到现有的noWeekendsOrHolidays方法中：

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

这里是更新的JSFIDLE