我想知道函数AddWeekDays（）在我的编码中是如何工作的

Question

这是我的编码： -

<script type="text/javascript">
var $cal = jQuery.noConflict();

$cal(function () {  
        var dateMin     = new Date();
        var weekDays    = AddWeekDays(4);
        dateMin.setDate(dateMin.getDate() + weekDays);
        var natDays = [
          [1, 1, 'uk'],
          [5,27,'uk'],
          /*[6,25,'uk'],
          [6,26,'uk'],
          [7,15,'uk'],
          [7,23,'uk'],
          [7,11,'uk'],*/
          [12, 25, 'uk'],
          [12, 26, 'uk']
        ];

        function noWeekendsOrHolidays(date) {
            var noWeekend = $cal.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 daysToAdd   = 0
            var mydate      = new Date()
            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;
        }

        $cal('#required_ship_date').datepicker(
        {
            inline: true,
            beforeShowDay: noWeekendsOrHolidays,
            altField: '#required_ship_date',
            //showOn: "button",
            dateFormat: "mm/dd/yy",
           // firstDay: 1,
            changeFirstDay: false,
            minDate: dateMin,
            //buttonImageOnly: true,
           // buttonImage: "http://www.papercraftproducts.com/skin/frontend/default/default/images/grid-cal.gif",
        });
        $cal("#required_ship_date").datepicker("setDate", weekDays);
      });  
</script>
<li>
                <label for="custom[required_ship_date]"><?php echo $this->__('Required Ship Date (Optional)')?></label>
                <div class="input-box" >
                   <div id="datepicker"> </div>
                    <input type="text" value="" class="input-text input-text1" title="custom[required_ship_date]" name="custom[required_ship_date]" id="required_ship_date" autocomplete="off">
                </div>
           </li>

这是我的截图： -

我希望事情是今天是13-6-2013假设 然后4天后将于19-6-2013。 那么在文本框中将显示19-6-2013，并且应该在日历之前的最后一天启用日期，即18-6-2013。 在我的情况下，在文本框19-6-2013即将到来，但在日历显示18被禁用：（

如果假设今天的日期是14-6-2013，那么在4天后，2013年6月20日将进入文本框，并且应该启用日历19-6-2013 ..

请尽快帮助我  提前谢谢

Answer 1

我并不完全明白你要做什么，但是如果你希望范围在所选日期前一天开始，只需减去1你计算的开始日期：

dateMin.setDate(dateMin.getDate() + weekDays-1);

Here is an example

Answer 2

在这个日历中如果你选择任何日期然后它完美地工作，因为我需要plz see.and别忘了接受我的答案.. :) 这是我的编码： -

<link type="text/css" rel="stylesheet" href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.7.1/themes/base/jquery-ui.css" /><script type="text/javascript" src="http://www.papercraftproducts.com/skin/frontend/default/theme380/js/jquery-ui.js"></script> 
<script type="text/javascript">
var $cal = jQuery.noConflict();

$cal(function () {  
        var dateMin     = new Date();
        var weekDays    = AddWeekDays(5);
        var day         = dateMin.getDay()
        if(weekDays==9 && day==6)
        {
             dateMin.setDate(dateMin.getDate() + weekDays-12);
        }
        if(weekDays==7 && day==1)
        {
            dateMin.setDate(dateMin.getDate() + weekDays-3);
        }
        else
        {
            dateMin.setDate(dateMin.getDate() + weekDays-1);
        }

        var natDays = [
          [1, 1, 'uk']
        ];

        function noWeekendsOrHolidays(date) {
            var noWeekend = $cal.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 daysToAdd   = 0
            var mydate      = new Date()
            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;
        }

        $cal('#required_ship_date').datepicker(
        {
            inline:         true,
            beforeShowDay:  noWeekendsOrHolidays,
            altField:       '#required_ship_date',
            //showOn: "button",
            dateFormat:     "mm/dd/yy",
           // firstDay: 1,
            changeFirstDay: false,
            minDate:        dateMin,

            //buttonImageOnly: true,
           // buttonImage: "http://www.papercraftproducts.com/skin/frontend/default/default/images/grid-cal.gif",
        });
        var mydate      = new Date();
        var day         = mydate.getDay();
        if(day==6)
        {
            $cal("#required_ship_date").datepicker("setDate", 6);
        }
        else if(day==2)
        {
            $cal("#required_ship_date").datepicker("setDate", 7);
        }
        else
        {
            $cal("#required_ship_date").datepicker("setDate", weekDays);
        }
      });   
</script>
 <li>
                <label for="custom[required_ship_date]"><?php echo $this->__('Required Ship Date (Optional)')?></label>
                <div class="input-box" >
                   <div id="datepicker"> </div>
                    <input type="text" value="" class="input-text input-text1" title="custom[required_ship_date]" name="custom[required_ship_date]" id="required_ship_date" autocomplete="off">
                </div>
           </li>

这是截图： - 如果今天的日期是13-6-2013（星期四）

如果今天的日期是14-6-2013（星期五）