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我想在编码中知道函数AddWeekDays()的工作原理

[英]I want to know function AddWeekDays() in my coding how it works

This is my coding:- 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>

Here is my screenshot:- 这是我的屏幕截图: 在此处输入图片说明

i want the thing is that today is 13-6-2013 suppose then after 4 days it will 19-6-2013. 我要说的是,今天是2013年6月13日,假设4天后会是2013年6月19日。 then in textbox 19-6-2013 will be display and day before last day should be enable in calender ie 18-6-2013. 然后将在文本框19-6-2013中显示,并应在日历中启用前一天的前一天,即18-6-2013。 In my case,in textbox 19-6-2013 is coming but in calender display 18 is disabled :( 以我为例,在19-6-2013文本框中,但在日历显示18中已禁用:(

if suppose today date is 14-6-2013 then after 4 days 20-6-2013 will come in textbox and in calendar 19-6-2013 should be enable.. 如果假设今天的日期是2013年6月14日,则在4天后的20-6-2013将出现在文本框中,并应启用日历19-6-2013。

plz help me in this as soon as possible 请尽快帮助我 在此处输入图片说明 thanks in advance 提前致谢

I did not completely understood what you are trying to do, but if you want the range to start one day before the selected day, just minus 1 your calculated start day: 我不完全了解您要执行的操作,但是如果您希望范围从所选日期的前一天开始,请减去计算出的开始日期1:

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

Here is an example 这是一个例子

IN this calendar if you select any date then it works perfectly as i needed plz see.and dont forget to accept my answer..:) Here is my coding:- 在这个日历中,如果您选择任何日期,那么它会完美地工作,因为我需要我看。不要忘了接受我的回答.. :)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>

here is the screenshot:- if today's date is 13-6-2013(thursday) 这是屏幕截图:-如果今天的日期是2013年6月13日(星期四) 在此处输入图片说明

If today's date is 14-6-2013(friday) 如果今天的日期是2013年6月14日(星期五) 在此处输入图片说明

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