jquery - jQuery Datepicker minDate表示一天中的时间

我在我的网站上使用Jquery datepicker，以便客户选择所需的交货日期。目前，我已经设置了minDate为4（不包括周末或节假日）。我只是意识到我需要考虑一天中的时间，这样，如果有人在下午2点之后下订单，那么该日期就不会算在minDate中（因为发货太晚了）。

我可以找到一些有关它的帖子，特别是其中一篇似乎适用：

How do I restrict dates on datepicker after a certain time

但是，我在将其应用于当前脚本时遇到了非常困难的时间（如下）。如果有人能告诉我如何适应这一点，那将非常酷。

非常感谢您的时间和帮助！ 〜苏珊

<script type="text/javascript">
   $(document).ready(function() {

   //holidays
   var natDays = [
     [1, 1, 'uk'],
     [12, 25, 'uk'],
     [12, 26, 'uk']
   ];

   var dateMin = new Date();
   var weekDays = AddBusinessDays(4);

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

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

   }

   function noWeekendsOrHolidays(date) {
       var noWeekend = $.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, ''];
    }
     $('#datepicker').datepicker({
        inline: true,
        beforeShowDay: noWeekendsOrHolidays,
        showOn: "both",
        firstDay: 0,
        dateformat: "dd/mm/yy",
        changeFirstDay: false,
        showButtonPanel: true,
        minDate: dateMin
    });
 });
 </script>

<p>
<label for="datepicker">Desired Delivery Date: </label>
  <input class="input-medium" type="text" id="datepicker" placeholder="ex. 01/01/2013" name="properties[Delivery Date]" readonly />
  <label><font size=1>Need it faster? Please call us! [phone]</font>
  </label></p>
<style>
  #datepicker { height: 20px; }
  #datepicker {-webkit-border-radius: 0 3px 3px 0; -moz-border-radius: 0 3px 3px 0; border-radius: 0 3px 3px 0;}
</style>

最佳答案

我终于解决了这个问题，所以我发布了我自己问题的答案。希望它可以帮助某人。以下是完整代码，其中minDate为4天（不包括美国国定假日和周末），以及minDate调整以排除当前日期（如果在下午2点之后）。

    <script type="text/javascript">
    $(document).ready(function() {

    //holidays
    var natDays = [
      [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, 14, 'Columbus Day'], //2013
      [11, 11, 'Veterans Day'], //2013
      [11, 28, 'Thanks Giving Day'], //2013
      [12, 25, 'Christmas'] //2013
];

    var dateMin = new Date();
    dateMin.setDate(dateMin.getDate() + (dateMin.getHours() >= 14 ? 1 : 0));
    AddBusinessDays(dateMin, 4);

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

    function noWeekendsOrHolidays(date) {
        var noWeekend = $.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, ''];
    }
      $('#datepicker').datepicker({
            inline: true,
            beforeShowDay: noWeekendsOrHolidays,
            showOn: "both",
            firstDay: 0,
            dateformat: "dd/mm/yy",
            changeFirstDay: false,
            showButtonPanel: true,
            minDate: dateMin
    });
  });
  </script>

<p>
<label for="datepicker">Desired Delivery Date: </label>
  <input class="input-medium" type="text" id="datepicker" placeholder="ex. 01/01/2013" name="properties[Delivery Date]" readonly />
  <label><font size=1>Need it faster? Please call us! [phone]</font>
  </label></p>
<style>
  #datepicker { height: 20px; }
  #datepicker {-webkit-border-radius: 0 3px 3px 0; -moz-border-radius: 0 3px 3px 0; border-radius: 0 3px 3px 0;}
</style>

关于jquery - jQuery Datepicker minDate表示一天中的时间，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/18993639/