请给jQuery datepicker自定义规则提供帮助

Question

目前我有一个日期选择器:

如果在中午12:00之后排除今天.

不包括星期日.

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


            // get today's date
            var myDate = new Date();
            // add one day if after 12:00
            if (myDate.getHours() > 12) {
                        myDate.setDate(myDate.getDate()+1);
            } else {
                myDate.setDate(myDate.getDate()+0);

            };

           $("#delivery-date-textbox").datepicker({
            dateFormat: 'dd-mm-yy',
            minDate: myDate,
            beforeShowDay: function(date){ return [date.getDay() != 0,""]}

            });
            });
</script>

我怎样才能让它排除一系列公众假期？

Answer 1

它可能对你有帮助.... 点击这里

/* create an array of days which need to be disabled */
var disabledDays = ["2-21-2010","2-24-2010","2-27-2010","2-28-2010","3-3-2010","3-17-2010","4-2-2010","4-3-2010","4-4-2010","4-5-2010"];

/* utility functions */
function nationalDays(date) {
  var m = date.getMonth(), d = date.getDate(), y = date.getFullYear();
  //console.log('Checking (raw): ' + m + '-' + d + '-' + y);
  for (i = 0; i < disabledDays.length; i++) {
    if($.inArray((m+1) + '-' + d + '-' + y,disabledDays) != -1 || new Date() > date) {
      //console.log('bad:  ' + (m+1) + '-' + d + '-' + y + ' / ' + disabledDays[i]);
      return [false];
    }
  }
  //console.log('good:  ' + (m+1) + '-' + d + '-' + y);
  return [true];
}
function noWeekendsOrHolidays(date) {
  var noWeekend = jQuery.datepicker.noWeekends(date);
  return noWeekend[0] ? nationalDays(date) : noWeekend;
}
/* create datepicker */
jQuery(document).ready(function() {
  jQuery('#date').datepicker({
    minDate: new Date(2010, 0, 1),
    maxDate: new Date(2010, 5, 31),
    dateFormat: 'DD, MM, d, yy',
    constrainInput: true,
    beforeShowDay: noWeekendsOrHolidays
  });
});

(David Walsh博客)