試圖弄清楚如何從jquery datepicker中刪除假期

Question

問題是我希望能夠從我的jquery datepicker中剝離某些日期（假日）。 noWeekends可以正常工作，而其他自定義變量也可以確定我的new_day變量，但是由於某些原因，我無法使假期變量正常工作，並且無法使用戶單擊假期日期。 任何幫助將不勝感激。

額外信息

應用使用

Rails 4 + jQuery

這是我的application.js

$(document).ready(function(){

     var holidays = ["12-25-2015","2015-04-03", "2015-05-25", "2015-07-03", "2015-09-07", "2015-11-26", "2015-12-25", "2016-01-01"];
     function editDays(date) {
         for (var i = 0; i < holidays.length; i++) {
            if (new Date(holidays[i]).toString() == date.toString()) {
                return true;
            }
         }
           return false;
     }

     $.datepicker.setDefaults({ dateFormat: 'D, dd M yy', numberOfMonths: 3, showButtonPanel: true, beforeShowDay: editDays });


     $('#leave_start').datepicker();

     $('#leave_end').datepicker();

     $('#leave_start_half').datepicker();

     $('#leave_end_half').datepicker();

     $('#future').blur(function() {
        var selected = $('.future').val();
        var touched = $('#indirect_id').val();
        var new_day = $('.anv').val();
        if (selected == 'YES') {
           $('#leave_start').datepicker( "option", "minDate", new_day );
           $('#leave_start').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );
           $('#leave_end').datepicker( "option", "minDate", new_day );
           $('#leave_end').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );
           $('#leave_start_half').datepicker( "option", "minDate", new_day );
           $('#leave_start_half').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );
           $('#leave_end_half').datepicker( "option", "minDate", new_day );
           $('#leave_end_half').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );

        } else if (selected == 'NO') {
           $('#leave_start').datepicker( "option", "minDate", 0 );
           $('#leave_start').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );
           $('#leave_end').datepicker( "option", "minDate", 0 );
           $('#leave_end').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );
           $('#leave_start_half').datepicker( "option", "minDate", 0 );
           $('#leave_start_half').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );
           $('#leave_end_half').datepicker( "option", "minDate", 0 );
           $('#leave_end_half').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );

        }

     });
 });

我嘗試了給出的答案，但是當用戶為我選擇的變量選擇“是”或“否”時，它拿走了假日變量並重新格式化為日期選擇器

      $('#leave_start').datepicker( "option", "minDate", new_day);                             
      $('#leave_start').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );

但是當他們離開我選擇的變量時，此方法有效，但是一旦選擇“是”或“否”，假日變量就會消失。

這就是我現在可以使用的東西，但是僅當選擇的var為空且不是YES或NO時才起作用

$(document).ready(function(){
    var holidays = ["12-25-2015","2015-04-03", "2015-05-25", "2015-07-03", "2015-09-07", "2015-11-26", "2015-12-25", "2016-01-01"];
     $.datepicker.setDefaults({ dateFormat: 'D, dd M yy', numberOfMonths: 3, showButtonPanel: true, beforeShowDay: function(date){
        var string = jQuery.datepicker.formatDate('yy-mm-dd', date);
        return [ holidays.indexOf(string) == -1 ]
      }});

     $'#leave_start').datepicker();

     $('#leave_end').datepicker();

     $('#leave_start_half').datepicker();

     $('#leave_end_half').datepicker();

     $('#future').blur(function() {
         var selected = $('.future').val();
         var touched = $('#indirect_id').val();
         var new_day = $('.anv').val();

         if (selected == 'YES') {
            $('#leave_start').datepicker( "option", "minDate", new_day );
            $('#leave_start').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );
            $('#leave_end').datepicker( "option", "minDate", new_day );
            $('#leave_end').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );
            $('#leave_start_half').datepicker( "option", "minDate", new_day );
            $('#leave_start_half').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );
            $('#leave_end_half').datepicker( "option", "minDate", new_day );
            $('#leave_end_half').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );

       } else if (selected == 'NO') {
            $('#leave_start').datepicker( "option", "minDate", 0 );
            $('#leave_start').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );
            $('#leave_end').datepicker( "option", "minDate", 0 );
            $('#leave_end').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );
            $('#leave_start_half').datepicker( "option", "minDate", 0 );
            $('#leave_start_half').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );
            $('#leave_end_half').datepicker( "option", "minDate", 0 );
            $('#leave_end_half').datepicker( "option", "beforeShowDay", $.datepicker.noWeekends );

       }

     });
 });

Answer 1

單個輸入日期的解決方法示例：

此處演示

$('input').datepicker({
    beforeShowDay: function(date){
        var string = jQuery.datepicker.formatDate('yy-mm-dd', date);
        return [ holidays.indexOf(string) == -1 ]
    }
});

資源