试图弄清楚如何从 jquery datepicker 中删除假期

Question

问题是我希望能够从我的 jquery datepicker 中删除某些日期（假期）。 noWeekends 有效，我的其他自定义变量也可以确定我的 new_day 变量，但由于某种原因，我无法让我的假期变量工作并禁止用户点击假期日期。任何帮助将不胜感激。

额外信息

应用使用

导轨 4 + jQuery

这是我的 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 );

        }

     });
 });

我尝试了给出的答案，但是当用户为我选择的变量选择“是”或“否”时，它带走了假期变量并将日期选择器重新格式化回此

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

但是当他们离开我选择的变量时，这有效，但是一旦他们选择是或否，假期变量就会消失。

这就是我现在所拥有的，但只有在选择的 var 为空白且不是 YES 或 NO 时才有效

$(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

单个输入日期的示例解决方法：

DEMO HERE

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

SOURCE