我有一个选择框,用户可以选择3个不同的商店.不应该选择2号和3号店的周末,而对于商店1,你应该只能选择周一至周六.

以下javascript仅适用于第一个选择.如果您之后选择另一家商店,它将坚持使用旧选项.

我尝试过使用$(“#datepicker”).datepicker(“refresh”); (见how to refresh datepicker?)但没有任何成功.我开始认为问题出在其他地方.

使用Javascript：

$(function() {

  var setting, currentShop = 0;

  /* Select box */

  $('select#shop').change(function() {
    (currentShop = $(this).val() == 1) ? loadDatePicker(setting = 'noSunday') : loadDatePicker(setting = 'noWeekends');
  });

  /* Datepicker */

  function noSunday(date){ 
    var day = date.getDay(); 
    return [(day > 0), '']; 
  }

  function loadDatePicker(setting) {
    if(setting == 'noWeekends') {
      $( "#datepicker" ).datepicker({ beforeShowDay: $.datepicker.noWeekends, minDate: +2, maxDate: "+1M" }); 
    }
    if(setting == 'noSunday') {
      $( "#datepicker" ).datepicker({ beforeShowDay: noSunday, minDate: +2, maxDate: "+1M" }); 
    }
    $( "#datepicker" ).datepicker("refresh");
  }
});

HTML：

  <select id="shop" name="shop">
    <option value="0" selected="selected">Choose a shop</option>
    <option value="1">1 (closed sundays)</option>
    <option value="2">2 (closed weekends)</option>
    <option value="3">3 (closed weekends)</option>
  </select>
  <label for="datepicker">Datepicker</label><input type="text" name="date" id="datepicker" value="" readonly="readonly" />

的jsfiddle：
http://jsbin.com/ajavek/1/edit

如何使用datepicker正确刷新/应用设置？

解决方法:

见：DEMO

  function loadDatePicker(setting) {
    $("#datepicker").datepicker("destroy");
    if(setting == 'noWeekends') {
      $( "#datepicker" ).datepicker({ beforeShowDay: $.datepicker.noWeekends, minDate: +2, maxDate: "+1M" }); 
    }
    else if(setting == 'noSunday') {
      $( "#datepicker" ).datepicker({ beforeShowDay: noSunday, minDate: +2, maxDate: "+1M" }); 
    }
    $( "#datepicker" ).datepicker("refresh");
  }

你需要把$(“#datepicker”).datepicker(“destroy”);每次更改设置之前…