jQuery datepicker-beforeShowDay问题 [英] jquery datepicker - beforeShowDay problem

问题描述

我有以下代码:

$(document).ready(function(){

   $(".datepicker").each(function() {

     $(this).datepicker({
    beforeShowDay: function(date){ return [$(this).parent().find("span.days").html(), ""] },

     });
   });

  });

使用此正文代码...

With this body code...

<div>
<span class="days">(date.getDay() == 1 || date.getDay() == 4)</span>
<input type="text" class="datepicker">
</div>


<div>
<span class="days">(date.getDay() == 2 || date.getDay() == 5)</span>
<input type="text" class="datepicker">
</div>

我想做的是让每个日期选择器有不同的可采日期.

What I want to be able to do is have each datepicker have different days pickable.

我尝试执行此操作的方式(在跨度中放入beforeShowDay代码的不同部分)可能不是最优雅的，所以请随时将我的代码拆开并在必要时进行更改.

The way i'm trying to do this (putting the varying part of the beforeShowDay code in a span) may not be the most elegant so feel free to take my code apart and change it if necessary.

推荐答案

您可以创建一个保留可挑选日期的对象，如下所示:

You could create an object keeping the pickable days, like this:

var pickable = { dp1: [1, 4], dp2: [2, 5] };
$('.datepicker').each(function() {
  $(this).datepicker({
    beforeShowDay: function(date){ 
        var day = date.getDay(), days = pickable[this.id];
        return [$.inArray(day, days) > -1, ""];
    }
  });
});​

您可以在此处尝试. dp1和dp2是控件的ID ..或您可以真正映射的任何属性，例如:

You can give it a try here. The dp1 and dp2 are IDs of the controls..or any attribute you can map really, like this:

<input type="text" class="datepicker" id="dp1">
<input type="text" class="datepicker" id="dp2">

这个概念非常简单，获取ID以获取该选择器的天数，然后使用 $.inArray() 来查看我们所在的日子是否在该数组中.如果这些天的选择可以共享，请不要使用ID，而是对属性进行data-pickerType之类的操作，并在上面的代码中将this.id替换为$(this).attr("data-pickerType").

The concept is pretty simple, take the ID to get the array of days for this picker, then use $.inArray() to see if the day we're on is in that array. If these day selections may be shared, don't use an ID, instead do something like data-pickerType for the attribute, and replace this.id with $(this).attr("data-pickerType") in the code above.

注意:我完好无损地保留了 .each() 您出于当前问题之外的原因需要它.

Note: I left the .each() intact because I know you need it for reasons outside the current question.

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