jQuery Datepicker日数 [英] jQuery Datepicker day count

问题描述

我有两个jQuery UI日期选择器，当他们都选择了日期时，我希望这些日期之间的区别显示在一个单独的输入[type =text]中

此外，我理想地喜欢我的计数减去任何周末天数，只计算几天

我的相关代码如下：

JS：

  $（'＃firstday，#lastday'）。datepicker（{
 dateFormat：'dd / mm / yy'
 }）; 
  

XHTML：

 code>< label for =firstday>离开的第一天< / label> 
< input type =textname =firstdayid =firstday/> 
 
< label for =lastday>离开的最后一天< / label> 
< input type =textname =lastdayid =lastday/> 
 
< label for =totaldays>总天数< / label> 
< input type =textname =totaldaysid =totaldays/> 
  

很多搜索都带给我很多不同的解决方案，我没有一个可以上班我希望任何想法都不会被赞赏。

解决方案

这样的事情应该可以工作：

  $（＃firstday，#lastday）。datepicker（{
 onSelect：function（）{
 
 //日期将以毫秒为单位，这就是为什么我们用1000 * 60 * 60 * 24 
 
 var firstday = new Date（$（＃firstday）。val（）。split /）。reverse（）。join（，））; 
 var lastday = new Date（$（＃lastday）val（）。split（/）。 （，）; 
 $（＃totaldays）。val（（lastday  -  firstday）/ 86400000）; 
} 
}）; 
  

在节点控制台中，它给出：

  > x = new Date（18/5/2010.split（/）。reverse（）。join（，））
 2010年5月17日星期一22:00:00 GMT 
> y = new Date（18/5/2015.split（/）。reverse（）。join（，））
 Sun，2015年5月17日22:00:00 GMT 
> x-y 
 -157766400000 
> y-x 
 157766400000 
> （yx）/ 86400000 
 1826 
  

- 编辑 -

当您有开始日期和结束日期时，可以使用从星期日0返回的getDay（）轻松计算周末天数，星期一为1，星期六为6。 >

Yo可以使用.getMonth（）和.getDate（）结合少数其他{}条件的假期。

  var weekend_count = 0; （i = firstday.valueOf（）; i< = lastday.valueOf（）; i + = 86400000）
 {
 var temp = new Date（i）; 
 if（temp.getDay（）== 0 || temp.getDay（）== 6）{
 weekend_count ++; 
} 
} 
  

最后你只是做

  $（＃totaldays）。val（（lastday  -  firstday）/ 86400000  -  weekend_count）; 
  

只是在结尾注释。你应该在一个单独的函数中推断这个代码（尽可能多的），以便保持你的代码更容易维护，以及在其他地方需要相同的功能。

祝你好运。

I've got two jQuery UI datepickers and when they've both had dates chosen, I'd like the difference between these dates to be displayed in a separate input[type="text"] as soon as the second date is selected.

Also, I'd ideally like my count to subtract any weekend days and just count days from Monday - Friday.

My (relevant) code is as follows:

JS:

$('#firstday, #lastday').datepicker({
  dateFormat: 'dd/mm/yy'
});

XHTML:

<label for="firstday">First Day of Leave</label>
<input type="text" name="firstday" id="firstday" />

<label for="lastday">Last Day of Leave</label>
<input type="text" name="lastday" id="lastday" />

<label for="totaldays">Total Days</label>
<input type="text" name="totaldays" id="totaldays" />

Lots of searching has led me to lots of different solutions, none of which I can get to work as I'd like so any ideas would be appreciated.

解决方案

Something like this should work:

$("#firstday, #lastday").datepicker({
  onSelect: function (){

    // Date will give time difference in miliseconds, that is why we divide with 1000*60*60*24

    var firstday = new Date($("#firstday").val().split("/").reverse().join(","));
    var lastday = new Date($("#lastday").val().split("/").reverse().join(",");
    $("#totaldays").val((lastday - firstday) / 86400000);        
  }
});

In node console it gives:

> x = new Date("18/5/2010".split("/").reverse().join(","))
Mon, 17 May 2010 22:00:00 GMT
> y = new Date("18/5/2015".split("/").reverse().join(","))
Sun, 17 May 2015 22:00:00 GMT
> x-y
-157766400000
> y-x
157766400000
> (y-x)/86400000
1826

-- EDIT --

When you have starting date and ending date number of days that are weekend is easily calculated using getDay() which returns from 0 for Sunday, 1 for Monday ... 6 for Saturday.

Yo could use .getMonth() and .getDate() combined with few else{} conditions for holidays as well.

var weekend_count = 0;
for (i = firstday.valueOf(); i <= lastday.valueOf(); i+= 86400000){
 var temp = new Date(i);
 if (temp.getDay() == 0 || temp.getDay() == 6) {
   weekend_count++;
 }
}

and in the end you just do

$("#totaldays").val( (lastday - firstday) / 86400000 - weekend_count);

Just to make a note at the end. You should probably extrapolate this code (as much of it as you can) in a separate function in order to keep your code easier to maintain and in case you need that same function on some other place.

Good luck.

这篇关于jQuery Datepicker日数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！