在JavaScript日期计算中排除周末 [英] exclude weekends in javascript date calculation
问题描述
这段代码让我有两个日期之间的区别。功能完美：
function test（）{
var date1 = new Date（txtbox_1.value）;
var date2 = new Date（txtbox_2.value）;
var diff =（date2 - date1）/ 1000;
var diff = Math.abs（Math.floor（diff））;
var days = Math.floor（diff /（24 * 60 * 60））;
var leftSec = diff - days * 24 * 60 * 60;
var hrs = Math.floor（leftSec /（60 * 60））;
var leftSec = leftSec - hrs * 60 * 60;
var min = Math.floor（leftSec /（60））;
var leftSec = leftSec - min * 60;
txtbox_3.value =天+。 +小时}
@cyberfly下面的代码似乎有排除sat和sun这是我需要的答案。 来源。但是，它在jquery和上面的代码是在JS。因此，需要帮助组合，因为我缺乏这些知识：（
< script type =text / javascript>
$（＃startdate，#enddate）。change（function（）{
var d1 = $（＃startdate）。val（）;
var d2 = $（＃enddate）val（）;
var minutes = 1000 * 60;
var hours = minutes * 60;
var day = 24;
var startdate1 = getDateFromFormat（d1，dmy）;
var enddate1 = getDateFromFormat（d2，dmy）;
var days = calcBusinessDays （new Date（startdate1），new Date（enddate1））;
if（days> 0）
{$（＃noofdays）val（days）;}
else
{$（＃noofdays）。val（0）;}
}）;
< / script>
编辑
尝试组合代码，这是我的示例，获取对象预期错误。
function test（）{
var date1 = new Date（startdate.value）;
var date2 = new Date（enddate.value）;
var diff =（date2 - date1）/ 1000;
var diff = Math.abs（Math.floor（diff））;
var days = Math.floor（diff /（24 * 60 * 60））;
var leftSec = diff - days * 24 * 60 * 60;
var hrs = Math.floor（leftSec /（60 * 60））;
var leftSec = leftSec - hrs * 60 * 60;
var min = Math.floor（leftSec /（60））;
var leftSec = leftSec - min * 60;
var startdate1 = getDateFromFormat（startdate，dd / mm / yyyy hh：mm）;
var enddate1 = getDateFromFormat（enddate，dd / mm / yyyy hh：mm）;
days = calcBusinessDays（new Date（startdate1），new Date（enddate1））;
noofdays.value = days +。 +小时}
start：< input type =textid =startdatename =startdatevalue =02/03/2015 00:00>
end：< input type =textid =enddatename =enddatevalue =02/03/2015 00:01>
< input type =textid =noofdaysname =noofdaysvalue =>
当确定两个日期之间的天数时，有很多关于什么是一天的决定。例如，2月1日至2月2日期间一般为一天，所以2月1日至2月1日为零日。
当增加计数工作日的复杂性时，事情变得更加艰难例如。 2015年2月2日星期一至5月6日星期五为4日（星期一至星期二为1，星期一至星期三为2等），但星期一至星期五一般被视为5个工作日，持续时间为2月2日到2月7日星期六也应该是4个工作日，但星期日到星期六应该是5。
所以这里是我的算法：
- 获取两个日期之间的整天的总数
- 除以7以获取整周的数量
- 将星期数乘以2，以获得周末天数
- 从整体减去周末天数以获得工作日
- 如果总天数不是偶数周，则将星期* 7的数字添加到开始日期以获取临时日期
- 当临时日期较少时超过结束日期：
- 如果临时日期不是星期六或星期日，则在工作日添加一个
- 将一个添加到临时日期
- 就是这样。
被其他一些算法所取代，但它永远不会循环超过6天，所以这是一个简单而且合理有效的解决问题的不平衡周。
以上：
- 周一至周五为4个工作日
- 不同的星期是偶数星期，因此甚至多个5，例如星期一2二月至星期一9月2日和星期日1月2日至星期日8月2日为5个工作日
- 星期五6月6日至3月7日7月2日零工作日
- 5月6日星期五到星期一9月2日是一个工作日
- 星期日2月8日星期日到2月15日星期日2月14日星期四和2月13日星期五全部5个工作日
以下是代码：
//预期开始日期到结束日期之前
//开始和结束是Date对象
函数dateDifference（开始，结束）{
//复制日期对象，因此不要修改原始文件
var s = new Date（+ start）;
var e = new Date（+ end）;
//设置时间到中午以避免dalight保存和浏览器怪癖
s.setHours（12,0,0,0）;
e.setHours（12,0,0,0）;
//获取整天的差异
var totalDays = Math.round（（e-s）/ 8.64e7）;
//获取整周的差额
var wholeWeeks = totalDays / 7 | 0;
//估计工作天数为整周数* 5
var days = wholeWeeks * 5;
//如果没有偶数周数，则剩余周末天数
if（totalDays％7）{
s.setDate（s.getDate（）+ wholeWeeks * 7） ;
while（s< e）{
s.setDate（s.getDate（）+ 1）;
//如果日不是星期日或星期六，则添加到工作日
if（s.getDay（）！= 0&& s.getDay（）！= 6 ）{
++ days;
}
}
}
返回天数;
}
Dunno如何比较jfriend00的答案或您引用的代码，如果你想期限是包容性的，如果开始或结束日期是工作日，只需添加一个。
I have two sets of codes that work. Needed help combining them into one.
This code gets me the difference between two dates. works perfectly:
function test(){
var date1 = new Date(txtbox_1.value);
var date2 = new Date(txtbox_2.value);
var diff = (date2 - date1)/1000;
var diff = Math.abs(Math.floor(diff));
var days = Math.floor(diff/(24*60*60));
var leftSec = diff - days * 24*60*60;
var hrs = Math.floor(leftSec/(60*60));
var leftSec = leftSec - hrs * 60*60;
var min = Math.floor(leftSec/(60));
var leftSec = leftSec - min * 60;
txtbox_3.value = days + "." + hrs; }
The code below by @cyberfly appears to have the answer of excluding sat and sun which is what i needed. source. However, its in jquery and the above code is in JS. Therefore, needed help combining as i lacked that knowledge :(
<script type="text/javascript">
$("#startdate, #enddate").change(function() {
var d1 = $("#startdate").val();
var d2 = $("#enddate").val();
var minutes = 1000*60;
var hours = minutes*60;
var day = hours*24;
var startdate1 = getDateFromFormat(d1, "d-m-y");
var enddate1 = getDateFromFormat(d2, "d-m-y");
var days = calcBusinessDays(new Date(startdate1),new Date(enddate1));
if(days>0)
{ $("#noofdays").val(days);}
else
{ $("#noofdays").val(0);}
});
</script>
EDIT Made an attempt at combining the codes. here is my sample. getting object expected error.
function test(){
var date1 = new Date(startdate.value);
var date2 = new Date(enddate.value);
var diff = (date2 - date1)/1000;
var diff = Math.abs(Math.floor(diff));
var days = Math.floor(diff/(24*60*60));
var leftSec = diff - days * 24*60*60;
var hrs = Math.floor(leftSec/(60*60));
var leftSec = leftSec - hrs * 60*60;
var min = Math.floor(leftSec/(60));
var leftSec = leftSec - min * 60;
var startdate1 = getDateFromFormat(startdate, "dd/mm/yyyy hh:mm");
var enddate1 = getDateFromFormat(enddate, "dd/mm/yyyy hh:mm");
days = calcBusinessDays(new Date(startdate1),new Date(enddate1));
noofdays.value = days + "." + hrs; }
start: <input type="text" id="startdate" name="startdate" value="02/03/2015 00:00">
end: <input type="text" id="enddate" name="enddate" value="02/03/2015 00:01">
<input type="text" id="noofdays" name="noofdays" value="">
When determining the number of days between two dates, there are lots of decisions to be made about what is a day. For example, the period 1 Feb to 2 Feb is generally one day, so 1 Feb to 1 Feb is zero days.
When adding the complexity of counting only business days, things get a lot tougher. E.g. Monday 2 Feb 2015 to Friday 6 February is 4 elapsed days (Monday to Tuesday is 1, Monday to Wednesday is 2, etc.), however the expression "Monday to Friday" is generally viewed as 5 business days and the duration Mon 2 Feb to Sat 7 Feb should also be 4 business days, but Sunday to Saturday should be 5.
So here's my algorithm:
- Get the total number of whole days between the two dates
- Divide by 7 to get the number of whole weeks
- Multiply the number of weeks by two to get the number of weekend days
- Subtract the number of weekend days from the whole to get business days
- If the number of total days is not an even number of weeks, add the numbe of weeks * 7 to the start date to get a temp date
- While the temp date is less than the end date:
- if the temp date is not a Saturday or Sunday, add one the business days
- add one to the temp date
- That's it.
The stepping part at the end can probably be replaced by some other algorithm, but it will never loop for more than 6 days so it's a simple and reasonably efficient solution to the issue of uneven weeks.
Some consequences of the above:
- Monday to Friday is 4 business days
- Any day to the same day in a different week is an even number of weeks and therefore an even mutiple of 5, e.g. Monday 2 Feb to Monday 9 Feb and Sunday 1 Feb to Sunday 8 Feb are 5 business days
- Friday 6 Feb to Sunday 7 Feb is zero business days
- Friday 6 Feb to Monday 9 Feb is one business day
- Sunday 8 Feb to: Sunday 15 Feb, Sat 14 Feb and Fri 13 Feb are all 5 business days
Here's the code:
// Expects start date to be before end date
// start and end are Date objects
function dateDifference(start, end) {
// Copy date objects so don't modify originals
var s = new Date(+start);
var e = new Date(+end);
// Set time to midday to avoid dalight saving and browser quirks
s.setHours(12,0,0,0);
e.setHours(12,0,0,0);
// Get the difference in whole days
var totalDays = Math.round((e - s) / 8.64e7);
// Get the difference in whole weeks
var wholeWeeks = totalDays / 7 | 0;
// Estimate business days as number of whole weeks * 5
var days = wholeWeeks * 5;
// If not even number of weeks, calc remaining weekend days
if (totalDays % 7) {
s.setDate(s.getDate() + wholeWeeks * 7);
while (s < e) {
s.setDate(s.getDate() + 1);
// If day isn't a Sunday or Saturday, add to business days
if (s.getDay() != 0 && s.getDay() != 6) {
++days;
}
}
}
return days;
}
Dunno how it compares to jfriend00's answer or the code you referenced, if you want the period to be inclusive, just add one if the start or end date are a business day.
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