jQuery Datepicker - 禁用周末/节假日和接下来的三个工作日
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时间：2020-08-26 12:37:44 来源:igfitidea点击:
jQuery Datepicker - Disable weekends / holidays and the next three working days combined
提问by Rebecca
Using this rather neat approachI can disable weekends and holidays from the datepicker.
使用这种相当简洁的方法，我可以从日期选择器中禁用周末和假期。
However, I want to combine this with the disabling of the next three business days from today's date. Simply setting the minimum date is relatively easy:
但是，我想将此与从今天起禁用接下来的三个工作日结合起来。简单地设置最小日期相对容易：
var dateMin = new Date();
dateMin.setDate(dateMin.getDate() + 3);
$(function() {
$('#txtCollectionDate').datepicker(
{
beforeShowDay: noWeekendsOrHolidays,
showOn: "both",
dateFormat: "dd/mm/yy",
firstDay: 1,
changeFirstDay: false,
minDate: dateMin
});
});
However, what I really need to a function that calculates the business days:
但是，我真正需要一个计算工作日的函数：
var dateMin = new Date();
dateMin.setDate(AddBusinessDays(3));
Anyone able to convert thisto JavaScript?
任何人都能够转换这为JavaScript？
回答by Rebecca
Found the solution here.
在这里找到了解决方案。
Code (apologies for the ASP):
代码（为 ASP 道歉）：
<html>
<head>
<title>Collection Date</title>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js"></script>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jqueryui/1.7.1/jquery-ui.min.js"></script>
<link type="text/css" rel="stylesheet" href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.7.1/themes/base/jquery-ui.css" />
</head>
<body>
<h1>jQuery Datepicker Test</h1>
<%
If Request.Form("Submit") <> "" then
%>
<h2>Form Post Confirmation</h2>
<table border="1">
<tr>
<td><B>Form Variable</B></td>
<td><B>Value</B></td>
</tr>
<%
Dim Item
For Each Item In Request.Form
%>
<tr>
<td><%=Item %></td>
<td><%=Request.Form(Item) %></td>
</tr>
<% Next %>
</table>
<%
End If
%>
<form name="Form1" method="post" action="TestDatePicker.asp" id="Form1">
<h2>Collection Form</h2>
<fieldset>
<legend>Choose the collection date</legend>
<div id="datepicker"></div>
<input type="text" id="txtCollectionDate" name="txtCollectionDate" class="requiredField" style="display: none;" />
</fieldset>
<input type="submit" name="submit" value="Submit" />
</form>
<script type="text/javascript">
$(document).ready(function() {
var dateMin = new Date();
var weekDays = AddWeekDays(3);
dateMin.setDate(dateMin.getDate() + weekDays);
var natDays = [
[1, 1, 'uk'],
[12, 25, 'uk'],
[12, 26, 'uk']
];
function noWeekendsOrHolidays(date) {
var noWeekend = $.datepicker.noWeekends(date);
if (noWeekend[0]) {
return nationalDays(date);
} else {
return noWeekend;
}
}
function nationalDays(date) {
for (i = 0; i < natDays.length; i++) {
if (date.getMonth() == natDays[i][0] - 1 && date.getDate() == natDays[i][1]) {
return [false, natDays[i][2] + '_day'];
}
}
return [true, ''];
}
function AddWeekDays(weekDaysToAdd) {
var daysToAdd = 0
var mydate = new Date()
var day = mydate.getDay()
weekDaysToAdd = weekDaysToAdd - (5 - day)
if ((5 - day) < weekDaysToAdd || weekDaysToAdd == 1) {
daysToAdd = (5 - day) + 2 + daysToAdd
} else { // (5-day) >= weekDaysToAdd
daysToAdd = (5 - day) + daysToAdd
}
while (weekDaysToAdd != 0) {
var week = weekDaysToAdd - 5
if (week > 0) {
daysToAdd = 7 + daysToAdd
weekDaysToAdd = weekDaysToAdd - 5
} else { // week < 0
daysToAdd = (5 + week) + daysToAdd
weekDaysToAdd = weekDaysToAdd - (5 + week)
}
}
return daysToAdd;
}
$('#datepicker').datepicker(
{
inline: true,
beforeShowDay: noWeekendsOrHolidays,
altField: '#txtCollectionDate',
showOn: "both",
dateFormat: "dd/mm/yy",
firstDay: 1,
changeFirstDay: false,
minDate: dateMin
});
});
</script>
</body>
</html>
回答by Konstantin Petrukhnov
My case was very close, but idea is that number of business days should be added to script.
I replaced AddWeekDays() to AddBusinessDays() from Junto example. So when calculation minday it skip all non-business days.
我的案例非常接近，但想法是应该将工作日数添加到脚本中。我从 Junto 示例中将 AddWeekDays() 替换为 AddBusinessDays()。因此，当计算 minday 时，它会跳过所有非工作日。
//holidays
var natDays = [
[1, 1, 'uk'],
[12, 25, 'uk'],
[12, 26, 'uk']
];
var dateMin = new Date();
var weekDays = AddBusinessDays(3);
dateMin.setDate(dateMin.getDate() + weekDays);
function AddBusinessDays(weekDaysToAdd) {
var curdate = new Date();
var realDaysToAdd = 0;
while (weekDaysToAdd > 0){
curdate.setDate(curdate.getDate()+1);
realDaysToAdd++;
//check if current day is business day
if (noWeekendsOrHolidays(curdate)[0]) {
weekDaysToAdd--;
}
}
return realDaysToAdd;
}
function noWeekendsOrHolidays(date) {
var noWeekend = $.datepicker.noWeekends(date);
if (noWeekend[0]) {
return nationalDays(date);
} else {
return noWeekend;
}
}
function nationalDays(date) {
for (i = 0; i < natDays.length; i++) {
if (date.getMonth() == natDays[i][0] - 1 && date.getDate() == natDays[i][1]) {
return [false, natDays[i][2] + '_day'];
}
}
return [true, ''];
}
$('#datepicker').datepicker(
{
inline: true,
beforeShowDay: noWeekendsOrHolidays,
altField: '#txtCollectionDate',
showOn: "both",
dateFormat: "dd/mm/yy",
firstDay: 1,
changeFirstDay: false,
minDate: dateMin
});
//holidays
var natDays = [
[1, 1, 'uk'],
[12, 25, 'uk'],
[12, 26, 'uk']
];
var dateMin = new Date();
var weekDays = AddBusinessDays(3);
dateMin.setDate(dateMin.getDate() + weekDays);
function AddBusinessDays(weekDaysToAdd) {
var curdate = new Date();
var realDaysToAdd = 0;
while (weekDaysToAdd > 0){
curdate.setDate(curdate.getDate()+1);
realDaysToAdd++;
//check if current day is business day
if (noWeekendsOrHolidays(curdate)[0]) {
weekDaysToAdd--;
}
}
return realDaysToAdd;
}
function noWeekendsOrHolidays(date) {
var noWeekend = $.datepicker.noWeekends(date);
if (noWeekend[0]) {
return nationalDays(date);
} else {
return noWeekend;
}
}
function nationalDays(date) {
for (i = 0; i < natDays.length; i++) {
if (date.getMonth() == natDays[i][0] - 1 && date.getDate() == natDays[i][1]) {
return [false, natDays[i][2] + '_day'];
}
}
return [true, ''];
}
$('#datepicker').datepicker(
{
inline: true,
beforeShowDay: noWeekendsOrHolidays,
altField: '#txtCollectionDate',
showOn: "both",
dateFormat: "dd/mm/yy",
firstDay: 1,
changeFirstDay: false,
minDate: dateMin
});
回答by Jorge Arreola
Modifying the above code, for use with different days in different years, you can use...
修改上面的代码，针对不同年份不同天使用，可以使用...
<script>
//holidays
var natDays = [
[11,1,2012, 'mx'],[11,2,2012, 'mx'],[11,19,2012, 'mx'],
[12, 12, 2012, 'mx'],[12,20,2012, 'mx'],[12,25,2012, 'mx'],[12,31,2012, 'mx'],
[1,2,2013, 'mx'],[1,3,2013, 'mx'],[1, 4, 2013, 'mx'],[1,1,2014, 'mx'],[1,2,2014, 'mx']
];
var dateMin = new Date();
var weekDays = AddBusinessDays(3);
dateMin.setDate(dateMin.getDate() + weekDays);
function AddBusinessDays(weekDaysToAdd) {
var curdate = new Date();
var realDaysToAdd = 0;
while (weekDaysToAdd > 0){
curdate.setDate(curdate.getDate()+1);
realDaysToAdd++;
//check if current day is business day
if (noWeekendsOrHolidays(curdate)[0]) {
weekDaysToAdd--;
}
}
return realDaysToAdd;
}
function noWeekendsOrHolidays(date) {
var noWeekend = $.datepicker.noWeekends(date);
if (noWeekend[0]) {
return nationalDays(date);
} else {
return noWeekend;
}
}
function nationalDays(date) {
for (i = 0; i < natDays.length; i++) {
if (date.getMonth() == natDays[i][0] - 1 && date.getDate() == natDays[i][1] && date.getFullYear() == natDays[i][2]) {
return [false, '', 'No laboral ' + natDays[i][3]+'']; /* 'Holiday in ' + natDays[i][3] */
}
}
return [true, ''];
}
$(function() {
$( "#datepicker" ).datepicker({
beforeShowDay: noWeekendsOrHolidays,
altField: '#FI',
dateFormat: "dd/mm/yy",
defaultDate: '-0y',
changeMonth: true,
changeYear: true,
minDate: new Date(2012, 1 - 1, 5),
//minDate: "-1Y",
maxDate: "0Y",
showWeek: true,
firstDay: 0,
showOn: "both",
buttonImage: "images/calendar.gif",
buttonText: "Seleccionar Fecha", /*Select date text */
buttonImageOnly: true,
});
$(this).focus();
});
</script>
