I use a datepicker for choosing an appointment day. I already set the date range to be only for the next month. That works fine. I want to exclude Saturdays and Sundays from the available choices. Can this be done? If so, how?
13 Answers
There is the beforeShowDay
option, which takes a function to be called for each date, returning true if the date is allowed or false if it is not. From the docs:
beforeShowDay
The function takes a date as a parameter and must return an array with [0] equal to true/false indicating whether or not this date is selectable and 1 equal to a CSS class name(s) or '' for the default presentation. It is called for each day in the datepicker before is it displayed.
Display some national holidays in the datepicker.
$(".selector").datepicker({ beforeShowDay: nationalDays})
natDays = [
[1, 26, 'au'], [2, 6, 'nz'], [3, 17, 'ie'],
[4, 27, 'za'], [5, 25, 'ar'], [6, 6, 'se'],
[7, 4, 'us'], [8, 17, 'id'], [9, 7, 'br'],
[10, 1, 'cn'], [11, 22, 'lb'], [12, 12, 'ke']
];
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, ''];
}
One built in function exists, called noWeekends, that prevents the selection of weekend days.
$(".selector").datepicker({ beforeShowDay: $.datepicker.noWeekends })
To combine the two, you could do something like (assuming the nationalDays
function from above):
$(".selector").datepicker({ beforeShowDay: noWeekendsOrHolidays})
function noWeekendsOrHolidays(date) {
var noWeekend = $.datepicker.noWeekends(date);
if (noWeekend[0]) {
return nationalDays(date);
} else {
return noWeekend;
}
}
Update: Note that as of jQuery UI 1.8.19, the beforeShowDay option also accepts an optional third paremeter, a popup tooltip
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2Thanks for this, couldn't find this method in the docs anywhere. Apr 13, 2010 at 14:55
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2excellent solution. particularly like how concise your explanation is. @Neil: dev.jqueryui.com/ticket/5851– KG -Oct 28, 2010 at 11:36
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With this i get
Uncaught TypeError: Cannot read property 'noWeekends' of undefined
the national days function works as expected though Mar 28, 2017 at 8:36 -
If you have already instantiated it, you can update the option: $('#selector'). datepicker('option', 'beforeShowDay', $.datepicker.noWeekends); Oct 18, 2019 at 16:57
If you don't want the weekends to appear at all, simply:
CSS
th.ui-datepicker-week-end,
td.ui-datepicker-week-end {
display: none;
}
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1To hide the days completely but show the headers and blank spaces:
td.ui-datepicker-week-end { visibility: hidden; }
May 9, 2016 at 21:29 -
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Thanks for this. I prefer the $.datepicker.noWeekends ... but I had a datepicker in a jquery-ui Dialog window and couldn't get this to work without Jquery TypeError. There this CSS-Hiding was useful– leoleAug 10, 2020 at 11:35
The datepicker has this functionality built in!
$( "#datepicker" ).datepicker({
beforeShowDay: $.datepicker.noWeekends
});
These answers were very helpful. Thank you.
My contribution below adds an array where multiple days can return false (we're closed every Tuesday, Wednesday and Thursday). And I bundled the specific dates plus years and the no-weekends functions.
If you want weekends off, add [Saturday], [Sunday] to the closedDays array.
$(document).ready(function(){
$("#datepicker").datepicker({
beforeShowDay: nonWorkingDates,
numberOfMonths: 1,
minDate: '05/01/09',
maxDate: '+2M',
firstDay: 1
});
function nonWorkingDates(date){
var day = date.getDay(), Sunday = 0, Monday = 1, Tuesday = 2, Wednesday = 3, Thursday = 4, Friday = 5, Saturday = 6;
var closedDates = [[7, 29, 2009], [8, 25, 2010]];
var closedDays = [[Monday], [Tuesday]];
for (var i = 0; i < closedDays.length; i++) {
if (day == closedDays[i][0]) {
return [false];
}
}
for (i = 0; i < closedDates.length; i++) {
if (date.getMonth() == closedDates[i][0] - 1 &&
date.getDate() == closedDates[i][1] &&
date.getFullYear() == closedDates[i][2]) {
return [false];
}
}
return [true];
}
});
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3You could just edit your old answer, instead of posting a brand new one - include both in the same answer, given how similar they are, or simply delete the old one and leave the better version– Yi JiangOct 8, 2010 at 10:15
The solution here that everyone likes seems to very intense... personally I think it's much easier to do something like this:
var holidays = ["12/24/2012", "12/25/2012", "1/1/2013",
"5/27/2013", "7/4/2013", "9/2/2013", "11/28/2013",
"11/29/2013", "12/24/2013", "12/25/2013"];
$( "#requestShipDate" ).datepicker({
beforeShowDay: function(date){
show = true;
if(date.getDay() == 0 || date.getDay() == 6){show = false;}//No Weekends
for (var i = 0; i < holidays.length; i++) {
if (new Date(holidays[i]).toString() == date.toString()) {show = false;}//No Holidays
}
var display = [show,'',(show)?'':'No Weekends or Holidays'];//With Fancy hover tooltip!
return display;
}
});
This way your dates are human readable. It's not really that different it just makes more sense to me this way.
You can use noWeekends function to disable the weekend selection
$(function() {
$( "#datepicker" ).datepicker({
beforeShowDay: $.datepicker.noWeekends
});
});
This version of code will make u to get the holiday dates from the sql database and disable the specified date in the UI Datepicker
$(document).ready(function (){
var holiDays = (function () {
var val = null;
$.ajax({
'async': false,
'global': false,
'url': 'getdate.php',
'success': function (data) {
val = data;
}
});
return val;
})();
var natDays = holiDays.split('');
function nationalDays(date) {
var m = date.getMonth();
var d = date.getDate();
var y = date.getFullYear();
for (var i = 0; i ‘ natDays.length-1; i++) {
var myDate = new Date(natDays[i]);
if ((m == (myDate.getMonth())) && (d == (myDate.getDate())) && (y == (myDate.getFullYear())))
{
return [false];
}
}
return [true];
}
function noWeekendsOrHolidays(date) {
var noWeekend = $.datepicker.noWeekends(date);
if (noWeekend[0]) {
return nationalDays(date);
} else {
return noWeekend;
}
}
$(function() {
$("#shipdate").datepicker({
minDate: 0,
dateFormat: 'DD, d MM, yy',
beforeShowDay: noWeekendsOrHolidays,
showOn: 'button',
buttonImage: 'images/calendar.gif',
buttonImageOnly: true
});
});
});
Create a Database in sql and put you holiday dates in MM/DD/YYYY format as Varchar Put the below contents in a file getdate.php
[php]
$sql="SELECT dates FROM holidaydates";
$result = mysql_query($sql);
$chkdate = $_POST['chkdate'];
$str='';
while($row = mysql_fetch_array($result))
{
$str .=$row[0].'';
}
echo $str;
[/php]
Happy Coding !!!! :-)
$("#selector").datepicker({ beforeShowDay: highlightDays });
...
var dates = [new Date("1/1/2011"), new Date("1/2/2011")];
function highlightDays(date) {
for (var i = 0; i < dates.length; i++) {
if (date - dates[i] == 0) {
return [true,'', 'TOOLTIP'];
}
}
return [false];
}
In this version, month, day, and year determines which days to block on the calendar.
$(document).ready(function (){
var d = new Date();
var natDays = [[1,1,2009],[1,1,2010],[12,31,2010],[1,19,2009]];
function nationalDays(date) {
var m = date.getMonth();
var d = date.getDate();
var y = date.getFullYear();
for (i = 0; i < natDays.length; i++) {
if ((m == natDays[i][0] - 1) && (d == natDays[i][1]) && (y == natDays[i][2]))
{
return [false];
}
}
return [true];
}
function noWeekendsOrHolidays(date) {
var noWeekend = $.datepicker.noWeekends(date);
if (noWeekend[0]) {
return nationalDays(date);
} else {
return noWeekend;
}
}
$(function() {
$(".datepicker").datepicker({
minDate: new Date(d.getFullYear(), 1 - 1, 1),
maxDate: new Date(d.getFullYear()+1, 11, 31),
hideIfNoPrevNext: true,
beforeShowDay: noWeekendsOrHolidays,
});
});
});
In the latest Bootstrap 3 version (bootstrap-datepicker.js) beforeShowDay
expects a result in this format:
{ enabled: false, classes: "class-name", tooltip: "Holiday!" }
Alternatively, if you don't care about the CSS and tooltip then simply return a boolean false
to make the date unselectable.
Also, there is no $.datepicker.noWeekends
, so you need to do something along the lines of this:
var HOLIDAYS = { // Ontario, Canada holidays
2017: {
1: { 1: "New Year's Day"},
2: { 20: "Family Day" },
4: { 17: "Easter Monday" },
5: { 22: "Victoria Day" },
7: { 1: "Canada Day" },
8: { 7: "Civic Holiday" },
9: { 4: "Labour Day" },
10: { 9: "Thanksgiving" },
12: { 25: "Christmas", 26: "Boxing Day"}
}
};
function filterNonWorkingDays(date) {
// Is it a weekend?
if ([ 0, 6 ].indexOf(date.getDay()) >= 0)
return { enabled: false, classes: "weekend" };
// Is it a holiday?
var h = HOLIDAYS;
$.each(
[ date.getYear() + 1900, date.getMonth() + 1, date.getDate() ],
function (i, x) {
h = h[x];
if (typeof h === "undefined")
return false;
}
);
if (h)
return { enabled: false, classes: "holiday", tooltip: h };
// It's a regular work day.
return { enabled: true };
}
$("#datePicker").datepicker({ beforeShowDay: filterNonWorkingDays });
This work for me:
$( "#datepicker" ).datepicker({
beforeShowDay: function( date ) {
var day = date.getDay();
return [ ( day > 0 && day < 6 ), "" ];
}
});
For Saturday and Sunday You can do something like this
$('#orderdate').datepicker({
daysOfWeekDisabled: [0,6]
});
To Disable the Weekends the API has a built-in feature
$('#data_1 .input-group.date').datepicker({
daysOfWeekDisabled: [0,6],
});
0 = Sunday
6 = Sunday