Asked  7 Months ago    Answers:  5   Viewed   29 times

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?



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:


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

Tuesday, June 1, 2021
answered 7 Months ago

You are getting this problem as all rows have


You need to give them different id. Keep a counter variable and append it to id. Increase it in loop after appending.

Hope this helps.

Saturday, May 29, 2021
answered 7 Months ago

rather than changing the source it's best to use the existing events

onSelect: function() {
    $(this).data('datepicker').inline = true;                               
onClose: function() {
    $(this).data('datepicker').inline = false;
Monday, June 14, 2021
answered 6 Months ago

When you write

$(document).ready(function () {

This fragment is getting executed right after the page has loaded. Therefore, your dynamic datepickers are not there yet. You need to call $(aSuitableSelector).datepicker(...) on each newly-inserted element. First, use a var to hold your options:

var datePickerOptions = {
    dateFormat: 'yy/m/d',
    firstDay: 1,
    changeMonth: true,
    changeYear: true,
    // ...

This allows you to write

 $(document).ready(function () {

and to write

 // right after appending dateFrom to the document ...


 // right after appending dateTo ...

You can also use JQuery's ability to listen to DOM changes to avoid having to apply JS magic to newly-inserted elements -- but I do not think it is worth it.

Sunday, August 1, 2021
answered 4 Months ago

After digging through the datepicker source a bit, I realized what was going on. Apparently whenever you click on a date, or cycle through months, it empties the entire div and re-creates it.

Since the picker is further down in the DOM, it handles the click event first. By the time the document handler gets called, the target element is no longer in the DOM, and is therefore no longer a descendant of the container.

My hasty hack was to check the target's more direct parents, to see whether any of them is a datepicker table or header. Since this widget is thus far my only use of datepicker, I can assume that a click on such an element should not count as a click outside the container.

I'll update this answer once I come up with a real solution.

Saturday, August 14, 2021
answered 4 Months ago
Only authorized users can answer the question. Please sign in first, or register a free account.
Not the answer you're looking for? Browse other questions tagged :