jQuery UI Datepicker可以在周六和周日(和节假日)禁用吗?

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?

我用一个数据表来选择约会日期。我已经把日期范围设定为下个月。工作的很好。我想把星期六和星期天排除在可供选择之外。这个可以做吗?如果是这样,如何?

11 个解决方案

#1

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:

前面有一个选项，它为每个日期调用一个函数，如果允许，返回true;如果不允许，返回false。从文档:

---

beforeShowDay

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.

函数接受一个日期作为参数，并且必须返回一个包含[0]等于true/false的数组，该数组指示这个日期是否可以选择，1等于CSS类名或“用于默认表示”。在它显示之前，它在datepicker中被调用。

Display some national holidays in the datepicker.

在datepicker中显示一些国家假日。

$(".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.

有一个内置的函数叫noWeekends，它可以阻止你选择周末。

$(".selector").datepicker({ beforeShowDay: $.datepicker.noWeekends })

---

To combine the two, you could do something like (assuming the nationalDays function from above):

要把这两者结合起来，你可以做一些类似的事情(假设国家国庆日从上面开始起作用):

$(".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

更新:注意，在jQuery UI 1.8.19中，前面的选项还接受可选的第三个参数，一个弹出工具提示

#2

If you don't want the weekends to appear at all, simply:

如果你根本不想让周末出现，那就简单地说:

CSS

CSS

th.ui-datepicker-week-end,
td.ui-datepicker-week-end {
    display: none;
}

#3

These answers were very helpful. Thank you.

这些答案很有帮助。谢谢你！

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.

我在下面添加了一个数组，在这个数组中，多个天数可以返回false(我们每周二、周三和周四关闭)。我把具体的日期加上年份和无周末的功能捆绑在一起。

If you want weekends off, add [Saturday], [Sunday] to the closedDays array.

如果你想要周末休假，可以在closedDays数组中添加[Saturday]和[Sunday]。

$(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];
    }




});

#4

The datepicker has this functionality built in!

datepicker内置了这个功能!

$( "#datepicker" ).datepicker({
  beforeShowDay: $.datepicker.noWeekends
});

http://api.jqueryui.com/datepicker/#utility-noWeekends

http://api.jqueryui.com/datepicker/ utility-noWeekends

#5

The solution here that everyone likes seems to very intense... personally I think it's much easier to do something like this:

这里每个人都喜欢的解决方案似乎非常强烈……我个人认为做这样的事情要容易得多:

       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.

这样，你的约会对象就可以读懂了。这并没有什么不同，这对我来说更有意义。

#6

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

这个版本的代码将使u从sql数据库获取假日日期，并禁用UI Datepicker中的指定日期

$(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

在sql中创建一个数据库，并将假期日期以MM/DD/YYYY格式保存，Varchar将下面的内容放在getdate.php文件中

[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 !!!! :-)

编码快乐! ! ! !:-)

#7

$("#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];

}

#8

You can use noWeekends function to disable the weekend selection

您可以使用noWeekends函数来禁用周末选择

  $(function() {
     $( "#datepicker" ).datepicker({
     beforeShowDay: $.datepicker.noWeekends
     });
     });

#9

In this version, month, day, and year determines which days to block on the calendar.

在这个版本中，月、日和年确定日历上要阻塞的日期。

$(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,
     });
  });
});

#10

for disabling days you can do something like this. <input type="text" class="form-control datepicker" data-disabled-days="1,3"> where 1 is Monday and 3 is Wednesday

在禁用的日子里，你可以做这样的事情。，其中1为周一，3为周三

#11

In the latest Bootstrap 3 version (bootstrap-datepicker.js) beforeShowDay expects a result in this format:

在上述最新的Bootstrap 3版本(Bootstrap -datepicker.js)中，预计会出现这种格式的结果:

{ 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.

另外，如果您不关心CSS和工具提示，那么只需返回一个boolean false，使日期不可选。

Also, there is no $.datepicker.noWeekends, so you need to do something along the lines of this:

此外，没有$.datepicker。现在，你需要做一些类似的事情

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 });