可以使jQuery UI Datepicker禁用星期六和星期日(和节假日)吗?


Answers:


243

有一个beforeShowDay选项,该选项需要为每个日期调用一个函数,如果允许日期,则返回true,否则,则返回false。从文档:


ShowShow之前

该函数以日期作为参数,并且必须返回一个数组,该数组的[0]等于true / false,表示此日期是否可选,默认表示形式为1,该数组等于CSS类名或。在日期选择器中的每一天都会调用它,然后再显示它。

在日期选择器中显示一些国定假日。

$(".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, ''];
}

存在一个内置函数,称为noWeekends,它可以防止选择周末。

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

要将两者结合起来,您可以做类似的事情(假设nationalDays上面的函数):

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

function noWeekendsOrHolidays(date) {
    var noWeekend = $.datepicker.noWeekends(date);
    if (noWeekend[0]) {
        return nationalDays(date);
    } else {
        return noWeekend;
    }
}

更新:请注意,从jQuery UI 1.8.19开始,beforeShowDay选项还接受可选的第三个参数,即弹出工具提示


2
谢谢您,无法在任何文档中找到此方法。
尼尔·艾特肯

2
优秀的解决方案。特别喜欢您的解释多么简洁。@Neil:dev.jqueryui.com/ticket/5851
Kaushik Gopal,2010年

这样,我就得到Uncaught TypeError: Cannot read property 'noWeekends' of undefined了国庆日功能正常的工作
LewisJWright

如果已实例化它,则可以更新选项:$('#selector')。datepicker('option','beforeShowDay',$ .datepicker.noWeekends);
Marcaum54

42

如果您根本不希望周末出现,只需:

的CSS

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

1
要完全隐藏日期,但要显示标题和空白:td.ui-datepicker-week-end { visibility: hidden; }
Vael Victus '16

这是排除周末的一种非常巧妙的方法。
itsdarrylnorris

谢谢,非常好:)
福克斯

26

这些答案非常有帮助。谢谢。

我在下面的贡献中添加了一个数组,其中多天可能返回false(我们每个星期二,星期三和星期四都关闭)。而且我捆绑了特定的日期,年份和无周末功能。

如果要休假,请将[Saturday],[Sunday]添加到closeDays数组中。

$(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];
    }




});

3
您可以编辑旧答案,而不是发布全新的答案-考虑到两者的相似程度,将它们都包含在同一个答案中,或者直接删除旧答案并留下更好的版本
江一民


13

每个人都喜欢的解决方案似乎非常激烈...就我个人而言,我认为这样做很容易:

       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;
            }
        });

这样,您的日期就易于阅读。其实并没有什么不同,这样对我来说更有意义。


6

此版本的代码将使您能够从sql数据库获取假期日期,并在UI Datepicker中禁用指定的日期


$(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
     });
  });
});

在sql中创建一个数据库,然后以MM / DD / YYYY格式将您的假期日期作为Varchar放入以下内容到文件getdate.php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]

快乐编码!!!! :-)


5

您可以使用noWeekends功能禁用周末选择

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

4
$("#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];

}

2

在此版本中,月,日年确定日历上要阻止的日期。

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

0

对于残疾的日子,你可以做这样的事情。 <input type="text" class="form-control datepicker" data-disabled-days="1,3">其中1是星期一,3是星期三


0

在最新的Bootstrap 3版本(bootstrap-datepicker.js)中,beforeShowDay预期结果采用以下格式:

{ enabled: false, classes: "class-name", tooltip: "Holiday!" }

另外,如果您不关心CSS和工具提示,则只需返回一个布尔值false以使日期不可选择。

另外,没有$.datepicker.noWeekends,因此您需要按照以下步骤进行操作:

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 });

0

对于周六和周日,您可以执行以下操作

             $('#orderdate').datepicker({
                               daysOfWeekDisabled: [0,6]
                 });
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