确定 maxDate 时排除 beforeShowDay

Posted 2023-05-08

【中文标题】确定 maxDate 时排除 beforeShowDay

【英文标题】：Exclude beforeShowDay when determining maxDate

【发布时间】：2019-08-02 12:35:52

【问题描述】：

我正在“尝试”创建一个工作系统，让我的学生可以预订设备。我有两个日期输入，用于借入日期和归还日期。目前它是如何设置的，我有一堆日期，周末被禁用，而且学生不能在借来的那天归还装备。我正在尝试的下一步是只为学生提供下一个可用日期的回报

例如，一个学生想在周末借用装备，但星期一也是学生放假日，所以归还日期只需在星期二。

maxDate: 1 //不考虑我在 BeforeShowDay 中禁用的所有日期

我希望这是有道理的，这是我当前的混乱代码。

< script type = "text/javascript" >
  jQuery(document).ready(function($) 
      $(function() 

        var publicd = ["4,3", "3,6", ];
        var term1 = ["15,4", "16,4", "17,4", "18,4", "19,4", "22,4", "23,4", "24,4", "25,4", "26,4", ];
        var term2 = ["7,7", "8,7", "9,7", "10,7", "11,7", "12,7", "15,7", "16,7", "17,7", "18,7", "19,7", ];
        var term3 = ["29,9", "30,9", "1,10", "2,10", "3,10", "4,10", "7,10", "8,10", "9,10", "10,10", "11,10"];
        var chrissy = ["2,12", "3,12", "4,12", "5,12", "6,12", "9,12", "10,12", "11,12", "12,12", "13,12", "16,12", "17,12", "18,12", "19,12", "20,12", "23,12", "24,12", "25,12", "26,12", "27,12", "30,12", "31,12", "1,1", "2,1", "3,1", "6,1", "7,1", "8,1", "9,1", "10,1", "13,1", "14,1", "15,1", "16,1", "17,1", "20,1", "21,1", "22,1", "23,1", "24,1", "27,1", "28,1", "29,1", "30,1", "31,1"];
        var pupil = ["8,3", "22,7", "15,11", "9,4", "27,5"];




        var combinedholidays = publicd.concat(term1, term2, term3, chrissy, pupil);

        function TotalHolidays(date) 
          var m = date.getMonth();
          var d = date.getDate();
          var currentdate = d + "," + (m + 1);
          for (var i = 0; i < combinedholidays.length; i++) 
            if ($.inArray(currentdate, combinedholidays) != -1) 
              return [false];
            
          
          var noWeekend = $.datepicker.noWeekends(date);
          return !noWeekend[0] ? noWeekend : [true];
        


        $("#field_borrow").datepicker(
          dateFormat: 'yy/mm/dd',
          minDate: new Date(),
          beforeShowDay: TotalHolidays,
          showOn: "button",
          buttonImage: "https://jccamediaarts.edublogs.org/files/2018/06/if_calendar_285670-copy-2c5zvzr-2o5mcij-e1529025491810.png",
          buttonImageOnly: true,
          buttonText: "Select date",
          onSelect: function(date) 

            var selectedDate = new Date(date);
            var msecsInADay = 86400000;

            var endDate = new Date(selectedDate.getTime() + msecsInADay);


            $("#field_return").datepicker("option", "minDate", endDate);


          
        );

        $("#field_return").datepicker(
          dateFormat: 'yy/mm/dd',
          beforeShowDay: TotalHolidays,
          showOn: "button",
          buttonImage: "https://jccamediaarts.edublogs.org/files/2018/06/if_calendar_285670-copy-2c5zvzr-2o5mcij-e1529025491810.png",
          buttonImageOnly: true,
          buttonText: "Select date"
        );

      ); <
      /script>

【参考方案1】：

我没有看到您显示的代码有任何具体问题。我已经对其进行了一些更新，它似乎可以按预期工作。

$(function() 
  var publicd = ["4,3", "3,6"];
  var term1 = ["15,4", "16,4", "17,4", "18,4", "19,4", "22,4", "23,4", "24,4", "25,4", "26,4", ];
  var term2 = ["7,7", "8,7", "9,7", "10,7", "11,7", "12,7", "15,7", "16,7", "17,7", "18,7", "19,7", ];
  var term3 = ["29,9", "30,9", "1,10", "2,10", "3,10", "4,10", "7,10", "8,10", "9,10", "10,10", "11,10"];
  var chrissy = ["2,12", "3,12", "4,12", "5,12", "6,12", "9,12", "10,12", "11,12", "12,12", "13,12", "16,12", "17,12", "18,12", "19,12", "20,12", "23,12", "24,12", "25,12", "26,12", "27,12", "30,12", "31,12", "1,1", "2,1", "3,1", "6,1", "7,1", "8,1", "9,1", "10,1", "13,1", "14,1", "15,1", "16,1", "17,1", "20,1", "21,1", "22,1", "23,1", "24,1", "27,1", "28,1", "29,1", "30,1", "31,1"];
  var pupil = ["8,3", "22,7", "15,11", "9,4", "27,5"];
  var dtFt = 'yy/mm/dd';

  var combinedHolidays = publicd.concat(term1, term2, term3, chrissy, pupil);

  function isHoliday(day, holidays) 
    var result = false;
    if ($.inArray(day, holidays) >= 0) 
      result = true;
    ;
    return result;
  

  function isWeekend(dt) 
    var result = $.datepicker.noWeekends(dt);
    return !result[0];
  

  function TotalHolidays(date) 
    var m = date.getMonth();
    var d = date.getDate();
    var currentDate = d + "," + (m + 1);
    var result = [true, "weekDay"];
    if (isHoliday(currentDate, combinedHolidays) || isWeekend(date)) 
      result = [false, ""];
    
    return result;
  

  $("#borrow, #return").datepicker(
    dateFormat: dtFt,
    beforeShowDay: TotalHolidays,
    showOn: "button",
    buttonImage: "https://jccamediaarts.edublogs.org/files/2018/06/if_calendar_285670-copy-2c5zvzr-2o5mcij-e1529025491810.png",
    buttonImageOnly: true,
    buttonText: "Select date"
  );

  $("#borrow").datepicker("option", 
    minDate: new Date(),
    onSelect: function(date) 
      var selectedDate = new Date(date);
      var msecsInADay = 86400000;
      var endDate = new Date(selectedDate.getTime() + msecsInADay);
      $("#return").datepicker("option", "minDate", endDate);
    
  );
);

<link rel="stylesheet" href="//code.jquery.com/ui/1.12.1/themes/base/jquery-ui.css">
<script src="https://code.jquery.com/jquery-1.12.4.js"></script>
<script src="https://code.jquery.com/ui/1.12.1/jquery-ui.js"></script>
<p>Borrow Date: <input type="text" id="borrow" /></p>
<p>Return Date: <input type="text" id="return" /></p>

希望对您有所帮助。