Jquery三个datepickers在一起

Question

我在左侧创建了2个datepicker输入字段，第一个是From日期，第二个是TO日期，我还创建了一个带有类datepicker的div，它在右侧显示了datepicker日历，因此前两个输入字段我使用了这个

$(function() {
  $('#datepickerSickFrom').datepicker({
    dateFormat: 'dd/mm/yy',
    onSelect: function(dateText, inst) {
      $('#dateTo').datepicker('option', 'minDate', new Date(inst.selectedYear, inst.selectedMonth, inst.selectedDay));
    }
  });

  $('#datepickerSickTo').datepicker({
    dateFormat: 'dd/mm/yy',
    onSelect: function(dateText, inst) {
      $('#dateFrom').datepicker('option', 'maxDate', new Date(inst.selectedYear, inst.selectedMonth, inst.selectedDay));
    }
  });
});

现在我希望将这两个输入连接到我创建的第三个datepicker div中，我想如何做到这一点？

这是我的HTML

<div class="group">
  <p class="inputIcon"><i class="fa fa-calendar" aria-hidden="true"></i></p>
  <input type="text" class="setLabel datepicker" placeholder="11/30/2017" id="datepickerSickFrom">
  <i class="fa fa-caret-down" style="position: absolute;right: 0;top: 17px;" aria-hidden="true"></i>
</div>
<div class="group">
  <p class="inputIcon"><i class="fa fa-calendar" aria-hidden="true"></i></p>
  <input type="text" class="setLabel datepicker" placeholder="11/30/2017" id="datepickerSickTo">
  <i class="fa fa-caret-down" style="position: absolute;right: 0;top: 17px;" aria-hidden="true"></i>
</div>
<div class="col-md-8">
  <div class="datepicker" id="sickPanelDatepicker"></div>
</div>

Answer 1

此代码应该执行您要执行的操作。

$(function() {

  $('#datepickerSickFrom').datepicker({
    dateFormat: 'dd/mm/yy',
    onSelect: function(dateText, inst) {
      $('#sickPanelDatepicker').datepicker('option', 'minDate', new Date(inst.selectedYear, inst.selectedMonth, inst.selectedDay));
    }
  });
  $('#datepickerSickTo').datepicker({
    dateFormat: 'dd/mm/yy',
    onSelect: function(dateText, inst) {
      $('#sickPanelDatepicker').datepicker('option', 'maxDate', new Date(inst.selectedYear, inst.selectedMonth, inst.selectedDay));
    }
  });
  $('#sickPanelDatepicker').datepicker({
    dateFormat: 'dd/mm/yy',

  });
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script src="https://code.jquery.com/ui/1.12.1/jquery-ui.min.js"></script>
<link rel="stylesheet" href="https://code.jquery.com/ui/1.12.1/themes/base/jquery-ui.css">
<div class="group">
  <p class="inputIcon"><i class="fa fa-calendar" aria-hidden="true"></i></p>
  <input type="text" class="setLabel datepicker" placeholder="11/30/2017" id="datepickerSickFrom">
  <i class="fa fa-caret-down" style="position: absolute;right: 0;top: 17px;" aria-hidden="true"></i>
</div>
<div class="group">
  <p class="inputIcon"><i class="fa fa-calendar" aria-hidden="true"></i></p>
  <input type="text" class="setLabel datepicker" placeholder="11/30/2017" id="datepickerSickTo">
  <i class="fa fa-caret-down" style="position: absolute;right: 0;top: 17px;" aria-hidden="true"></i>
</div>
<div class="col-md-8">
  <br><br><br>
  <div class="datepicker" id="sickPanelDatepicker"></div>
</div>