javascript - 如何显示 Check In 的 jquery datepicker 值加上 1 in Check Out 的值？

Question

我的代码是这样的:

jQuery("#datepicker_checkin").datepicker({ 
    minDate: 0, 
    numberOfMonths: 2,
    showButtonPanel: true,
    changeMonth: true,
    changeYear: true,
    dateFormat: "yy-mm-dd"
}); 
jQuery("#datepicker_checkout").datepicker({ 
    beforeShow : function()
    {
        jQuery( this ).datepicker('option','minDate', jQuery('#datepicker_checkin').val() );
    } , 
    numberOfMonths: 2,
    showButtonPanel: true,
    changeMonth: true,
    changeYear: true,
    dateFormat: "yy-mm-dd"
}); 
jQuery("#datepicker_checkin").on('change', function() {
    jQuery("#datepicker_checkout").val(jQuery('#datepicker_checkin').val());
    jQuery("#datepicker_checkout").datepicker({ 
        beforeShow : function()
        {
            jQuery( this ).datepicker('option','minDate', jQuery('#datepicker_checkin').val() );
        } , 
        numberOfMonths: 2,
        showButtonPanel: true,
        changeMonth: true,
        changeYear: true,
        dateFormat: "yy-mm-dd"
    }); 
});

演示是这样的:DEMO

我是这样描述的:

如果我选择 checkin 的文本字段，那么 checkout 的文本字段将显示 checkout 的值 + 1

例如:

如果 checkin = 2016-02-26 则 checkout = 2016-02-27

非常感谢任何帮助

干杯

Answer 1

试试这段代码

$(function() {
  $( "#check-in" ).datepicker({
    numberOfMonths: 2,
    showButtonPanel: true,
    changeMonth: true,
    changeYear: true,
    dateFormat: "yy-mm-dd",
    onSelect: function() {
        var checkOutDate = $(this).datepicker('getDate');
        checkOutDate.setDate(checkOutDate.getDate() + 1);
        $( "#check-out" ).datepicker( "option", "minDate", checkOutDate );
        $('#check-out').datepicker('setDate', checkOutDate);
    }
  });

  $( "#check-out" ).datepicker({
    numberOfMonths: 2,
    showButtonPanel: true,
    changeMonth: true,
    changeYear: true,
    dateFormat: "yy-mm-dd"
  });
});