Question

使用日期选择器将日期添加到日期时出现问题。我知道这已被多次询问，但我们有不同的代码实现，这就是为什么我需要帮助谁知道这一点。添加天数时，必须排除星期六和星期日。我有这段代码：

<script type="text/javascript">
        $(document).ready(function () {
            $("#dt1").datepicker({
                dateFormat: "dd-M-yy",
                minDate: 0,
                onSelect: function (date) {
                    var date2 = $('#dt1').datepicker('getDate');
                    date2.setDate(date2.getDate() + 2);
                    if (date2.getDay() != 0 && date2.getDay() != 6) // Skip weekends
                   { $('#dt2').datepicker('setDate', date2);
                    //sets minDate to dt1 date + 1
                    $('#dt2').datepicker('option', 'minDate', date2);
                }
                }
            });
            $('#dt2').datepicker({
                dateFormat: "dd-M-yy",
                onClose: function () {
                    var dt1 = $('#dt1').datepicker('getDate');
                    console.log(dt1);
                    var dt2 = $('#dt2').datepicker('getDate');
                    if (dt2 <= dt1) {
                        var minDate = $('#dt2').datepicker('option', 'minDate');
                        $('#dt2').datepicker('setDate', minDate);
                    }
                }
            });
        });
    </script>

使用<script src="../js/jquery-1.9.1.js"></script>

这是我的输入字段：

<input type="text" id="dt1" name="date_borrow" pattern="alpha" required/>
<input type="text" id="dt2" name="date_will_return" pattern="alpha" required/>

因此，当我点击this或this时，结果必须为this。

我上面尝试添加的天数不起作用。我从那里错过了什么？

Answer 1

$(document).ready(function () {
    $("#dt1").datepicker({
        dateFormat: 'dd-M-yy',
        minDate: 0,
        beforeShowDay: function (date) {
            var day = date.getDay();
            return [(day != 0 && day != 6), '']
        },
        onSelect: function (date) {
            var date2 = $("#dt1").datepicker('getDate');
            if (date2.getDay() == 5) {
                date2.setDate(date2.getDate() + 3);
                $("#dt2").datepicker('setDate', date2);
            } else if (date2.getDay() == 3) {
                date2.setDate(date2.getDate() + 2);
                $("#dt2").datepicker('setDate', date2);
            } else if (date2.getDay() == 4) {
                date2.setDate(date2.getDate() + 4);
                $("#dt2").datepicker('setDate', date2);
            } else {
                date2.setDate(date2.getDate() + 2);
                $("#dt2").datepicker('setDate', date2);
            }

        }
    })
    $('#dt2').datepicker({
        dateFormat: 'dd-M-yy',
        minDate: 0,
        beforeShowDay: function (date) {
            var day = date.getDay();
            return [(day != 0 && day != 6), '']
        },
    });
});

Answer 2

如果有人想跳过某些工作日，我写了这段代码。可能这对任何人都有用。

function AddDays(date, offset, skipSunday, skipMonday, skipTueday, skipWednesday, skipThursday, skipFriday, skipSaturday) {

// If user wants to skip all days then return the same date because this is not possible
if (skipSunday && skipMonday && skipTueday && skipWednesday && skipThursday && skipFriday && skipSaturday) {
    return date;
}

// Create array for all day numbers
var skipDays = [];
if (skipSunday) skipDays.push(0);
if (skipMonday) skipDays.push(1);
if (skipTueday) skipDays.push(2);
if (skipWednesday) skipDays.push(3);
if (skipThursday) skipDays.push(4);
if (skipFriday) skipDays.push(5);
if (skipSaturday) skipDays.push(6);

while (offset > 0) {
    date.setDate(date.getDate() + 1);
    if ($.inArray(date.getDay(), skipDays) <= -1) {
        offset--
    }
}
return date;

}

添加日期选择器JS

2 个答案: