Javascript 日历在闰年期间跳过几天？

Question

我用 javascript 和 jquery 编写了一个日历(改编自 "Simple Calendar with Javascript" )，它允许我一次前后移动几个月。

它通过每 4 年的 2 月增加一天来计算闰年，但似乎因此导致了错误。

当您向后或向前移动时，它会更改日期并根据新日期运行 calendar() 函数。通常，它有 5 行 7 列，每列包含一天。如果 5 行没有足够的空间，例如有 31 天，其中第一天是星期六，那么它将额外添加第 6 行来容纳。

但是，在闰年期间，如果发生上述相同情况，该函数似乎不会添加额外的行。

这是一个 jsFiddle:http://jsfiddle.net/charlescarver/DrMWM/

calendar() 的要点:

function calendar(d){

                $("#calendar").remove();
                $("body").append("<table border='1' id='calendar'></table>");

                // Date vars
                var today = new Date(d);
                var month = today.getMonth();
                var day = today.getDay();
                var dayN = today.getDate();
                var week = ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"];
                var monthdays = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
                var days = monthdays[month];
                var monthNames = ["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"];

                // Determine if leap year, if not, set to current year
                if (month > 0) {
                    year = today.getYear();
                    if (year % 4 == 0){
                        days = 29;
                    }
                } else {
                    year = today.getYear();
                }

                // Parse vars
                var jumped = 0;
                var inserted = 1;
                var start = day - dayN%7 + 1;
                if (start < 0) start += 7;
                var weeks = parseInt((start + days)/7);
                if ((start + days)%7 != 0){
                    weeks++;
                }

                $("#calendar").append("<tr><td class='month' colspan='7'>" + monthNames[month] + " - " + today.getFullYear() + "</td></tr>");

                $("#calendar").append("<tr class='day'></tr>");
                for (var i=0; i<7; i++) {
                    $(".day").append("<td>" + week[i][0] + "</td>");
                }

                function datDate(m, n, y){
                    date = new Date(" " + monthNames[m] + " " + n + " " + y);
                    return date;
                }

                for (var i=weeks; i>0; i--) {
                    if (i === i){
                        $("#calendar").append("<tr class=" + i + "></tr>");
                        for (var j=0; j<7; j++) {
                            console.log(inserted);
                            if (jumped<start || inserted>days) {
                                $("." + i).append("<td> </td>");
                                jumped++;
                            }
                            else {
                                m = month;
                                n = inserted;
                                y = today.getFullYear();
                                date = datDate(m, n, y);
                                if (inserted == dayN) {
                                    $("." + i).append("<td><a class='calendar_date current_date' data-date='" + date + "'>[" + n + "]</a></td>");
                                } else {
                                    $("." + i).append("<td><a class='calendar_date' data-date='" + date + "'>" + n + "</a></td>");
                                }
                                inserted++;
                            }
                        }
                    }
                }   
            }

Answer 1

            // Determine if leap year, if not, set to current year
            if (month == 1) {
                year = today.getYear();
                if (year % 4 == 0){
                    days = 29;
                }
            } else {
                year = today.getYear();
            }