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调用函数代码举例

                        //PS :这里需要注意的是 周日 是 0 ！！！！！console.log(getBeforeDate(-10000)) //一万天之后的日期console.log(getWeek('2019-07-03',[2,3],20))//从'2019-07-03' 开始获取之后 20 个 周二、周三的日期

JS代码

/*** 获取某个时间开始 之后的 N次[周几,周几]* @param {Object} begin 开始时间* @param {Object} week_arr 需要获取的周几数组* @param {Object} Max 需要获取数组最大长度*/function getWeek (begin, week_arr,Max){var dateArr = new Array();var stimeArr = begin.split("-");//=>["2018", "01", "01"]var etimeArr = getBeforeDate(-10000).split("-");//=>["2018", "01", "30"] 这里我给了一个10000天后的一起可更具需求更改var stoday = new Date();stoday.setUTCFullYear(stimeArr[0], stimeArr[1]-1, stimeArr[2]);var etoday = new Date();etoday.setUTCFullYear(etimeArr[0], etimeArr[1]-1, etimeArr[2]);var unixDb = stoday.getTime();//开始时间的毫秒数var unixDe = etoday.getTime();//结束时间的毫秒数for (var k = unixDb; k <= unixDe;) {//达到最大次数时,结束循环if(dateArr.length==Max){break;}else{let needJudgeDate = msToDate(parseInt(k)).withoutTime;//不加这个if判断直接push的话就是已知时间段内的所有日期$.each(week_arr, function(i,o) {if (new Date(needJudgeDate).getDay() === o) {dateArr.push(needJudgeDate);}})k = k + 24*60*60*1000;}}return dateArr;}//根据毫秒数获取日期
function msToDate (msec) {let datetime = new Date(msec);let year = datetime.getFullYear();let month = datetime.getMonth();let date = datetime.getDate();let hour = datetime.getHours();let minute = datetime.getMinutes();let second = datetime.getSeconds();let result1 = year +'-' +((month + 1) >= 10 ? (month + 1) : '0' + (month + 1)) +'-' +((date + 1) < 10 ? '0' + date : date) +' ' +((hour + 1) < 10 ? '0' + hour : hour) +':' +((minute + 1) < 10 ? '0' + minute : minute) +':' +((second + 1) < 10 ? '0' + second : second);let result2 = year +'-' +((month + 1) >= 10 ? (month + 1) : '0' + (month + 1)) +'-' +((date + 1) < 11 ? '0' + date : date);let result = {hasTime: result1,withoutTime: result2};return result;
}//-----[获取前n天的日期]
function getBeforeDate(n) {var n = n;var d = new Date();var year = d.getFullYear();var mon = d.getMonth() + 1;var day = d.getDate();if(day <= n) {if(mon > 1) {mon = mon - 1;} else {year = year - 1;mon = 12;}}d.setDate(d.getDate() - n);year = d.getFullYear();mon = d.getMonth() + 1;day = d.getDate();s = year + "-" + (mon < 10 ? ('0' + mon) : mon) + "-" + (day < 10 ? ('0' + day) : day);return s;
}
//-----[获取前n天的日期END]