調用函數代碼舉例數組
//PS :這裏須要注意的是 週日 是 0 !!!!! console.log(getBeforeDate(-10000)) //一萬天以後的日期 console.log(getWeek('2019-07-03',[2,3],20))//從'2019-07-03' 開始獲取以後 20 個 周2、週三的日期
JS代碼函數
/** * 獲取某個時間開始 以後的 N次[周幾,周幾] * @param {Object} begin 開始時間 * @param {Object} week_arr 須要獲取的周幾數組 * @param {Object} Max 須要獲取數組最大長度 */ 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"] 這裏我給了一個10000天后的一塊兒可更具需求更改 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();//開始時間的毫秒數 var unixDe = etoday.getTime();//結束時間的毫秒數 for (var k = unixDb; k <= unixDe;) { //達到最大次數時,結束循環 if(dateArr.length==Max){ break; }else{ let needJudgeDate = msToDate(parseInt(k)).withoutTime; //不加這個if判斷直接push的話就是已知時間段內的全部日期 $.each(week_arr, function(i,o) { if (new Date(needJudgeDate).getDay() === o) { dateArr.push(needJudgeDate); } }) k = k + 24*60*60*1000; } } return dateArr; } //根據毫秒數獲取日期 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; } //-----[獲取前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; } //-----[獲取前n天的日期END]