Calculate Fortnight Date

The following code will calculate a recurring fortnightly date from a given date, i.e. a star-date. The date format should be 'Y-m-d' for the following code. However, you can make updates in it as per requirements.

For example: The start date for the fortnight cycle date was set to be as "2018-01-10". Then, as per current cycle, next will be 2018-02-20.



/* Fortnight Calculation in JavaScript */ function getFortNightString(sDate) { var param = []; var objDate = new Date(); param.todayDate = objDate.getFullYear() + '-' + ("0" + (objDate.getMonth() + 1)).slice(-2) + '-' + ("0" + (objDate.getDate())).slice(-2); param.todayDate = new Date(param.todayDate).getTime(); param.sDate = new Date(sDate).getTime(); param.timeDiff = Math.abs(param.todayDate - param.sDate); param.diffDays = Math.ceil(param.timeDiff / (1000 * 3600 * 24)); if (param.diffDays > 0 && param.sDate > param.todayDate) { param.diffDays = (14 + (param.diffDays % 14)); } else { param.diffDays = (14 - (param.diffDays % 14)); } param.dayCount = param.diffDays - 1; objDate.setDate(objDate.getDate() + param.dayCount); objDate = objDate.getFullYear() + '-' + ("0" + (objDate.getMonth() + 1)).slice(-2) + '-' + ("0" + (objDate.getDate())).slice(-2); return " Start Date: " + sDate + " | Next Date: " + objDate; } /* Fortnight Calculation in PHP */ $sDate = "2018-02-05"; $sDate = date_create($sDate); $today = date_create(date("Y-m-d")); // or your date as well $dateDiff = date_diff($sDate, $today)->format("%a"); if($dateDiff > 0 && $sDate > $today){ $dateDiff = (14 + ($dateDiff % 14)); }else{ $dateDiff = (14 - ($dateDiff % 14)); } $dateDiff--; echo date('Y-m-d', strtotime("$dateDiff day", strtotime(date('Y-m-d'))));