我试图将几天和/或几周添加到从数据库中提取的日期中。我得到的只是它无法正确输出的1969年12月31日的默认日期。这是我的代码:
$lastFeed = "6-25-2013"; //pulled from database last feed date
$feedSchedule = "2"; //pulled from database number of weeks.
$nextFeeding = date("m-d-Y", strtotime($lastFeed . ' + ' . $feedSchedule. ' week'));
我还尝试将天数乘以$ feedSchedule变量,然后将周替换为天。
最佳答案
这是将起作用并说明无效的Date Time String的代码
function nextFeeding($lastFeed,$feedSchedule){
//fix date format
$correctedDate = explode("-",$lastFeed);
//pad month to two digits may need to do this with day also
if($correctedDate[0] < 10 && strlen($correctedDate[0])!==2){
$correctedDate[0] = "0".$correctedDate[0];
}
$correctedDate = $correctedDate[2]."-".$correctedDate[0]."-".$correctedDate[1];
//get the next feeding date
$nextFeeding = date("m-d-Y", strtotime($correctedDate . ' + ' . $feedSchedule. ' week'));
//return value
return $nextFeeding;
}
$lastFeed = "6-25-2013"; //pulled from database last feed date
$feedSchedule = "2"; //pulled from database number of weeks.
$nextFeeding = nextFeeding($lastFeed,$feedSchedule);
echo $nextFeeding;
退货
07-09-2013