$date1 = date(‘Y-m-d’); // yyyy-mm-dd format
$date2 = ’2006-07-01′;
$days = (strtotime() – strtotime()) / (60 * 60 * 24);
echo “Number of days since ’2006-07-01′: $days”;
Where strtotime
int strtotime ( string $time [, int $now] )
The function expects to be given a string containing a US English date format and will try to parse that format into a UNIX timestamp (the number of seconds since January 1 1970 00:00:00 GMT), relative to the timestamp given in now, or the current time if now is not supplied.
Comments