I needed to find the number of months between 2 dates. I don't know of a function in PHP, so I wrote this.

PHP Code:

<?php
define('DATE_MONTH',1);
define('DATE_DAY',2);

function dateDifference($date_1, $date_2, $type=DATE_MONTH) {
  $start_date = strtotime($date_1);
  $end_date = strtotime($date_2);
  $sign = ((int)($end_date-$start_date)/abs($end_date-$start_date));

  $period_offset = 0;
  if ($sign > 0) {
    switch ($type) {
      case DATE_MONTH:
        while (strtotime('+'.$period_offset.' months',$start_date) <= $end_date) {
          $period_offset++;
        }
        break;
      case DATE_DAY:
        while (strtotime('+'.$period_offset.' days',$start_date) <= $end_date) {
          $period_offset++;
        }
        break;
    }
  } else {
    switch ($type) {
      case DATE_MONTH:
        while (strtotime('-'.$period_offset.' months',$start_date) <= $end_date) {
          $period_offset++;
        }
        break;
      case DATE_DAY:
        while (strtotime('-'.$period_offset.' days',$start_date) <= $end_date) {
          $period_offset++;
        }
        break;
    }
  }
  return $period_offset;
}
echo dateDifference('5/1/2008','11/1/2008');
?>

Anyone know of a better way?

I'm not in a position where I can test this myself but it makes since in my head:

PHP Code:

function numberOfMonths($date1, $date2)
{
     $diff = abs(strtotime($date2) - strtotime($date1));
     $month = date('n', $diff);
     $year = date('o', $diff);
     $deltaYears = $year - 1970; 

     $deltaMonths = $deltaYears * 12 + ($month - 1);
     return $deltaMonths;
} 


I might be missing something but I think that will work, provided you don't use any dates prior to 1970. If it is the case that you are working with dates before 1970 just add the difference to each date (ie 1968 and 1969 become 1970 and 1971)

Thanks. That code doesn't actually work b/c of how $diff is calculated (it's meaning is distorted for the date function. I see your difference in logic, however, and recoded it like this:

PHP Code:

<?php
function numberOfMonths($date1, $date2)
{
     $diff = abs(strtotime($date2) - strtotime($date1));
     $month = date('n', $diff);
     $year = date('o', $diff);
     $deltaYears = $year - 1970;

     $deltaMonths = $deltaYears * 12 + ($month - 1);
     return $deltaMonths;
}
echo numberOfMonths('11/15/2008','2/1/2009').'<br /><br />';

function numMonths($date_1, $date_2) {
  $smallest_date = min(strtotime($date_1),strtotime($date_2));
  $smallest_year = date('Y',$smallest_date);
  $smallest_month = date('m',$smallest_date);

  $largest_date = max(strtotime($date_1),strtotime($date_2));
  $largest_year = date('Y',$largest_date);
  $largest_month = date('m',$largest_date);

  $years_between = $largest_year - $smallest_year - 1;
  $months_from_smallest = 13 - $smallest_month;

  if ($largest_year == $smallest_year) {
    return $largest_month - $smallest_month + 1;
  } else {
    return $years_between*12 + $months_from_smallest + $largest_month;
  }
}
echo numMonths('11/15/2008','2/1/2009').'<br />';
echo numMonths('2/15/2008','11/1/2008').'<br />';
?>

The numMOnths function appears to work correctly, using your logic, and is more efficient than my original routine.

Thanks a lot!

I need to find the number of days between 2 dates.
could you please help me??

Here are some date functions I made that I use in many of my scripts:

PHP Code:

//convert unix to month day year
//returns array [0]->month [1]->day [2]->year
function unixToDate($unix) { 
    $date = date('m-d-y',$unix);
    $date_array = explode("-", $date);
    return $date_array;
}


//converts month day year to unix stamp
//sample usage:  $unix = dateToUnix(0, 0, 0, $_POST['month'], $_POST['day'], $_POST['year']);
function dateToUnix($month,$day,$year) {
    $unix = mktime(0,0,0,ltrim($month, '0'),ltrim($day, '0'),ltrim($year, '0'));
    return $unix;
}

//returns the number of days difference between two dates
//dates must be sent to function in unix!
//returns number of days ie 1 2 ... 12 13 etc...
function dateDiff($earlier_date, $later_date) { 
    $diff_in_secs=$later_date - $earlier_date;
    $diff_in_days=$diff_in_secs / 86400;
    return $diff_in_days;
} 


I know that doesn't help find the number of months between two dates but maybe it would be useful somehow.... I see abs0lut could use the dateDiff function

OK. I sat down to redo the day calculation based off of Null's more intelligent implication and came up with this:

PHP Code:

<?php
function numDays($date_1,$date_2) {
  $start_day = strtotime(date('m/d/Y 00:00:00',strtotime($date_1)));
  $end_day = strtotime(date('m/d/Y 23:59:59',strtotime($date_2)));

  $difference = abs($end_day - $start_day);
  return ceil($difference/(60*60*24));
}
echo numDays('1/1/08','1/3/08').'<br />';
echo numDays('2/28/08','3/1/08').'<br />';
echo numDays('1/1/08','12/31/08').'<br />';
echo numDays('1/1/07','12/31/08').'<br />';
?>

The idea, modified from the month one, is to start at the first second of the first day and end on the last second of the last day, so that seconds-based calculations are accurate.

Originally Posted by Inet411

Here are some date functions I made that I use in many of my scripts:

PHP Code:

//convert unix to month day year
//returns array [0]->month [1]->day [2]->year
function unixToDate($unix) { 
    $date = date('m-d-y',$unix);
    $date_array = explode("-", $date);
    return $date_array;
}


//converts month day year to unix stamp
//sample usage:  $unix = dateToUnix(0, 0, 0, $_POST['month'], $_POST['day'], $_POST['year']);
function dateToUnix($month,$day,$year) {
    $unix = mktime(0,0,0,ltrim($month, '0'),ltrim($day, '0'),ltrim($year, '0'));
    return $unix;
}

//returns the number of days difference between two dates
//dates must be sent to function in unix!
//returns number of days ie 1 2 ... 12 13 etc...
function dateDiff($earlier_date, $later_date) { 
    $diff_in_secs=$later_date - $earlier_date;
    $diff_in_days=$diff_in_secs / 86400;
    return $diff_in_days;
} 


I know that doesn't help find the number of months between two dates but maybe it would be useful somehow.... I see abs0lut could use the dateDiff function

thank you very much