[RESOLVED] Date Difference Calculation Help

condoug

I need someone to help me in achieving the following..

I have an integer which represents the number of days:

$days = 3;

Then I have 2 dates:

$first_date = "2010-04-01";

$last_date = "2010-04-07";

I need code which will check to see if the difference between the $first_date and $last_date variables is greater than $days.

Based on the outcome I will redirect/display messages etc. I will be grateful for any help offered.

Many Thanks for reading
Doug

bradgrafelman

Are these dates coming from a database, by any chance?

anakadote

$difference = strtotime($last_date) - strtotime($first_date); // Difference between dates (in seconds)

if($difference/(60*60*12) >= $days)
  echo 'Greater than or equal to  $days';
else
  echo 'Less than';

bradgrafelman

Or, if the dates are indeed coming from a SQL database..

$sql = 'SELECT DATEDIFF(date2, date1) AS daysDiff ...';
$exec = mysql_query($sql);

while($row = mysql_fetch_assoc($exec)) { // or something similar

// Check difference in dates:
if($row['daysDiff'] >= $days)
  echo 'Greater than or equal to';
else 
  echo 'Less than'; 

echo " $days days";

}

condoug

That's a great help and weight off my shoulders - thank you for your prompt replies.

Doug

condoug

Hi anakadote

I have tried your method but unfortunately I get mixed results.

If I have date1 as '2010-04-01' and date2 as '2010-04-03', (this is 3 nights/days difference) but the output using your code says $difference is 4:

$difference = strtotime('2010-04-01') - strtotime('2010-04-03');
echo $difference/(60*60*12);

If I change date2 to be '2010-04-02' however it works and $difference is equal to 2 days which is correct.

Any thoughts??

Doug

condoug

This is sorted now thanks. I used the following functionality:

function daysDifference($endDate, $beginDate) {
$beginDate = strtotime($beginDate);
$endDate = strtotime($endDate);
$diff = $endDate - $beginDate;
$diff = ceil($diff / (606024)) ;
return $diff;
}