difference of dates and then add days problem

[dethfire]

Let's say:
$termdate = 9/10/11
$addeddate = 8/11/11
$nowdate = 9/23/11

The result is calculated at 13 days diff and $newtermdate will be 09/23/11. That is fine.

However let's say:
$termdate = 12/09/11
$addeddate = 8/11/11
$nowdate = 9/23/11

The result is still calculated at 13 days diff and $newtermdate will be 12/22/11. That is obviously wrong. Any ideas?

$termdate = $row['termdate'];
$addeddate = $row['date'];
$nowdate = date("m/d/y");

$diff = abs(strtotime($nowdate) - strtotime($addeddate));

$years = floor($diff / (365*60*60*24));
$months = floor(($diff - $years * 365*60*60*24) / (30*60*60*24));
$days = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24)/ (60*60*24));

$newtermdate = strtotime('+'.$days.' day', strtotime($termdate));
$newtermdate = date('m/d/y', $newtermdate); 

[ParkinT]

Why are you not using the PHP function date_diff?
PHP offers a great set of functions, already built, to manipulate dates and times.

[dethfire]

I guess because I am using PHP 5.2 still

[Cups]

your question might be clearer to me if you gave an example of exact dates you have in your database, and exactly it is you want to derive from those 2 dates - a US type mm/dd/yyyy date? or a mysql ready date? And what is computation has to be done to get this date?

Dont describe them - paste the dates here. Give 2 or more examples if you want.

[dethfire]

your question might be clearer to me if you gave an example of exact dates you have in your database, and exactly it is you want to derive from those 2 dates - a US type mm/dd/yyyy date? or a mysql ready date? And what is computation has to be done to get this date?

Dont describe them - paste the dates here. Give 2 or more examples if you want.

I'm sorry for not being clear. Those are actual examples of dates in my database.

I'm trying to find the difference in days between the $nowdate and $dateadded. The difference in days should be added to the $termdate.

[StarLion]

$newdate = date('d/m/Y',strtotime($termdate) + abs(strtotime($nowdate) - strtotime($addeddate)));

Dont bother doing the conversions. Slap everything into seconds, do the math, and then translate THAT.

[dethfire]

I made it so difficult on myself! Thanks StarLion!

[Michael Morris]

Even faster, since this is coming from the database use MySQL's date functions.

[Cups]

From what the OP has shown I understand the OP is storing dates as mm/dd/yyyy and not yyyy-mm-dd, and that is one thing which is confusing me.

Never mind, PO has got a sensible reply and is happy.