date() + 30 days = xxx

paulOr · 11-22-2008, 11:05 PM

Ok so iv been trying to figure out how to do this little thing:

i have todays date: 22 November 2008 and i want to add 30 days onto it, and end up with another date, which would be 22 December 2008.

so like:

PHP Code:


			
date('j F Y') + 30 = new_date('j F Y')

Is there a easy way of doing this? Im so stuck!

codefreek · 11-22-2008, 11:24 PM

PHP Code:


			
$d = mktime(0,0,0,$month,$day,$year);

$end_date = date(”Y m d”,strtotime(”+2 days”,$d));



//To add days to current date

date(’Y-m-d’, strtotime(”+2 days”));

paulOr · 11-22-2008, 11:26 PM

well, the date is just coming from date() but new_date will be stored in mysql :P

Tanax · 11-22-2008, 11:29 PM

Quote:

Originally Posted by codefreek

is it for mysql ? then i know :P

Doesn't matter? xD

Just use it like this:

php Code:

$endDate = date("D F d Y", strtotime("+30 days"));

paulOr · 11-22-2008, 11:31 PM

PHP Code:


			
date('j F Y', strtotime('+30 days'));

got it :)

codefreek · 11-22-2008, 11:35 PM

i was thinking on something else sorry but i posted my way but tanax was faster oh well ;)

Wildhoney · 11-23-2008, 01:56 AM

I created this class to modify the dates using the strtotime function. Use it if you wish!

php Code:

class Date_Calculate
{
    private $m_iFromTimestamp;
    
    public function __call($szCall, $aArgs)
    {
        if (preg_match('~^to(Minus|Plus)(\d+)(.+)$~i', $szCall, $aMatches))
        {
            $szOperator = $aMatches[1] == 'Minus' ? '-' : '+';
            $iTimestamp = !empty($this->m_iFromTimestamp) ? $this->m_iFromTimestamp : time();
            return date($aArgs[0], strtotime($szOperator . $aMatches[2] . ' ' . $aMatches[3], $iTimestamp));
        }
        
        throw new Exception('Invalid date to syntax specified');
    }
    
    public function from($iTimestamp)
    {
        if (!is_integer($iTimestamp))
        {
            $iTimestamp = strtotime($iTimestamp);
        }
        
        $this->m_iFromTimestamp = $iTimestamp;
        return $this;
    }
}

$pDate = new Date_Calculate();

echo $pDate->from(time())->toMinus4Days('jS M Y');
echo $pDate->toPlus5Years('jS M Y');

codefreek · 11-23-2008, 07:27 AM

looks great, Wildhoney :D
thanks for sharing will probably learn something from it will,
look at it some more later..

Kalle · 11-23-2008, 09:25 AM

Damns some sweet logic Adam :)