how to add 7 days to order date

[Guest]

I have an issue

 

I want to add +7days to order date so:

 

i get the date value DB

 

i cant manage to add 7days now i get: 2010.gads 01.maijs (gads - means year and maijs - means may)

but i need: 2010.gads 08.maijs, i`m getting an error

 

Parse error: syntax error, unexpected '(', expecting ')' in /var/www/sapnudarbnica/includes/application_top.php  on line 298

 

function replaceMon3($paydate = date( 'Y-m-d', strtotime( '+7 day', $Orderdp ) )) {

 // replace English 3 letter month abbreviation with full any languge - month name
 switch(substr($paydate,8,3)) {
case 'Jan': $paydate = substr($paydate,0,4) . YEAR . substr($paydate,4,3) . MONTH_JAN; break;
       case 'Feb': $paydate = substr($paydate,0,4) . YEAR . substr($paydate,4,3) . MONTH_FEB; break;
       case 'Mar': $paydate = substr($paydate,0,4) . YEAR . substr($paydate,4,3) . MONTH_MAR; break;
case 'Apr': $paydate = substr($paydate,0,4) . YEAR . substr($paydate,4,3) . MONTH_APR; break;
case 'May': $paydate = substr($paydate,0,4) . YEAR . substr($paydate,4,3) . MONTH_MAY; break;
case 'Jun': $paydate = substr($paydate,0,4) . YEAR . substr($paydate,4,3) . MONTH_JUN; break;
case 'Jul': $paydate = substr($paydate,0,4) . YEAR . substr($paydate,4,3) . MONTH_JUL; break;
case 'Aug': $paydate = substr($paydate,0,4) . YEAR . substr($paydate,4,3) . MONTH_AUG; break;
case 'Sep': $paydate = substr($paydate,0,4) . YEAR . substr($paydate,4,3) . MONTH_SEP; break;
case 'Oct': $paydate = substr($paydate,0,4) . YEAR . substr($paydate,4,3) . MONTH_OCT; break;
case 'Nov': $paydate = substr($paydate,0,4) . YEAR . substr($paydate,4,3) . MONTH_NOV; break;
case 'Dec': $paydate = substr($paydate,0,4) . YEAR . substr($paydate,4,3) . MONTH_DEC; break;
 }
 return $paydate;
}

 

this is original function

 

function replaceMon3($Orderdp) {
 // replace English 3 letter month abbreviation with full Latvian month name
 switch(substr($Orderdp,8,3)) {
case 'Jan': $Orderdp = substr($Orderdp,0,4) . YEAR . substr($Orderdp,4,3) . MONTH_JAN; break;
   case 'Feb': $Orderdp = substr($Orderdp,0,4) . YEAR . substr($Orderdp,4,3) . MONTH_FEB; break;
   case 'Mar': $Orderdp = substr($Orderdp,0,4) . YEAR . substr($Orderdp,4,3) . MONTH_MAR; break;
case 'Apr': $Orderdp = substr($Orderdp,0,4) . YEAR . substr($Orderdp,4,3) . MONTH_APR; break;
case 'May': $Orderdp = substr($Orderdp,0,4) . YEAR . substr($Orderdp,4,3) . MONTH_MAY; break;
case 'Jun': $Orderdp = substr($Orderdp,0,4) . YEAR . substr($Orderdp,4,3) . MONTH_JUN; break;
case 'Jul': $Orderdp = substr($Orderdp,0,4) . YEAR . substr($Orderdp,4,3) . MONTH_JUL; break;
case 'Aug': $Orderdp = substr($Orderdp,0,4) . YEAR . substr($Orderdp,4,3) . MONTH_AUG; break;
case 'Sep': $Orderdp = substr($Orderdp,0,4) . YEAR . substr($Orderdp,4,3) . MONTH_SEP; break;
case 'Oct': $Orderdp = substr($Orderdp,0,4) . YEAR . substr($Orderdp,4,3) . MONTH_OCT; break;
case 'Nov': $Orderdp = substr($Orderdp,0,4) . YEAR . substr($Orderdp,4,3) . MONTH_NOV; break;
case 'Dec': $Orderdp = substr($Orderdp,0,4) . YEAR . substr($Orderdp,4,3) . MONTH_DEC; break;
 }
 return $Orderdp;
}

 

help!

[burt]

http://www.google.com/search?hl=en&safe=off&rlz=1B3GGGL_en___GB242&q=php+add+days+to+date&aq=f&aqi=&aql=&oq=&gs_rfai=

 

should give all the info you need

[Guest]

http://www.google.com/search?hl=en&safe=off&rlz=1B3GGGL_en___GB242&q=php+add+days+to+date&aq=f&aqi=&aql=&oq=&gs_rfai=

 

should give all the info you need

 

cant get it work propertly...

[candycanes]

cant get it work propertly...

 

 

Not sure if it matters, but you wrote +7 day not +7 days

 

Other than that, not sure!

[Guest]

Not sure if it matters, but you wrote +7 day not +7 days

 

Other than that, not sure!

 

i have managed with this:

 

<?php echo date('Y d M', strtotime("+7 days"));?>

 

to out put current date + 7days, but cannot manage out put date_purchased + 7 days

[Guest]

Actually that should look something like this:

 

$orderdp = tep_date_long($order->info['date_purchased'] , strtotime("+7 days"));

 

I`m right?

[candycanes]

i have managed with this:

 

<?php echo date('Y d M', strtotime("+7 days"));?>

 

to out put current date + 7days, but cannot manage out put date_purchased + 7 days

 

Try: date('Y d M', strtotime("+7 days", strtotime(&Orderdp)))

[Guest]

Try: date('Y d M', strtotime("+7 days", strtotime(&Orderdp)))

 

this:

 

<?php echo date('Y d M', strtotime("+7 days", strtotime($orderdp)))  ?>

 

it`s starts to count the days from year 1970 01 Jan and i have displayed - 1970 08 Jan

[Guest]

Solved this:

 

if someone needs it here is the solution for +7 days:

 

$orderdp = date('Y d M', ( strtotime($order->info['date_purchased']) +604800 ) );