$days = 35;
$day = $day + $days;
//calculate the new date
$calcDate = date($output, mktime (0,0,0,$month,$day,$year));

No need for a 2nd query...jus' insert the variable $calcDate into a column
in your table

Hope this helps,

Jimmy Brock

"Shannon Doyle" <[EMAIL PROTECTED]> wrote in message
> Hi People,
> Need some assistance in the following scenario:-
> Inserting a date into the database that is entered into a form by the site
> visitor. - This is easy enough.
> However I now need to use the same date that has been entered by the site
> visitor add 35 days and then insert into another table.
> My question, how do I get the date entered into the form add 35days to it
> and then include that into the same sql query as the first one. Or do I
> to use a second sql query? If the second query how would I get the date
> add 35days??
> Cheers,
> Shannon

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