RE: [PHP] Help with " Function " parameter
| From: | Martin Towell | Date: | Fri, 05 Apr 2002 00:39:06 +0000 |
| Subject: | RE: [PHP] Help with " Function " parameter | ||
| Groups: | php.general | ||
| Request: | Send a blank email to php-general+get-91648@lists.php.net to get a copy of this message | ||
Your problem lies in these lines
Return $realday;
Return $month;
Return $year;
and
Datereturn();
A function can only return one "value" - that value can be an array (see
later)
To use the returned value, you need to assign it to something (or use it
directly)
So, change your return to this:
Return array($realday, $month, $year);
and then change this line
Datereturn();
to either
$tmp = Datereturn();
$realday = $tmp[0];
$month = $tmp[1];
$year = $tmp[2];
Or this (which I think is a better way)
list($realday, $month, $year) = Datereturn();
HTH
Martin
-----Original Message-----
From: Simonk [mailto:fong@btinternet.com]
Sent: Friday, April 05, 2002 1:21 AM
To: php-general@lists.php.net
Subject: [PHP] Help with " Function " parameter
I have programed a function for counting Date:
function Datereturn()
{
$monthorder = Array ( 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);
$day = Date("j");
$monthday = Date("t");
$month = Date("n");
$year = Date("Y");
$current = 42 - ($monthday - $day);
if ($current > $monthorder[$month])
{ $realday = $current - $monthorder[$month];
$month = $month + 2;
if ($month > 12)
{ $year = $year + 1;
$month = $month - 12; }
}
else
{ $realday = $current;
$month = $month + 1;
if ($month > 12)
{ $year = $year + 1;
$month = $month - 12; } }
Return $realday;
Return $month;
Return $year;
}
But when I want to echo out the result, I have typed:
Datereturn();
echo "$year, $month, $realday";
Then the server return:
Undefined variable year, month, realday
plz help!
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