Since you don't declare variables in PHP, they are preinitialized for you as 'false' or '0' So since $b is not set initially, it is 0 therefore second elseif passes. > -----Original Message----- > From: Choy, Wai Yew [mailto:waiyew.choy@xxxxxxxxxx] > Sent: Wednesday, November 08, 2006 9:34 PM > To: php-general@xxxxxxxxxxxxx > Subject: Simple logic but can't get it right > > Hi gurus, > > > > I've the following piece of PHP code....It is a simple logic > but I can't > get it right....The output result of $id is "b"!!...It should be > "outside", right?? > > > > I think it is the variable $b....Coz' it depend on the > previous check ( > if ($a == 0) ) for the value... If this is the case, how can > I get this > logic works?? > > > > Thanks a million, > > Choy > > > > > > > > > > <?php > > > > $a = 1; > > $id = "outside"; > > > > if ($a == 0) { > > > > $b = 1; > > $id = "a"; > > > > } > > > > > > elseif ($b == 0) { > > > > $c = 1; > > $id = "b"; > > > > } > > > > > > elseif ($c == 0){ > > > > $d = 1; > > $id = "c"; > > > > } > > > > > > echo "ID = $id<br>"; //Output is "b"...WHY?? It should be > "outside"... > > > > > > > > ?> > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
