Re: Sloppy coding?

Sorry... too much wine! I get it. So if(!isset($someval)) will execute if $someval = null or ""? -Shawn "Shawn McKenzie" <nospam@mckenzies.net> wrote in message news:20030507042931.3165.qmail@pb1.pair.com... > That would clear it up, except that I use the first if() all of time and it > doesn't execute if I haven't defined the var, such as a script that receives > post or get and that var was not included in the post or get. > > Thanks! > Shawn > > "Martin Towell" <martin.towell@world.net> wrote in message > > news:6416776FCC55D511BC4E0090274EFEF508207078@exchange.world.net... > > Supposing $somevar has one of the following values: unset, null, 0, "", > > false > > the first if() will execute > > while the second will only execute for "unset" > > > > hope that clears things up.... > > > > -----Original Message----- > > From: Shawn McKenzie [mailto:nospam@mckenzies.net] > > Sent: Wednesday, May 07, 2003 2:21 PM > > To: php-general@lists.php.net > > Subject: [PHP] Sloppy coding? > > > > > > I want to sidestep any pitfalls and try to be a good coder. So what is > the > > difference between these two: > > > > if(!$somevar){ > > ...do stuff > > } > > > > if(!isset($somevar)){ > > ...do stuff > > } > > > > If one is bad, then are there any other common pitfalls similar to this > that > > I should avoid??? > > > > Thanks! > > Shawn > > > > > > > > > > -- > > PHP General Mailing List (http://www.php.net/) > > To unsubscribe, visit: http://www.php.net/unsub.php > >