Thank you, that was it. I needed to test the variable against both an integer and a string, so I ended up using (string) when I need to compare a string. Rest of the time, it's an integer.
Robin Rasmus Lerdorf wrote: > > > If you set $qty=0; then $qty has no value. > > Of course it has a value. The value is 0. > Quite distinct from not having a value, or in proper terms, not being set. > Try this: > > var_dump($qty); > $qty = 0; > var_dump($qty); > > Well, I will save you the trouble, it outputs: > > NULL > int(0) > > > Type casting is irrelevant in php. > > No they aren't. Try this: > > $qty = 0; > if($qty=="test") echo "foo"; > if((string)$qty=="test") echo "bar"; > > This will only output "foo". Not "bar". You could also use the === > operator to force PHP to check not only the value but also the type. > > See http://www.php.net/manual/en/language.operators.comparison.php > > -Rasmus > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, e-mail: [EMAIL PROTECTED] > For additional commands, e-mail: [EMAIL PROTECTED] > To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
