Hello Neil,
Wednesday, January 10, 2001, 10:33:09 AM, you wrote:
>> : <?php if (! $a) print "Hello, World!"; // script 2 ?>
>>
>> $a evaluates to false, the ! reverses it, and it prints "Hello, World!"
NZ> What is bothering me is the following: if variables that are not assigned
NZ> a value were to evaluate to false then since false is the same as the
NZ> number 1 the following PHP script should print the number 1 but instead
NZ> prints nothing:
NZ> <?php if (! $a) print $a; ?>
NZ> How is this behavior justified?
The operation (! $a) must return something .. true or false. The only
other option would be to crash the system to a debugger with a nasty
message about the developer should have checked the variable cause it
wasn't assigned.
For this reason if you are unsure about the state try
If (isset($a) {
If (! $a) {
}
Else {
}
}
Something like this always works.
--
DSig `````
David Tod Sigafoos ( O O )
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