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Posted by Marcus Bointon on 10/31/05 22:53
On 31 Oct 2005, at 17:55, Richard Lynch wrote:
> You're wrong.
No. You're just missing what I'm on about. I'll agree it's very dull.
> isset() does not, under any circumstances, create an index nor a
> variable.
Quite right; I never said it did.
> Its entire purpose *IS* to tell you if something has been set to a
> value.
> That's why it's CALLED "isSet"
But that's not the question you're using isset to answer. You just
want to know if something exists - you probably don't even care what
its value is.
Take apart this operation:
$a = isset($myarray['a']);
Implicit in this simple line is a 'hidden' step which is to look up
the index 'a' in $myarray to get its value before testing if it is
set or not. The issue I have is that that step's existence is being
overlooked. That line could also be written:
$a = !is_null($myarray['a']);
Either way, if $myarray['a'] does not exist, its value will be
regarded as null (and as such, isset and !is_null would give correct
results as a consequence of this convenient side-effect), but I would
also fully expect to receive an undefined index notice as you have
explicitly looked up an array index that does not exist. If you used
other functions the same way you're using isset, you would see
nothing wrong with this:
$myarray = array();
print $myarray['a'];
but I would hope that you would have a problem with that. Why treat
isset differently?
Marcus
--
Marcus Bointon
Synchromedia Limited: Putting you in the picture
marcus@synchromedia.co.uk | http://www.synchromedia.co.uk
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