NYCPHP Meetup

[Darryle Steplight]

Kristina,

That code is not testing two conditions. It's testing one condition
and assigning one value to $result if $a['c'] is set or another value
to $result if $a['c'] is not set. It is the shorthand equivalent to
writing


if (isset($a['c'])
{
    $result = $a['c] ;
}
else
{
     $result = NULL;
}

. Both do the same thing, but one is a lot faster to write. So you can
either write all that out or just
 $result = isset($a['c']) ? $a['c'] : NULL;.

I hope that clarifies things.

On Sun, Aug 17, 2008 at 6:10 PM, Kristina Anderson
<ka at kacomputerconsulting.com> wrote:
> Hi all.  I have a question regarding the below syntax
>
> $result = isset($a['c']) ? $a['c'] : NULL;
>
>
> This is testing two conditions, first if $a['c'] is set, and then, if $a
> ['c'] is NULL...correct?
>
> What would $result be if $a['c'] is set but is not NULL?  The same as
> if $a['c'] was not set?  or...?
>
> I've seen similar syntax before, in some code Tedd & I were working
> with a couple of months back...and just want to make sure I fully
> understand what is actually happening with it.
>
> Thanks.
>
> -- Kristina
>
>> On Sun, Aug 17, 2008 at 4:49 PM, Michael B Allen <ioplex at gmail.com>
> wrote:
>> > Is it ok to access non-existent array elements?
>> >
>> > Will accessing an array element using a key that does not exist
> always
>> > return NULL?
>>
>> it depends on how you have warnings configured. you might be getting
>> them in your log, for example... i believe the result of accessing a
>> non-existent array element is undefined.
>>
>> if i were you, i'd prob'ly ask if it exists first:
>>
>> <?php
>>
>> $result = isset($a['c']) ? $a['c'] : NULL;
>>
>> /** OR **/
>>
>> $result = array_key_exists('c', $a) ? $a['c'] : NULL;
>>
>> ?>
>>
>> see:
>>
>> http://us2.php.net/function.isset
>> http://us2.php.net/function.array-key-exists
>>
>>
>> justin
>> --
>> http://justinhileman.com
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