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Posted by Jerry Stuckle on 11/18/40 11:44
Oli Filth wrote:
> Jerry Stuckle said the following on 09/04/2006 02:53:
>
>> Oli Filth wrote:
>>
>>> Jerry Stuckle said the following on 08/04/2006 21:51:
>>>
>>>> Oli Filth wrote:
>>>>
>>>>> function getWithDefault(&$array, $key, $default = NULL)
>>>>> {
>>>>> return (isset($array[$key])) ? $array[$key] : $default;
>>>>> }
>>>>>
>>>>>
>>>>
>>>> And if $array is empty you get a warning.
>>>>
>>>
>>> You do? I don't. Nor can I see any reason why one should get a
>>> warning. What version of PHP are you running?
>>>
>>
>> Try turning on all warnings.
>
>
>
> I have error level set to E_ALL | E_STRICT. The following code executes
> fine:
> <?php
>
> function getValueWithDefault(&$array, $key, $default = NULL)
> {
> return (isset($array[$key])) ? $array[$key] : $default;
> }
>
> $var = array();
> echo getValueWithDefault($var, "Roger", "Dodger") . "\n";
>
> ?>
>
> I see no reason why an error/warning should get thrown. isset() simply
> looks for an element with the specified key name in the associative
> array. In an empty array, the key doesn't exist, so isset() returns false.
>
That's not what I said. I said if $var is EMPTY.
Take out the $var=array() line and see what you get.
--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
==================
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