SM wrote:
> On May 8, 11:50 am, SM <servandomont...@gmail.com> wrote:
>> On May 8, 11:42 am, Piotr <s...@poczta.onet.pl> wrote:
>>
>>
>>
>>>>> $thumb = 1;
>>>>> $item = '$thumbs_cat_' . $thumb; //this outputs a string:
>>>>> $thumbs_cat_1
>>>>> $item_total = sizeof($item);
>>>>> output: Nothing!
>>>>> How do i pass a variable to the 'sizeof' function?
>>>> You should use variable variable:
>>>> $item_total = sizeof($$item);
>>>> Mark double $$
>>>> This is true in any situation, not only with sizeof
>>> Forgot about one small detail:
>>> $item = 'thumbs_cat_' . $thumb; //this outputs a string:
>>> ---------^ loose the '$';
>>> best regards
>>> Piotr N
>> I don't need the length of the string , just the actual variable. But
>> i did like the fact that sizeof is an alias of count. So i will use
>> count instead.
>> I didn't know about the $$ tip. Wow! so easy. Thanks, It Works!
>>
>> Heres the final code:
>>
>> $thumb = 1;
>> $item = 'thumbs_cat_' . $thumb;
>> $item_total = count($$item);
>>
>> Thanks again
>> Marco
>
> Maybe i should open a new post for this one, but here it goes:
>
> I'm stock! When i want to reference that variable in a echo statment,
> it does't work. Any suggestions?
> Thanks
>
> <li><a href="<?php echo $SERVER['PHP_SELF']; ?>?cat=<?php echo
> $category_this; ?>"><img src="images/<?php echo($$item[$i]);?>"
> alt="" /></a></li>

:)
It should look like this. Mark the { and }
echo ${$item[$i]};

As a side note, is this really best way to access needed data this way ?
Cant you keep original values in the array instead of the variable names ?

best regards
Piotr N