Question

currently I’m cleaning up some code which creates a lot of php warnings.
I’ve got the following code:

if(isset($info[1])) {
        list($artist, $song) = explode(" - ", trim(strip_tags($info[1])));
        $fullname = trim(strip_tags($info[1]));
    }

So I only want to execute this block is the index 1 is set. But when executing the the script I still get “Notice: Undefined offset: 1 in ….” on this line.

So what would be the right way to check if there is a value at a specified index in my array?

EDIT:
I tried your solutions but it is still not working as I expect.
I’ve added some debug stuff which just increased my confusion:
I’ve added the following for testing:

echo "<pre>";
print_r($info);
echo "</pre>";
echo ".";
die();

and the result is:

Array
(
   [0] => 
)
.
Played @
[1] => Song Title

Right now I don’t really know how to interprete the output… can you give me a hint!?

Answer 1

First, you might need to ensure that $info exists:

if (isset($info, $info[1]))

But my guess is that there’s no ‘ – ‘ match in $info[1] after you trim it, and it’s there that only one element is returned (and therefore, list() would complain that there’s no 1 index).