Re: Re: isset

Chris W. Parker wrote:

M. Sokolewicz <mailto:tularis@php.net>

    on Tuesday, February 15, 2005 8:25 AM said:

seems lengthy. is there a way around this? i tried using $cmd = @ $_POST['cmd']; to suppress errors but didnt seem to have ay effect.

still if(isset($_POST['cmd'])) { $cmd = $_POST['cmd']; } is the only (really) correct way. All others throw notices of undefined indices, which you *should not* ignore! You can't shorten it, you can't even wrap it in a function... it's simply tough luck

Please correct me if I'm wrong (and maybe this is what you mean by "(really)") but I think even Rasmus recommends: <?php if(!empty($_POST['cmd'])) {

    // stuff

} ?> I'm using the following code to test this: <?php

    error_reporting(E_ALL);

    if(!empty($_POST['cmd']))
    {
        echo "set and not empty<br />";
    }
    else
    {
        echo "empty and/or not set<br />";
    }

?> Chris. that's a different issue.

There are always at least 2 things you should do with your (expected) input: 1 - check if it *exists* (isset) 2 - check the validity (input-validation) for step #2 empty is very commonly used, and also a very useful function. However, you should never do #2 without #1, since that again raises issues (of security, problems, unexpected input, etc) Also note that empty($non_existent_var) will always throw an E_NOTICE error when the variable in question is not set. isset() is the only function/language-construct that can check for the existence of variables without throwing an E_NOTICE. - tul