php Undefined variable: article_id

Question

I keep getting this as a warning I want to avoid getting this warning when it is undefined without turning warnings off

here is the context

  $url_items = array("foo");
  $article_id = db_escape($url_items[1]);
  $article = get_article($article_id);

  function get_article($article_id = NULL) {.....}

Answer 1

put this before your code:

error_reporting(E_ALL ^ E_NOTICE);

PHP Manual: error_reporting

Answer 2

i think this would be a nicer way maybe there is a better way but i think just hiding the warning is the wrong way...

$article_id = db_escape($url_items[1]);
if(empty($article_id)){
    $article_id = null;
}

edit corrected the code

Answer 3

I don't know if this is related to the error, but you should know that PHP indexes arrays starting with 0, so the second line should be

$article_id = db_escape($url_items[0]);

Also, it is probably a typo, but the first line should be

var $url_items = array("foo");

Answer 4

You don't say exactly which line is causing the error, but you ought to use isset for any variables you are not sure exist. For example:

$url_items = array("foo");
if ( isset($url_items[1]) )
{
    $article_id = db_escape($url_items[1]);
    $article = get_article($article_id);
}

function get_article($article_id = NULL) {.....}

You'll also want to check the content of the db_escape method, in case that is also doing something with an undefined variable.

---

One other way around the problem is to pass the variable to the function by reference using &:

function get_article(&$article_id) {
    if ( $article_id == null ) {
        // handle null case here
    }
    else {
        // get the article
    }
}

Answer 5

I'm thinking that the easiest way to solve it is like this:

$url_items = array("foo");
$article = empty($url_items[1]) ? get_article() : get_article(db_escape($url_items[1]));

function get_article($article_id = NULL) {.....}

This should works because you give $article_id a default value in the function. However, you could just as easily change the middle ternary part to null if you don't want to execute at all if there is no $article_id.

Edit: If you have an article_id 0, you may want to change empty to !isset
Edit 2: Modified to avoid the undefined offset warning.

Answer 6

so it was pretty simple after all that and yes maybe I could have phrased my question better, but it still bothers me that there is not a simpler way to do this

    if (isset($url_items[1])){
        $article_id = db_escape($url_items[1]);
    }else{
        $article_id = null;
    }
    $article = get_article($article_id);