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

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.