PHP if not statements

You're saying "if it's not set or it's different from add or it's different from delete". You realize that a != x && a != y, with x != y is necessarily false since a cannot be simultaneously two different values.

Your logic is slightly off. The second || should be &&:

if ((!isset($action)) || ($action != "add" && $action != "delete"))

You can see why your original line fails by trying out a sample value. Let's say $action is "delete". Here's how the condition reduces down step by step:

// $action == "delete"
if ((!isset($action)) || ($action != "add" || $action != "delete"))
if ((!true) || ($action != "add" || $action != "delete"))
if (false || ($action != "add" || $action != "delete"))
if ($action != "add" || $action != "delete")
if (true || $action != "delete")
if (true || false)
if (true)

Oops! The condition just succeeded and printed "error", but it was supposed to fail. In fact, if you think about it, no matter what the value of $action is, one of the two != tests will return true. Switch the || to && and then the second to last line becomes if (true && false), which properly reduces to if (false).

There is a way to use || and have the test work, by the way. You have to negate everything else using De Morgan's law, i.e.:

if ((!isset($action)) || !($action == "add" || $action == "delete"))

You can read that in English as "if action is not (either add or remove), then".

No matter what $action is, it will always either not be "add" OR not be "delete", which is why the if condition always passes. What you want is to use && instead of ||:

(!isset($action)) || ($action !="add" && $action !="delete"))