PHP if not statements | 易学教程

问题

This may be the way my server is set up, but I'm banging my head against the wall. I'm trying to say that if $action has no value or has a value that is not "add" or "delete" then have an error, else keep running the script. However, I get an error no matter what $action is.

 $action = $_GET['a'];
 if((!isset($action)) || ($action != "add" || $action != "delete")){
     //header("location:index.php");
     echo "error <br>";
 }

$action is being set properly and if run something like if($action =="add") it works. This is on my local host, so it could be a settings issue.

回答1:

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".

回答2:

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"))

回答3:

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.

回答4:

You could also try:

if ((!isset($action)) || !($action == "add" || $action == "delete")) {
  // Do your stuff
}

回答5:

For future reference, you can quickly create a truth table to check if it evaluates the way you want... it's kind of like Sudoku.

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

Example:

column 1 is issetaction, column 2 and 3 evaluates !="add","delete" respectively

if($a=add) T && (F && T) => T && F => FALSE

if($a=delete) T && (T && F) => T && F => FALSE

if($a=nothing) T && (T && T) => T && T => TRUE

回答6:

I think this is the best and easiest way to do it:

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

回答7:

Not an answer, but just for the sake of code formatting

if((isset($_GET['a'])) $action=$_GET['a']; else $action ="";
if(!($action === "add" OR $action === "delete")){
  header("location: /index.php");
  exit;
}

Note the exit; statement after header(). That's the important thing. header() does not terminate script execution.

来源：https://stackoverflow.com/questions/3376945/php-if-not-statements

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