double output for variable

Question

I have this next function:

function checkLoggedIn($status, $redirect=TRUE){ switch($status){ case "yes": if(!isset($_SESSION["loggedIn"])){ if($redirect) { header("Location: login.php"); exit; } else { $authenticated = false; } } else { checkLoggedIn("no"); } break; case "no": if(isset($_SESSION["loggedIn"]) && $_SESSION["loggedIn"] === true ){ $authenticated = true; } break; } var_dump($authenticated); return $authenticated; }

The strange this is that when I enable var_dump($authenticated);, I get as output if true:

bool(true) NULL

and just

bool(false)

if false

Any ideas why?

Samy Dindane · Accepted Answer · 2012-05-27 20:26:38Z

3

That's because you are calling checkLoggedIn() inside itself.

answered May 27, 2012 at 20:26

Samy Dindane

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3 Comments

Alex Over a year ago

oh, I see.. then, any ideas how to avoid the double response?

Samy Dindane Over a year ago

I'd say put the case "no" code directly in the else statement instead of calling the function again.

Alex Over a year ago

bufff, almost impossible because of the way that it was implemented (not by me). I'll find a solution, eventually! Thanks

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