undefined variable in a function return

Question

Here is my function :

function check1NumSeries($no_serie){

$reponse = $bdd->query('SELECT * FROM produit');
$check = "false";
while(($produit = $reponse->fetch())AND($check == "false")){
    if ($produit[1] == $no_serie){
        $check = "true";
        $id_error = $produit[1];
    }
} return array($check,$id_error);

};

It says when i do the return array($check,$id_error), that i have an "Notice: Undefined variable: id_error in C:\wamp\www\fonction.php on line 90"

line 90 = the return array();

I don't get it ... what is the problem ?

And my code below is executed correctly, it doens't block but i can see a big orange warning box for this error :/

and when i do $id_error = $produit[1], it doesn't create the variable ? — Booba__2012
– Booba__2012, Commented Jul 18, 2014 at 13:02

John Conde · Accepted Answer · 2014-07-18 13:02:34Z

1

$id_error is only defined in your if statement. If you do not enter that control structure it is never defined. You should declare a default value for it so it is always defined before you try to use it:

function check1NumSeries($no_serie){

    $reponse = $bdd->query('SELECT * FROM produit');
    $check = "false";
    $id_error = null;  // default value for that variable
    while(($produit = $reponse->fetch())AND($check == "false")){
        if ($produit[1] == $no_serie){
            $check = "true";
            $id_error = $produit[1];
        }
    } 
    return array($check,$id_error);
};

answered Jul 18, 2014 at 13:02

John Conde

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1 Comment

Booba__2012 Over a year ago

yes i know that :D the problem is, when i access to the $check value it shows me "true" and the only way to get that is to be in the "if" condition. So i am in the "if" and soo the $id_error must be declared ? Oo

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