[SOLVED] include warning - maybe namespace issues

[solariaman]

Hi all,

 

I've encountered a curious problem, the variable $langs is defined in an include file, and used by a function defined in the same include file?  But it doesnt seem to recognize the variable...

 

index.php :

    include("lang_utils.inc");
    $lang=detect_lang();
    ...

 

lang_utils.inc :

$langs =  array (
    "fr" => "français",
    "en" => "English",
    "es" => "espanol",
    "pt" => "portugese"
);

function detect_lang(){
    ...
    if (!array_key_exists($lang, $langs)){
        $lang='en';
    }
    ...
    return $lang;
}

 

Straight forward right?  So i was surprised when a user reported this warning visible on the site

 

Warning: array_key_exists() [function.array-key-exists]: The second argument should be either an array or an object in .../lang_utils.inc on line 42

 

line 42 is the "if (!array_key_exists($lang, $langs)){" copied above.  So according to the warning, $langs didnt get defined. 

 

It might be worth noting tnat not everybody gets this warning...  it was encountered at least once with firefox on ubuntu. 

 

Anybody see where it goes wrong?

 

thanks

 

 

 

 

[Mark Baker]

Where is $lang being set:

 

function detect_lang(){

    ...

    if (!array_key_exists($lang, $langs)){

        $lang='en';

    }

    ...

    return $lang;

}

 

You're not passing it into the function as a parameter, so unless something in ... sets it, then it's not defined within the scope of the function

[jonsjava]

*head just exploded*. Um....did you define a variable to equal a function that checks against that variable that calls that function that defines that variable that calls that function that checks that variable.............

[solariaman]

Where is $lang being set:

 

You're not passing it into the function as a parameter, so unless something in ... sets it, then it's not defined within the scope of the function

 

yep, the "..." sets it.  i probably should have left it in even if its a little sloppy.  Anyway here's the whole function:

function detect_lang(){
   /**
    * Detect and include language, 
    *   returns variable $lang
    */
    $lang='';
    
    // take care of lang set by the user overide              
    if(isset($_POST["lang"])) {
        $_SESSION["lang"] = $_POST["lang"]; 
    }
    
    // then see if the language has already been determined
    if(isset($_SESSION["lang"])) {
        $lang = $_SESSION["lang"];
    } else {
        // try to guess the lang
        $accept_language = getenv('HTTP_ACCEPT_LANGUAGE');
        if (isset($accept_language)) {
            $lang = substr($accept_language, 0, 2);
            // note that firefox gives second and third choice lang info...
        
            if (!array_key_exists($lang, $langs)){
                $lang='en';
            }
        } else {
            $lang = $lang_default;
        }
        $_SESSION["lang"] = $lang;
    }
    return $lang;
}

 

it should work i reckon, and does...  with a warning§

 

[solariaman]

*head just exploded*. Um....did you define a variable to equal a function that checks against that variable that calls that function that defines that variable that calls that function that checks that variable.............

 

whao, slow down my head's exploding too!  its not supposed to be that complicated.   

 

I just include a file that does two things, it set the array $langs and defines a function detect_lang().

 

The variable $langs should be visible inside detect_lang() but i get a warning when i try to use it. 

the snippet

 

if (!array_key_exists($lang, $langs)){

                $lang='en';

            }

seems to always get executed.  The warning says that $langs is empty,  so array_key_exists returns FALSE and since its NOTted, the parser enters the if...

 

 

[Daniel0]

Can't you just pass the variables by argument?

[Mark Baker]

The array $langs isn't in function scope, only in global scope.

 

Either include the line:

global $langs

at the beginning of the function code,

 

of pass the $langs array to the function as a parameter

[solariaman]

The array $langs isn't in function scope, only in global scope.

 

include the line:

global $langs

at the beginning of the function code,

 

okay, i figured $langs would be in the scope.    This means that in general :

$x=2;
function f() {
   $x+=1;
   return $x;
}

f should return 1 and x should be 2.  I tested it and yep! 

 

On the other hand with global

$x=2;
function f() {
   global $x;
   $x+=1;
   return $x;
}

and f returns 3, and x is 3 afterward.  perfect i get it now  Thanks!!

 

[Daniel0]

This means that in general :

$x=2;
function f() {
   $x+=1;
   return $x;
}

f should return 1 and x should be 2.  I tested it and yep!

 

That is an erroneous operation. You are not allowed to manipulate or otherwise use uninitialized variables. It'll result in an E_NOTICE type error.