CODEWITHSUNDEEP
php
Brandon Lebedev
Brandon Lebedev

Reputation: 2797

PHP Function Call Returns Fatal Error

For some reason my PHP function call <? r(); ?> returns a fatal error. Any help?

<?php
//$r is previously assigned
function r() {
    echo ( $r );
};
?>

<html>
    <head>
        <link type="text/css" rel="stylesheet" href="<? r(); ?>rs/css/master.css">
    </head>

Upvotes: 0

Views: 88

Answers (2)

animuson
animuson

Reputation: 54729

Defining a function changes the scope, where $r won't be inside that function. Try sending in $r with the function in this manner:

<?php
function r($r) {
    echo ( $r );
}
?>
<link type="text/css" rel="stylesheet" href="<?php r($r); ?>rs/css/master.css">

or defining global $r; at the beginning of the function (not preferred).

Also, you shouldn't use <? to open PHP. If all the function does is echo the value of $r, it would make more sense to just do this:

<link type="text/css" rel="stylesheet" href="<?php echo ( $r ); ?>rs/css/master.css">

Upvotes: 4

Oliver Charlesworth
Oliver Charlesworth

Reputation: 272487

If you want to refer to a global object from inside an object, you need to explicitly declare it as global:

function r() {
    global $r;
    echo ( $r );
}

But in general, referring to global variables in this way is bad style. You should consider passing in prerequisites as function arguments:

function r($r) {
     echo ( $r );
}

...

r($r);

Upvotes: 1

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