PHP Function Call Returns Fatal Error

Question

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>

Unless something is turning an E_NOTICE into an Exception, what you've describe isn't sufficient to generate a fatal error. Please post more information about the error you're experiencing. — cmbuckley
– cmbuckley, Commented Mar 2, 2012 at 0:42

animuson · Accepted Answer · 2012-03-02 00:37:59Z

4

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">

answered Mar 2, 2012 at 0:37

animuson♦

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3 Comments

Nilpo Over a year ago

Agreed...except for defining $r as global. ;-)

animuson Over a year ago

@Nilpo: I don't like using global for a simple variable, which is why I more mentioned it as a side-note afterwards. I used my ninja-powers to utilize my grace period and make it smaller. ;)

Nilpo Over a year ago

Right on. I definitely agree with you. Using global variables is almost always a bad practice. Variables should be explicitly passed between scopes. Less mistakes, neater code that easier to follow, etc. I won't say definitively say they don't have their place, but I can only think of a single time I would condone using globals--when defining globally accessible constants. But even then, I personally resort to other means.

Oliver Charlesworth · Accepted Answer · 2012-03-02 00:37:14Z

1

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);

answered Mar 2, 2012 at 0:37

Oliver Charlesworth

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1 Comment

Brandon Lebedev Over a year ago

bad style still gets the job done and keeps the code short. props

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