My First Function returns undefined variable ($pageName) error on line #14

Question

Newb here.

I'm trying to create my first php function which will get the name of the file, stripe out the '.php' extension and return a string which is the name of the page file without the '.php' file. This if i was using the index page, the title tag would read 'Index'.

To create the function, first i wrote the following process which did what I hoped:

$fileName2 = basename($_SERVER['PHP_SELF']);
$pageName2 = str_replace(".php","", $fileName2);

However when i try to put it into the context/format of a function i get an error stating:

NOTICE: UNDEFINED VARIABLE: ITC240.PHP ON LINE 14

Here is what i believe to be the relevant code:

inc_function.php:

function GetPageName($pageName) {
$fileName = basename($_SERVER['PHP_SELF']);
$pageName = str_replace('.php','', $fileName);
}

I call the function between the title tags:

<?php echo GetPageName(); ?>

(I tried to put this in code block but every time i put anything in the title tags it didn't print for reasons i don't understand).

Any help or instruction is very much appreciated

I don't know why, but part of my code isn't showing - trying to resolve this — Chezshire
I just edited your question because of your missing indents and you overwrote it. I'm not re-editing. — Funk Forty Niner
I've indented it several times - i can't seem to get it all to format - i am on lots of nyQuil though so i'm sure the fault here is all mine. Sorry - i had no way of know you were trying to edit it or were editing it. Sorry — Chezshire
Thank you ... Daryl Gill's comment totally confuses me and has nothing to do with this right? I'm a real newb (on nyquil no less) so i'm just a twit now i guess. Thank you I will try again. — Chezshire

Daryl Gill Daryl Gill · Accepted Answer · 2014-02-02T00:11:53

Restructured Answer

You are being shown an error because your function is looking for data & using the information from your comment I don't know why but stack overflow kills anything i plop inside title tags: This is what is in the title tags: <?php echo GetPageName(); ?>

So using my best judgement from information provided, I'm assuming you want to have a function to automatically return something with no data specification. In this case, you would need to take note of 2 things

The return block in your function will essentially, do as the function says. Return something so it can be used
No requirements in your function definition braces

function TestFunc($RequestedData){ }

the above function definition is asking the developer for information to be provided to the function for data manipulation.

function TestFunc(){  }

The above function definition is asking for nothing from the developer apart from calling, doing this will enable you to perform exactly what you want.

So, to fix your function:

function GetPageName() {
    $fileName = basename($_SERVER['PHP_SELF']);
    $pageName = str_replace('.php','', $fileName);
    return $pageName;
}

var_dumps to show the power of return

Using the exact code above on my local development machine Web Addr of (http://127.0.0.1) returns:

string(5) "index"

Now, using the exact function as specified above, just without the return:

NULL

My First Function returns undefined variable ($pageName) error on line #14

2 Answers