Footer area

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Hello,
I am making a theme.
But in this theme i have three collums in the footer area. How can i make all the three footer area's editable. When i past this peace of code in each collum (code:
<?php $a = new Area('Footer'); $a->display($c); ?>
) i get the same text in each colum. Can enyone help me.
(sorry for my bad english i am form holland and 13 years old.)

 
madeforspace replied on at Permalink Best Answer Reply
madeforspace
Try changing the Area names to Footer1 Footer2 Footer3.
12345j replied on at Permalink Reply
12345j
yeah, the same area name calls the same block for that page, so if you change the names like made for space suggested it should work.
WantSomeCookies replied on at Permalink Reply
Thank you.
When i am home i gonna test it.
creatovisguru replied on at Permalink Reply
creatovisguru
I always give each new area object a new variable name as well.

<?php $fc1 = new Area('Footer Col 1'); ...?>
<?php $fc2 = new Area('Footer Col 2'); ...?>
<?php $fc3 = new Area('Footer Col 3'); ...?>

This helps me keep organized code easier to understand.
jordanlev replied on at Permalink Reply
jordanlev
Hey creatovisguru,
This is totally my personal opinion (so I'm not saying you're wrong), but to me I actually think the code is easier to understand when I use the same variable name for each area. Because to me, when I see different variable names I think that it means they will be used later on for some other purpose (otherwise why would they have different names?) -- so using the same variable name to me means "okay it's just a plain old concrete5 area, like everywhere else in my theme templates, so nothing I need to think about here".

I'm not trying to be argumentative -- I enjoy hearing different people's opinions on this kind of stuff (hope you do as well).

-Jordan
Mnkras replied on at Permalink Reply
Mnkras
Yea as jordan said, I do the same thing, eg:

<?php $a = new Area('Footer Col 1'); ...?>
<?php $a = new Area('Footer Col 2'); ...?>
<?php $a = new Area('Footer Col 3'); ...?>
WantSomeCookies replied on at Permalink Reply
It worked.
I used:
<?php
$a = new Area('Footer 1'); $a->display($c);
$a = new Area('Footer 2'); $a->display($c);
?>

All thank you for the reply's.
Greets from Holland.