PHP include pages error - php

Question

Still trying to get how PHP works :)
Kindly help me with one solution & idea, here is when i'm using:
<?php
$content = array(
'id01'=>'sub_id01.php',
'id02'=>'sub_id02.php'
);
if(in_array($_GET['show'], array_keys($content))) {
include($content[$_GET['show']]);
} else {
include('sub_id00.php');
}
?>
and:
<?php
$content = array(
'id00'=>'N/A',
'id01'=>'ID01',
'id02'=>'ID02',
);
if(!empty($_GET['show']) && isset($content[$_GET['show']])) {
echo $content[$_GET['show']];
} else {
echo $content['id00'];
}
?>
Where first example includes pages and second includes simple code inside ''
Problem is that if there is no ID set (index.php) ut shows default page/code. And if wrong ID set, it'll also show default page/code.
How to update it so without any ID set it would show default page and if some wrong ID set, it would show some error page/code?
Cheers!
UPDATE!
After a while reqding i've updated it with:
<?php
$content = array(
'id01'=>'sub_id01.php',
'id02'=>'sub_id02.php'
);
if (in_array($_GET['show'], array_keys($content)))
{
include($content[$_GET['show']]);
}
elseif (isset($_GET['show']))
{
include('sub_error.php');
}
else {
include('sub_id00.php');
}
?>
And:
<?php
$content = array(
'error'=>'error msg',
'id00'=>'N/A',
'id01'=>'ID01',
'id02'=>'ID02',
);
if(!empty($_GET['show']) && isset($content[$_GET['show']]))
{
echo $content[$_GET['show']];
}
elseif (isset($_GET['show']))
{
echo $content['error'];
}
else
{
echo $content['id00'];
}
?>
:)

Answer 1

$content = array(
'id01'=>'sub_id01.php',
'id02'=>'sub_id02.php'
);
if (isset($_GET['show']))
{
if (array_key_exists($_GET['show'], $content))
{
//$_GET id is set and it exists in content
include($content[$_GET['show']]);
}
else
{
//$_GET id is set but does not exist in content
//include whatever page you have for a wrong id here
}
}
else
{
//else no $_GET was set
//include default page
}