Passing href through php variable

[mcfmullen]

I spent the entire day trying to figure this out and I'm about to give up. FYI, I use wordpress hence some wordpress calls. The bloginfo('url'); simply refers to my domain url (I.e. www.cool.com/).

 

URL path example:

http://www.cool.com/TEST/wp-content/themes/arjuna-x/itemspec2.php?item=Heart&type=Recovery

 

Here is the code I have in my index.php file:

<?php include("variables.php"); ?>
<a href="<?php bloginfo('url'); echo $url.$item; ?>&type=<?php echo $table; ?>" title="<?php _e('Permalink to', 'Arjuna'); ?><?php echo $item; ?>"><?php echo $item; ?></a>

and this is inside my variables.php file:

<?php
$table = $_GET['type'];
$item = $_GET['item'];
$url='/wp-content/themes/arjuna-x/itemspec2.php?item=';
?>

 

The two files work together just fine but for the sake of simplicity, what I want to do is create a single variable (ideally $url) which will contain the whole of the code given from index.php but everthing I try results in errors. In essence, I need someone to help me correct this code:

<?php
$url = <a href="<?php bloginfo('url'); ?>/wp-content/themes/arjuna-x/itemspec2.php?item=.$item; ?>&type=<?php echo $table; ?>" title="<?php _e('Permalink to', 'Arjuna'); ?><?php echo $item; ?>"><?php echo $item; ?></a>
?>

 

which will be stored in variables.php so that my code in index.php will show as follows:

<?php include("variables.php"); ?>
<?php $url; ?>

[mcfmullen]

While we're at it, I also have this code:

$sql = "SELECT * FROM (.$table LEFT JOIN animalThemes2 USING (Name)) LEFT JOIN animalMethods2 USING (Name) WHERE .$table.Name = '{$item}'";

 

where I want animalThemes2 and animalMethods2 to resemble something like .$tableThemes2 and .$tableMethods2 if at all possible. (I.E. $table contains animal)