[SOLVED] farfetched variable question

[envexlabs]

alright, i have my function:

 

function select_table($table_id, $table_name){
$store_id = mysql_escape_string($store_id);
$store_info_query = mysql_query('SELECT * FROM `store` WHERE ID = ' . $store_id . '');
$store_info = mysql_fetch_row($store_info_query);
return $store_info;
}

 

i want to be able to use this for selecting any table, but i will be selecting multiple tables on one page. this means that there would be multiple $store_info arrays, thus making the page not work.

 

is there a way to use the $table_name variable to create a dynamic $($table_name)_info variable?

 

i hope this isn't just some random ramblings!

[envexlabs]

i figured it out!

 

${ $table_name }