Double Dollar Sign - Assign Array key=>value - php

Question

I would like to create 3 different arrays which can change depending on which fields are pulled from some mysql tables.
This is the code I have:
// Gather db rows with query
$compQuery['subCats'] = mysql_query("SELECT id,name,category FROM subcategories ORDER BY id DESC");
$compQuery['cats'] = mysql_query("SELECT id,name FROM categories ORDER BY id DESC");
$compQuery['mans'] = mysql_query("SELECT id,name,link FROM manufacturers ORDER BY id DESC");
// Place rows into arrays
foreach($compQuery as $name=>$query){
$name = array();
while($data = mysql_fetch_array($compQuery[$name])){
foreach($data as $key=>$value){
$$name[$key] = $value;
}
}
}
What I would expect this to produce is 3 arrays: subCats, cats and mans each with the value pairs. Instead of finding the $$name variable and assigning [$key] = $value to the array it is taking a single letter out of the $$name variable and assigning the $value to this new variable.
For example, if $name = 'subCats', $key = '3', and value = 'machine tools'
It would create the variable $C using the 3rd letter in subCats:
$C = 'machine tools'
What it should do is create:
$subCats[3] = 'machine tools';
Now, obviously there are easier ways to create separate arrays for each query, and I have modified my code to a lazier/easier version. However, I am still curious as to how I would go about assigning a key value pair to a double dollar sign variable.
Any help would be greatly appreciated! I have the simple code working now, but would like to implement this new code to be more dynamic and allow for more queries as needed without hassle.
Thanks!
Josh

Answer 1

${$name}[$key]
But I do not like to create 'unknown' variable, as in variables whose name is not in the code... What if the name is 'key','db','_GET','compQuery' etc? All hell breaks loose...

Answer 2

In this line:
$name = array();
You're overwriting $name. You probably meant to write to $$name:
$$name = array();