php mysql query query :D

This question has two parts. An answer to either part would be great :D

1) I am using one table to store links between values of another table. I am then searching through that table to find inheritance lists.

while($cont) {
        $rst = DB::query("SELECT parentID FROM code_inheritance WHERE childID=".$sid);
        if($rst->num_rows != 0) {
            $ob = $rst->fetch_object();
            if($list != '') $list .= ',';
            $list .= $ob->parentID;
            $sid = $ob->parentID;
        } else $cont = false;
    }

DB is my database class. However, this isn't a very efficient way of searching a database. Can anyone point me in a direction where I can do this in one database query please?

2) I then use the list to search for the names of the classes so that I can echo an inheritance list to the screen, however they are returned in ID numerical order I want them to be in the order they were found in part 1. I am currently doing this:

if($list != '') {
        $rst = DB::query("SELECT Name FROM code WHERE ID IN ($list)");
        $list = '';
        while($ob = $rst->fetch开发者_开发技巧_object()) {
            echo $ob->Name;
            if($list != '') $list .= ' -> ';
            $list .= '<a href="/code/'.$ob->Name.'/">'.$ob->Name.'</a>';
        }
        $list = "Inheritance -> ".$list;
    }

How can I get then list in the order I found them in part one so that the inheritance tree is in the correct order?

Sorry that this was a bit lengthy but I wanted to add as much information as I could.

A Sub-Query could be helpful so aviod the first part:

SELECT Name FROM code WHERE ID IN 
  (SELECT parentID FROM code_inheritance WHERE childID='$sid')

This selects all your parentID in the inner query and then use this to find the code.name. Using IN might not return the correct order as it is just a set which has no particular order. So you should add some "orderable" column.

To get all IDs within one query as list you can use GROUP_CONCAT, more information at MySQL manual about group by.