Function to get the path using a parent-child relationship

Question

Hello,
I have a table which uses parent-child relationship to store the options available. I need a function to give me the full path given the id of a particular option.
I have two different functions. One of them uses the Oracle built in function and the other uses simple queries with a loop.The code of the functions are given below.
Now, the problem is with their "performance". The difference in their performance is significant. The function using the Oracle function takes more than 2 hours to run a query whereas the other function takes less than 2 minutes.
I am having trouble trusting the other function. No matter how many tests I perform on the output of both the functions, it always comes out to be the same.
Any thoughts to help me understand this ??
Function 1
=====================
FUNCTION Gettree (opt_id IN NUMBER,i_app_id IN NUMBER)
RETURN VARCHAR2
IS
path VARCHAR2(32767);
application_no NUMBER;
BEGIN
SELECT ABC.APP_OPT_ID INTO application_no FROM ABC
WHERE ABC.APP_ID = i_app_id AND ABC.PARENT_ID IS NULL;
SELECT LPAD(' ', 2*LEVEL-1)||SYS_CONNECT_BY_PATH(app_opt_name, '=>') "Path" INTO path
FROM ABC
WHERE app_opt_id = opt_id
START WITH parent_id =application_no
CONNECT BY PRIOR app_opt_id =parent_id;
path := SUBSTR(path,INSTR(path,'>')+1,LENGTH(path));
RETURN path;
END Gettree ;
Function 2
======================
FUNCTION GetOptPath(opt_id NUMBER,app_id NUMBER)
RETURN VARCHAR2
IS
string VARCHAR2(900);
opt VARCHAR2(100);
pid NUMBER(38);
BEGIN
SELECT ABC.parent_id,ABC.app_opt_name INTO pid,string FROM ABC WHERE ABC.app_opt_id = opt_id;
IF pid IS NULL
THEN
RETURN 'root';
ELSIF pid IS NOT NULL
THEN
LOOP
SELECT ABC.app_opt_name,ABC.parent_id INTO opt, pid FROM ABC WHERE ABC.app_opt_id = pid;
EXIT WHEN pid IS NULL;
string := opt || '=>'|| string;
END LOOP;
RETURN string;
END IF;
END;

Answer 1

Hi,
user8653480 wrote:
Hello Frank,
The parameters taken by gettree & getoptpath are app_opt_id and app_id and both the functions return only one string i.e. the path of that particular option (app_opt_id) starting from its root and not all the descendants of that option/root node of that option.
So, does that mean that gettree first fetches all the descendants of the root node of the given option and then returns the required one ??Yes. It's a little like the situation where you need to meet with your co-worker Amy, so you send an e-mail to everyone in the department telling them to come to your office, and then, when they arrive, tell everyone except Amy that they can leave.
>
And if that is the case, then won't it be better to use the bottom-up approach to fetch the required path just for that particular option ?? 'coz my requirement is that only.. given an option_id get the full path starting from the root.Exactly!
I have used explain plan also for both the functions.. but since I did not know how to anlayze the output from plan_table so I just compared the value in the fields and they were exactly the same for both the queries.If you'd like help analyzing the plans, post them, as Centinul said.
I am attaching a sample data with the outputs of both the functions for the reference.
(tried attaching the file but could not find the option, so pasting the data here)
App_opt_ID     App_ID     Parent_ID     App_opt_name     "gettree(app_opt_id,app_id)"     "getoptpath(app_opt_id,app_id)"
1          1     NULL          application          NULL                    root
2          1     1          module1               module1                    module1
3          1     1          module2               module2                    module2
4          1     2          submod1               module1=>submod1          module1=>submod1
5          1     3          submod1               module2=>submod1          module2=>submod1
6          1     5          opt1               module2=>submod1=>opt1          module2=>submod1=>opt1
7          2     NULL          app2               NULL                    root
8          2     7          scr1               scr1                    scr1
9          2     8          opt1               scr1=>opt1               scr1=>opt1
10          2     7          scr2               scr2                    scr2Please help.The best solution is to do a bottom-up query, and write a function like reverse_path (described in my first message) to manipulate the string returned by SYS_CONNECT_BY_PATH. You seem to have all the PL/SQL skills needed for this.
Another approach is a revised form of gettree, like I posted earlier. Does it do what you want or not?
If you'd help, then post a little sample data in a form people can actually use, such as CREATE TABLE and INSERT statements. Post a few sets of parameters, and the results you need from each set, given the sample data posted.