Hierarchical difficult query.

Question

Hi guys,
I am trying to solve this issue in one select and still I did not manage to find the sollution.
I have a table with the following configuration:
DROP TABLE USERS_TAB;
create table users_tab (user_id number PRIMARY KEY, group_id number, USERTYPE VARCHAR2 (2));
--THESE USERS ARE IN THE SAME GROUP id:80 and have different types
INSERT INTO users_tab VALUES (1,80,'V');
INSERT INTO users_tab VALUES (2,80,'V');
INSERT INTO users_tab VALUES (3,80,'E');
--These are in another group (40) and have different types
INSERT INTO users_tab VALUES (4,40,'E');
INSERT INTO users_tab VALUES (5,40,'V');
INSERT INTO users_tab VALUES (6,40,'E');
insert into users_tab values (7,null, 'V');
commit;
create table links (v_type_users number, e_type_users number);In links table we define the links between users. A link appears only one time in this table (the links are one to one). the v type users appear in v_type_user column and e type users appear in e_type_user
**I have a procedure that has for input parameter one user_id. I need to selectal users that have the same group_id with the input and that are lnked togheter. if a contract belongs to a group, all memenbers of the group need to be included and the links between groups are made in links table (you will understand from example). **
For example:
1.
truncate table links;
insert into links values (1,4);user 1 of v type is linked with user 4 of e type.
if my select has for input user the id 1,
it will need to return all the users that are in the group of user 1. and for these users all the links with their
groups and so on.
1 is in group 80...so I will need to return 1,2,3 users. but 1 is linked with 4 that is in group 40 so I will also return 4,5,6 because they are in the same group
so in this link I will need to return 1,2,3,4,5,6. they are representing a bigger group
2.
truncate table links;
insert into links values (1,4);
insert into links values (5,7);--we created a link between 5(type v and 7 type e)if we input user 1, we will need to show all memember from group 80, and because of the link (1,4) we will need to show
4 and because 4 is in the group with 5 and 6 we will need to show them also and because 5 is in link with 7 we will need to show 7 also
the group will be: 1,2,3,4,5,6,7.
**you can see that the bigger group will be the same no matter what the input is. if the input is for example 7 or 5 or (4,3,2,1) we will output*
*1,2,3,4,5,6,7. because 7 is in link with 5 and 5 is in a group and in that group there is another link to another group**
3.
We can have multimple links between groups (I can see here a problem with hierarchical or maybe nocycle and distinct helps?)
truncate table links;
insert into links values (1,4);
insert into links values (5,7);--we created a link between 5(type v) and 7 type e
insert into links values (2,6);--we created a link between 2(type v) and 6 type ethis will not change the output, although we have multimple links.
we will need to output:
1
2
3
4
5
6
7
4.
truncate table links;
insert into links values (3,7);-- we have one link between 3(type v) and 7 (type e)if we have the input user 3. we will need to show all memenber from the group of user 3 and because of the link we will show the user 7.
we will have
1
2
3
7
5.
truncate table links;
if we have no link...we will need to show all the users from a group(if the input is a user from a group) and only that user if the input has no group.
I think this is a difficult one, any help is appreaciated.
Thanks

Answer 1

Hi,
Interesting problem!
The CONNECT BY should really be done by groups; it doesn't matter that userid 1 is linked to userid 4; all that matters is that group 80 is linked to group 40.
But if the CONNECT BY is done accrding to groups, what about user_id=7, who is not in a group? For purposes of this query, all rows in users_tab that have NULL group_id will be assigned a unique group_id.
Sub-query erverybody_in_group below makes up a group_id for rows that do not already have one. It assumes that userid > 0.
The CONNECT BY will go a lot easier if, for every row like (1, 4) that is in links, we also have the mirror-image (4, 1). The first two branches of the UNION in sub-query group_link do that.
The last branch of the UNION makes sure all groups are represented in group_link, even if they are not linked to anything. That ensures that all members of the source group get included in hte results, even if they are not connected to any other group. (This incluudes the special case where there are no rows at all in linkls.)
WITH     everybody_in_group     AS
     SELECT     user_id
     ,     COALESCE ( group_id
                , user_id + ( SELECT  MAX (group_id)
                              FROM    users_tab
                ) AS group_id
     FROM    users_tab
,     group_link          AS
     SELECT     v1.group_id     AS from_group_id
     ,     e1.group_id     AS to_group_id
     FROM     links               l1
     JOIN     everybody_in_group     e1     ON     l1.e_type_users     = e1.user_id
     JOIN     everybody_in_group     v1     ON     l1.v_type_users     = v1.user_id
     UNION
     SELECT     e2.group_id     AS from_group_id
     ,     v2.group_id     AS to_group_id
     FROM     links               l2
     JOIN     everybody_in_group     e2     ON     l2.e_type_users     = e2.user_id
     JOIN     everybody_in_group     v2     ON     l2.v_type_users     = v2.user_id
     UNION
     SELECT  group_id     AS from_group_id
     ,     NULL          AS to_group_id
     FROM     everybody_in_group
SELECT     user_id
FROM     everybody_in_group
WHERE     group_id     IN ( SELECT  from_group_id
                    FROM    group_link
                    START WITH      from_group_id = ( SELECT  group_id
                                                      FROM        everybody_in_group
                                      WHERE   user_id     = 1     -- Or use a bind variable
                    CONNECT BY NOCYCLE      from_group_id = PRIOR to_group_id
ORDER BY  user_id
;In the example above, the starting user_id (1) is hard-coded, about 6 lines from the end. This is the only place where that parameter is used.
The column usertype plays no role in this problem; is that right?
Thanks for posting the CREATE TABLE and INSERT statements! That really helps.