Start with Connect by: Showing the whole hierarchy even if child parameter

Question

11g
Hi There,
In our case the manager_id i= employee_id for the top level manager. So when I run the sql, it eliminates the top level manager and shows the output for the next level onwards.
For the regular start with connect by option on the employee table the query used is
select case when connect_by_isleaf = 1 then 0 when level = 1 then 1 else -1 end as status,
       level,
       First_name as title,
       NULL  as icon,
       EMPloyee_id as value,
       First_name as tooltip,
       NULL  as link
from EMPLOYEES
start with Manager_id is null
connect by prior EMPLOYEE_ID = Manager_id
order siblings by First_namenow this will show the hierarchy and level starting with managers. Now if we provide the "start with Manager_id = 171" then since 171 is not a manager no rows are returned. So for this we can use something like
start with manager_id =171 or employee_id = 171However, the output would be only the employee record since 171 is the last child.
The requirement we had was that, Irrespective of the value entered, whether ultimate parent or ultimate child the output should still show the top down hierarchy. starting with that persons ultimate manager.
So for example if there are two employees, 170 and 171 and 170 is the manager of 171
If in this query I use Start with manager = 170. It will show me the manager and the child records.
However, if I use start with manager = 171 or employee_id = 171 then it will only show me only the child record. I want it to show me both the manager and employee records. i.e all the levels
I hope that make sense!
Thanks,
Ryan
Edited by: ryansun on Dec 12, 2012 1:13 AM
Edited by: ryansun on Dec 13, 2012 1:59 AM

Answer 1

Hi Alberto,
I was using this query and the output is correct. Except for one thing, the nature of our data is such that the top most manager in this case has "Manager_Id" as null so we can use "start with Manager is null"
But in our case the manager id for the top most manager is the same as his employee id.
So If I modify the query, it then does not show the "top most managers record"
WITH entire_tree AS
    SELECT *
      FROM employees
     START WITH manager_id = 171
   CONNECT BY PRIOR employee_id = manager_id
   UNION
    SELECT *
      FROM employees
     START WITH employee_id = 171
   CONNECT BY employee_id = PRIOR manager_id
SELECT CASE WHEN CONNECT_BY_ISLEAF = 1 THEN 0 WHEN LEVEL = 1 THEN 1 ELSE -1 END AS status
      , LEVEL
      , first_name AS title
      , NULL AS icon
      , employee_id AS VALUE
      , first_name AS tooltip
      , NULL AS link
   FROM entire_tree
  START WITH manager_id = 100
CONNECT BY PRIOR employee_id = manager_id
  ORDER SIBLINGS BY first_name;the only change I made is from
Start with manager_id is null
to
start with manager_id = 100Basically, in this case the manager_id of the top most manager is the same as his employee id. So how can we have that condition incorporated instead of checking for null.
Thanks,
Ryan