Using CONNECT BY with DBA_CONSTRAINTS VIEW

Question

I am building a query using the Dba_Constraints view and CONNECT BY Operator.
I should fill some tables and I want to avoid FKs errors, but... I'll need to follow the right order to do that. So, I wrote a query but it's not working!. The first version stopped the recursion on the second level... I tried to fix it but I've only got an even worst result!
This is the 1st version:
SELECT SYS_CONNECT_BY_PATH(C.TABLE_name, ' >> ') "Path"
FROM dba_constraints c
WHERE owner = :anOwner
AND LEVEL > 1 CONNECT BY constraint_name = prior r_constraint_name
AND PRIOR owner = prior r_owner
ORDER BY LEVEL DESC,
TABLE_NAME ASC;
Can anyone help me with this query?
Thanks!
Martin

Answer 1

Hi, Martin,
Whenever you post a question, it's helpful to include
(1) some sample data (just enough to illustrate the problem),
(2) the results you expect to get from that sample data, and
(3) an explanation of what you want.
For this problem, the only relevant columns are owner, table_name, constraint_name and r_constraint_name. If the problem is that the hierarchy never goes beyond LEVEL=2, then you only need to include enough sample data to get to LEVEL=3. So if you have table_a which refers to table_b, and table_b refers to table_c, you can illustrate your problem with just the data connected to those tables.
So you should run a query, something like this:
SELECT     table_name
,     constraint_name
,     r_constraint_name
FROM     dba_constraints
WHERE     owner     = :anOwner
AND     table_name IN ('TABLE_A', 'TABLE_B', 'TABLE_C');and then post the results, which might look like this:
TABLE_NAME           CONSTRAINT_NAME      R_CONSTRAINT_NAME
TABLE_A              A_B_FK               B_PK
TABLE_A              A_PK
TABLE_B              B_C_FK               C_PK
TABLE_B              B_PK
TABLE_C              C_PKNot only does this help others solve the problem, it can help you solve the problem. If you had done this, you might have noticed that the constraints do form a chain, but that chain is never more than two links long, because the child of a foreign key constraint in table_a is the unique constraint of table_b, and unique constraints never have any children.
You probably want something like this query, which uses the parent-child relationship between tables (gotten from the self-join) rather than constraints:
WITH     constraint_tree     AS
     SELECT DISTINCT
          a.table_name     AS table_name
     ,     b.table_name     AS parent_table_name
     FROM          dba_constraints a
     LEFT OUTER JOIN     dba_constraints b
     ON          a.r_constraint_name     = b.constraint_name
     AND          a.owner               = b.owner
     WHERE          a.owner     = :anOwner
SELECT     SYS_CONNECT_BY_PATH (table_name, ' >> ')     AS path
FROM     constraint_tree
WHERE     LEVEL               > 1
AND     CONNECT_BY_ISLEAF     = 1
AND     table_name          IN ('TABLE_A', 'TABLE_B', 'TABLE_C')
START WITH          parent_table_name     IS NULL
CONNECT BY NOCYCLE     parent_table_name     = PRIOR table_name
ORDER BY path
;which produces:
PATH
>> TABLE_B >> TABLE_A
>> TABLE_C >> TABLE_B >> TABLE_A