Subject | Re: [firebird-support] Listing Table of Database in Order of Dependency |
---|---|
Author | LtColRDSChauhan |
Post date | 2016-03-29T12:30:16Z |
On Mon, Mar 28, 2016 at 11:21 PM, setysvar setysvar@... [firebird-support] <firebird-support@yahoogroups.com> wrote:>Hi,
Hi Rajiv!
>How can i list tables of a database in order of dependency ie
>
>TABLE_1 (no foreign key references)
>TABLE_2 (foreign key references, if any, in TABLE_1 )
>TABLE_3 (foreign key references, if any, in TABLE_1 / TABLE_2 )
>TABLE_4 (foreign key references, if any, in TABLE_1 / TABLE_2 / TABLE_3)
Ideally, this would be how you designed your database, i.e. have a
forced order of your tables and refuse foreign keys pointing to any
tables further down the list. I'm not certain whether or not things will
want to line up as nicely as you want them in an existing database.
Nevertheless, I suggest you create a new table:
CREATE TABLE TABLES_ORDERED
(
ID INTEGER NOT NULL,
RDB$RELATION_NAME CHAR( 31 ),
CONSTRAINT PK_TABLES_ORDERED PRIMARY KEY ( ID )
);
CREATE GENERATOR TABLES_ORDERED_GEN;
SET TERM ^^ ;
CREATE TRIGGER TABLES_ORDERED_ID FOR TABLES_ORDERED ACTIVE BEFORE INSERT
POSITION 0 AS
begin
if ( ( new.ID is null ) or ( new.ID = 0 ) ) then
new.ID = gen_id( TABLES_ORDERED_GEN, 1 );
end
^^
SET TERM ; ^^
commit;
fill it with those tables that have no foreign key (start by running
DELETE FROM TABLES_ORDERED if the table is not empty):
INSERT INTO TABLES_ORDERED(RDB$RELATION_NAME)
SELECT RDB$RELATION_NAME FROM RDB$RELATIONS r
WHERE RDB$SYSTEM_FLAG=0
AND NOT EXISTS(SELECT * FROM RDB$INDICES i
WHERE r.RDB$RELATION_NAME=i.RDB$RELATION_NAME
AND i.RDB$FOREIGN_KEY IS NOT NULL);
and then, repeatedly, try:
INSERT INTO TABLES_ORDERED(RDB$RELATION_NAME)
SELECT RDB$RELATION_NAME FROM RDB$RELATIONS r
WHERE RDB$SYSTEM_FLAG=0
AND NOT EXISTS(SELECT * FROM TABLES_ORDERED o /*Ignore tables already
inserted*/
WHERE r.RDB$RELATION_NAME = o.RDB$RELATION_NAME)
AND NOT EXISTS(SELECT * FROM RDB$INDICES i /*Only insert tables whose
foreign key tables are inserted already*/
JOIN RDB$INDICES i2 ON i.RDB$FOREIGN_KEY =
i2.RDB$INDEX_NAME
AND r.RDB$RELATION_NAME <>
i2.RDB$RELATION_NAME /*Omit this line if you don't want to include
tables pointing to themselves*/
LEFT JOIN TABLES_ORDERED o ON i2.RDB$RELATION_NAME =
o.RDB$RELATION_NAME
WHERE r.RDB$RELATION_NAME=i.RDB$RELATION_NAME
AND i.RDB$FOREIGN_KEY IS NOT NULL
AND o.ID IS NULL);
When you've done this enough times for no further table to be inserted,you can get your ordered list by simply running:
SELECT RDB$RELATION_NAME
FROM TABLES_ORDERED
ORDER BY ID
If you find that the list lacks two or more tables, investigate those
remaining tables - maybe you have some circular dependencies?All Tables got added to TABLES_ORDERED
Note that I've never tried doing anything similar myself (I don't even
think I've written a query that uses the RDB$FOREIGN_KEY before), and
that there may well be errors in what I've written above.
There were no errors, Set. Thanks a lot.This helped !Regards,RajivHTH,
Set