Re: SQL challenge to any gurus out there

From: Laly <lalystar_at_free.fr>
Date: Sat, 19 Jul 2003 20:49:32 +0200
Message-ID: <3f1992b8$0$2214$626a54ce@news.free.fr>

Hi

I got the query for getting the inheritance of one element :

if you want to get it for D :

select

   'D', parent.parent, glen_test.property from

   (

      select distinct parent
      from   glen_test
      where  parent is not null
      start with type='D'
      connect by prior parent=type

   ) parent,
   glen_test
where

   parent.parent = glen_test.type
union
select

   type, null, property
from

   glen_test
where

   type = 'D'

But I cannot get a query that will bring all the inheritance for all the elements.

HTH, Laly.

"GlenT" <glen.turner_at_bt.com> a écrit dans le message de news: bf8cpn$5sh$1_at_pheidippides.axion.bt.co.uk...

> Hi,
> I am trying to create a view which shows inheritence but cannot get this
> right. I have provided the table and data to assist anyone kind enough to
> help us plus details of what we are trying to achieve.
> CREATE TABLE GLEN_TEST (
> TYPE VARCHAR2 (10),
> PARENT VARCHAR2 (10),
> PROPERTY NUMBER);
>
> INSERT INTO GLEN_TEST ( TYPE, PARENT, PROPERTY ) VALUES ( 'A', NULL, 1);
> INSERT INTO GLEN_TEST ( TYPE, PARENT, PROPERTY ) VALUES ( 'A', NULL, 2);
> INSERT INTO GLEN_TEST ( TYPE, PARENT, PROPERTY ) VALUES ( 'B', 'A', 3);
> INSERT INTO GLEN_TEST ( TYPE, PARENT, PROPERTY ) VALUES ( 'B', 'A', 4);
> INSERT INTO GLEN_TEST ( TYPE, PARENT, PROPERTY ) VALUES ( 'C', 'B', 5);
> INSERT INTO GLEN_TEST ( TYPE, PARENT, PROPERTY ) VALUES ( 'D', 'B', 6);
> COMMIT;
>
> We can see that this is a simple child-parent relationship.
>
> If we do a select * we would get the following result:
> TYPE PARENT PROPERTY
> ------- ------ ----------
> A null 1
> A null 2
> B A 3
> B A 4
> C B 5
> D B 6
>
> What we are trying to achieve would look like this:
> TYPE PARENT PROPERTY
> ------ ------ ----------
> A null 1
> A null 2
> B A 1
> B A 2
> B null 3
> B null 4
> C A 1
> C A 2
> C B 3
> C B 4
> C null 5
> D A 1
> D A 2
> D B 3
> D B 4
> D null 6
>
> What we are trying to achieve is for every type to show all it's
inheritance
> i.e. type B inherits from type A because A is it's parent so the
properties
> for type B are 3 and 4 plus the properties for type A which are 1 and 2.
> Type C would have it's own properties (5) plus those from type B (3,4)
which
> is it's parent plus those of type A (1,2) which is B's parent.
>
> I have tried using the connect by prior but with no success as the
> inheritence could be n layers deep. If anyone out there likes a challenge
> and can help - we would be extremely grateful.
> Regards
> Glen
>
>
>
Received on Sat Jul 19 2003 - 13:49:32 CDT