Re: Object-Relational table design question
От | Josh Berkus |
---|---|
Тема | Re: Object-Relational table design question |
Дата | |
Msg-id | 200306170923.34974.josh@agliodbs.com обсуждение исходный текст |
Ответ на | Object-Relational table design question (Michael A Nachbaur <mike@nachbaur.com>) |
Ответы |
Re: Object-Relational table design question
|
Список | pgsql-sql |
Michael, > (BTW, if this isn't the correct forum to post this in, please let me know.) This is the right forum. > I thought of defining the different services in their tables, all inherited > from the base "Service" table, and then insert rows for the different > services of each (for instance "Basic Webhosting", "Advanced Webhosting", > etc). I'm uncertain how much mileage I'll get with this approach however. > > When querying for all services a customer is subscribed to, would I be able > to have it return -- on a row-by-row basis -- the separate columns of each > sub-table even if that row isn't available for another record? (sort of > like a left outer join would be; or would I be better off just doing a > plain-ol' left outer join across all my inherited service tables?) Personally, I'm not a fan of inherited tables; I think they muddy up the relationality of SQL without providing any additional functionality. However, my opinion is (obviously) not shared by everyone. Were I building your database, I would instead do it "relationally" rather than using inheritance (hopefully the ASCII art will transmit ...) Customer (name, id, phone, etc.) | one | V many Service (type, price, payment method, sign-up date, etc.) | | | V V V Webhosting Dial-up DSL Details Details Details | V Advanced Details This sort of partitioning of data is what the relational model is *for*; using the above, you should be able to access as much, or as little, of each customer's service details as you wish with left outer joins or by sub-selecting on service type. For example, to get webhosting details: SELECT ... FROM customer JOIN service ON (customer.id = service.customer AND service.type = 'web')JOIN webhosting ON service.id= webhosting.serviceLEFT OUTER JOIN web_advanced ON webhosting.id = web_advanced.webhosting Which would give you all customer, service, and basic hosting details, plus advanced hosting details of there are any. -- Josh Berkus Aglio Database Solutions San Francisco
В списке pgsql-sql по дате отправления: