Hi,
I had a look at "create type" docs and it seems somewhat complex, involving
creation of functions and etc. I hope there's a simpler way for the following:
How should one declare a new custom type, say, "AddressType" that corresponds
internally to "varchar(50)". In other words, all columns tha
--- ow <[EMAIL PROTECTED]> wrote:
> Hi,
[...]
> How should one declare a new custom type, say, "AddressType" that corresponds
> internally to "varchar(50)". In other words, all columns that are assigned
> "AddressType" would internally be "varchar(50)".
>
> Example:
> create type AddressType ... -
> I guess, ideally it'd be
> create type AddressType AS varchar(50) ;
> but it does not work.
Only one keyword off. SQL calls this a domain.
They're limited in 7.3, but much improved (not yet perfect) for 7.4.
http://www.postgresql.org/docs/7.3/interactive/sql-createdomain.html
signature.asc
[EMAIL PROTECTED] (ow) writes:
> I had a look at "create type" docs and it seems somewhat complex, involving
> creation of functions and etc. I hope there's a simpler way for the following:
>
> How should one declare a new custom type, say, "AddressType" that corresponds
> internally to "varchar(50
--- Rod Taylor <[EMAIL PROTECTED]> wrote:
> Only one keyword off. SQL calls this a domain.
>
> They're limited in 7.3, but much improved (not yet perfect) for 7.4.
>
> http://www.postgresql.org/docs/7.3/interactive/sql-createdomain.html
Excellent news! Thanks
__
On Tuesday, September 16, 2003, at 05:27 PM, Christopher Browne wrote:
What you want instead is CREATE DOMAIN.
flexreg=# create domain addresstype varchar(50);
The problem here is that you can't tell the difference between a
addresstype column and a varchar(50) column in the row description
i
