Re: How to create computed field/column in a table

From: Jesper S. Knudsen <jesper.stougaard.knudsen_at_omgroup.com>
Date: Tue, 20 Nov 2001 13:42:05 +0100
Message-ID: <u0qK7.214$%X5.3304821778@news.euroconnect.net>

Hi Richard

Thank you for your answer. But it has to be a computed field/column. Do you know how?

Gr.

Jesper

"Richard B" <someguyonthenet_at_hotmail.com> wrote in message news:3bfa4f8d_at_peer1.news.newnet.co.uk...
>
> "Jesper S. Knudsen" <jesper.stougaard.knudsen_at_omgroup.com> wrote in
message
> news:WxmK7.207$w04.3163937823_at_news.euroconnect.net...
> > Hi
> >
> > Does anyone know how create a computed column in a table?
> >
> > In MS SQL Server you do it like this:
> >
> > CREATE TABLE mytable (
> > key numeric not null primary key,
> > value1 numeric not null,
> > value2 numeric not null,
> > value3 as (value1*value2)
> > )
> >
> > How is it done in Oracle?
> >
> > Gr.
> >
> > Jesper
> >
> >
>
> The simplest way is to create a view against the table.
> Eg.
>
> CREATE TABLE mytable (
> value1 number not null,
> value2 number not null);
>
> CREATE VIEW v_mytable as
> SELECT value1, value2, (value1 * value2) as value3
> FROM mytable;
>
>
> The drawback with this method is that you can't update the value3 colum
> directly. But presumably Transact-SQL wouldn't let you do this either.
>
>
> -Richard
>

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> ====
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> "Propose to an Englishman any principle, or instrument, however admirable,
> and you will observe that the
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> -- Charles Babbage
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Received on Tue Nov 20 2001 - 06:42:05 CST