Re: Relational Databases and Their Guts

From: Heinz Huber <XhhuberX_at_no-racon-linz.at-no>
Date: Fri, 27 Jun 2003 09:08:35 +0200
Message-ID: <3efbed74$0$34346$91cee783@newsreader02.highway.telekom.at>

Marshall Spight wrote:

> "Todd Bandrowsky" <anakin_at_unitedsoftworks.com> wrote in message news:af3d9224.0306221504.68914683_at_posting.google.com...
> 

>>Assume I have two tables R1 and R2. The number of columns in R1 is
>>irrelevant, except that there exists in R1 a primary key, R1PK, and
>>the rest of the columns in R1 we will call R1O. R2 has a minimum of 3
>>columns, the first of which is a foreign key into R1, the second is a
>>unique identifier to satisfy a one to many relationship between R1 and
>>R2, and the third we will call RCOL.
>>
>>Let [] be used to refer to the specific row for the primary key of a
>>given table Rn. Thus, for any given R1[ R1PK ], there are 0..N of
>>RCOL. We will later refer to the set of names for a specific PK as
>>RCOL( R1PK ).
>>
>>Let () be used to refer, by index, or by name, to a specific column on
>>a view. Thus, one might say R1( R1PK ) refers to the column name, and
>>R1( 0..N ) refers to columns 0 through N, and, finally, R1( 0...N -
>>R1PK ) refers to all of the columns except for the primary key of R1.
>>
>>1) Is there any way using the basic relational algebra operators, to
>>arrive at a final view RResult such that the columns of RResult are as
>>follows:
>>
>>R1( R1PK + RCOL( R1PK ) + R1O )?

> 
> 
> ISTM that all that is necessary to do this is an aggregate operator that
> will aggregate a column across a relation into a list (or perhaps set.) Then
> one could use the aggregate operator to group the various RCOL values
> for each R1PK into a list. The resulting relation would have a list of
> RCOL's type as the associated column type. The case of *no* values
> would produce a zero-length list.
> 
> I'm not aware of any existing DBMS that has such an aggregate operator,
> but now that I've thought about it, it seems like a necessity.

Sybase Adaptive Server Anywhere (Not Enterprise!!) has an operator LIST that does just that.

Regards,
Heinz Received on Fri Jun 27 2003 - 02:08:35 CDT