Death to recursion! Re: Child vs parent records

[Ed Kleban]

On 12/11/05 9:31 AM, "Ruslan Zasukhin" <sunshine at public.kherson.ua> wrote:

> On 12/11/05 5:19 PM, "Ed Kleban" <Ed at Kleban.com> wrote:
> 
>> 
>> Unlike CreateForeignKeyLink, the CreateBinaryLink command can create 1:1,
>> 1:M, M:1, and M:M relations ships -- as you have frequently pointed out to
>> me.  In this more generalized definition the concepts of "Parent" and
>> "Child" are less clear, may be irrelevant, and I would argue only offer an
>> opportunity for more confusion.
> 
> IF BinaryLink is type 1 : M THEN we can use Parent-Child terminology.
> 
> No difference to FK.
> 
> 
> But for M : M this not works! Right.
> 
> 
> Also note, that 1:1 (binary) link and M : M link on default do not know
> which table is owner.
> 
>     * Usually in 99% any of them is NOT owner.
> 
>     * we have symmetry
>     
>     * But if your task require you still can specify OWNER
>     

Aha!  Now there's a dimension I've been missing, and it explains much.  You
are also trying to overload determination of which Table if either is the
owner when a link is defined.  That's a whole different matter.  And one I
hadn't really considered since in the apps I'm working with I never expect
to delete records.

I propose that the correct solution to this is to make ownership EXPLICIT
when you define the link.  Namely:

CreateBinaryLink( 
 inName as String, 
 inLeftTable as VTable,
 inRightTable as VTable,
 inLeftPower as EVLinkType = kOne,
 inRightPower as EVLinkType = kMany,
 inOnDelete as EVOnDelete = kSetNull,
 inStorageType as EVStorageType = kDefault,
 inTemporary as Boolean = false,
 inOwner as EVOwner = kLeftTable,
) as VBinaryLink 

where enurations for EVOwner are:
    kLeftTable, kRightTable, kNoOwner

Thus the left table is always the owner by default.

But as you say, for now it is always possible to simply state:

    aNewlyCreatedBinaryLink.Owner = TableA