Re: SQL question

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If someone knows this it would be great - because I'm still curious how to
solve it. However I just remodelled my db structure to eliminate the problem
(basically I pulled the several tables into one since each of the
table2/table3 tables only has 3 fields)
so now I do:
table2 (
uid int FK to table1,
luid int FK to table1,
is_in_table3 boolean,
is_in_table4 boolean,
.....
)

this just needs a simple select with a join against table1.
UC

On Saturday 06 November 2004 03:13 pm, Uwe C. Schroeder wrote:
> Here's a question for the SQL guru's out there, which I've been trying to
> solve for the last couple of hours. There's got to be a solution to this,
> but somehow I can't find it.
>
> Tables:
>
> table1 (
> uid int PK,
> uname varchar(64)
> )
>
> table2 (
> uid int FK to table1,
> xuid int FK to table 1
> )
>
> table3 (
> uid int FK to table1,
> yuid int FK to table1
> )
>
> There might be more tables of the type like table2 and table3, but I'd
> already be happy to solve the puzzle with the 3 tables above.
> Ok, assume table1 is the master table - in my case a table used for login
> authentication (some columns removed above)
> table2 and table3 are tables where the uid always references to the uid in
> table1. The second "uid" (xuid and yuid in this example) references to
> another uid record in table1. The problem is that there may or may not be
> entries in table2 (or table3) referencing a specific uid in their second
> uid field.
> Maybe some data:
>
> table1:
> 1 test1
> 2 test2
> 3 test3
>
> table2:
> 1 2
> 1 3
> 3 1
>
> table3:
> 1 2
> 2 3
> 3 2
>
> What I want to do in a view is the following resultset:
>
> uid uname xuid yuid
> 1 test1 2 2
> 1 test1 3
> 2 test2 3
> 3 test3 1
> 3 test3 2
>
>
> So basically I want to know which uid is connected to which uid, one
> relationship per row. So xuid and yuid shall be identical if records exist
> in both table2 and table3 or the value shall be NULL if a corresponding
> record can't be found in either table2 or table3.
>
> Can anyone here help me out?
>
> Thanks a lot
>
> UC

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