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Posted by Hilarion on 10/13/21 11:25
Anon <Ao@anonc.com> wrote:
> Since putting this online I have come across this code
> Find all the children of AAA. In order to avoid running forever, stop after four levels.
>
> WITH parent (pkey, ckey, lvl) AS
> (SELECT DISTINCT
> pkey
> ,pkey
> ,0
> FROM trouble
> WHERE pkey = ’AAA’
> UNION ALL
> SELECT C.pkey
> ,C.ckey
> ,P.lvl + 1
> FROM trouble C
> ,parent P
> WHERE P.ckey = C.pkey
> AND P.lvl + 1 < 4
> )
> SELECT *
> FROM parent;
>
> However I do not have a lvl attribute and so I fear I would be creating an infinite loop.
What kind of DBMS (SQL engine) do you use? Where did you get this
example from? For what database structure it's given?
Hilarion
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