Re: Quote of the Week

>> [How would you compare two of these organization structure
relations for equality?] Aggregated total? <<

I never thought of that! I feel like an idiot! Why do you keep coming up with good ideas after the book is printed??

But there is no need for recursion to sum the subtrees:

SELECT O2.emp, SUM(S1.weight)

   FROM OrgChart AS O1, OrgChart AS O2,

        Salaries AS S1
  WHERE O1.lft BETWEEN O2.lft AND O2.rgt     AND O1.emp = S1.emp
  GROUP BY O2.emp;

>> Finally, tree structure are identical if they have identical weight
at the root. <<

Not true; consider the set of weights (1,2,3,4)= 10 which could be split up as ((1,4), (2,3))= 10 or ((1,2,3),(4)) = 10. What we need is two tables with the sums from the trees which are identical, weight for weight. Easy enough:

 ((<<above query for Table A>>) EXCEPT ((<<above query for Table B>>) IS NULL Received on Wed May 19 2004 - 18:33:05 CEST