Hierarchical SQL

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Post #169,172 by ChrisR 8/13/04 11:13:17 AM Reply	Hierarchical SQL Yet another article by Celko on [link\|http://www.onlamp.com/pub/a/onlamp/2004/08/05/hierarchical_sql.html\|Nested Sets] for those interested. Not much new here. What I really wanted to ask, though, is whether there's a more efficient method for calculating the level of nodes within a nested set. The following query is taking way more longer than I'd expect: UPDATE #Tree\n SET lvl = (SELECT COUNT(*) \n FROM #Tree tbl \n WHERE (tab.lft BETWEEN tbl.lft AND tbl.rgt))\n FROM #Tree tab
Post #169,179 by admin 8/13/04 11:48:12 AM Reply	Pre-calculate it. I do that in the stored proc that inserts the new nodes by adding one to the level of the parent node. Regards, -scott anderson "Welcome to Rivendell, Mr. Anderson..."
Post #169,192 by ChrisR 8/13/04 12:41:34 PM Reply	That's a better way to do it. Unfortunately, the tree I'm dealing with can take different shapes at run time, based on access rights and another tree that gets blended in.
Post #169,293 by deSitter 8/14/04 3:52:19 AM Reply	Re: Pre-calculate it. That's interesting, it's differentiation. -drl
Post #169,216 by admin 8/13/04 3:30:18 PM Reply	Those are different nested sets. He's using strings with the navigation information encoded within. Gumpy way to do things, IMO. Regards, -scott anderson "Welcome to Rivendell, Mr. Anderson..."

Hierarchical SQL - (ChrisR) - (4) - Aug. 13, 2004, 11:13:17 AM EDT

Pre-calculate it. - (admin) - (2) - Aug. 13, 2004, 11:48:12 AM EDT

That's a better way to do it. - (ChrisR) - Aug. 13, 2004, 12:41:34 PM EDT

Re: Pre-calculate it. - (deSitter) - Aug. 14, 2004, 03:52:19 AM EDT

Those are different nested sets. - (admin) - Aug. 13, 2004, 03:30:18 PM EDT

No manual entry for lrpd
40 ms