> > # explain update account.cust set prodid = tempprod.prodid
> > where tempprod.did = origid;
> > Merge Join (cost=0.00..232764.69 rows=4731410 width=252)
> > Merge Cond: (("outer".origid)::text = ("inner".did)::text)
> > -> Index Scan using ix_origid on cust (cost=0.00..94876.83
> > rows=4731410 width=244)
> > -> Index Scan using ix_did on tempprod (cost=0.00..66916.71
> > rows=4731410 width=18)
> I'm going to hazard a guess and say you have a number of foreign keys
> that refer to account.cust.prodid? This is probably the time consuming
> part -- perhaps even a missing index on one of those keys
> that refers to
> this field.
Actually, there are no foreign keys to those columns. Once they're
populated, I'll apply a foreign key constraint and they'll refer to the
appropriate row in the prod and subprod tables, but nothing will
reference account.cust.[sub]prodid. There are, of course, several foreign
keys referencing account.cust.custid.
> Going the other way should be just as good for your purposes, and much
> faster since you're not updating several foreign key'd fields bound to
> account.cust.prodid.
> UPDATE tempprod.prodid = prodid
> FROM account.cust
> WHERE temprod.did = cust.origid;
Not quite. Without this update, acount.cust.[sub]prodid are null. The
data was strewn across multiple tables in MS SQL; we're normalizing it
into one, hence the need to populate the two columns independently.
/rls
--
Rosser Schwarz
Total Card, Inc.
