David McAfee
davidmcafee at gmail.com
Fri Sep 17 13:51:22 CDT 2010
Have you tried it this way? UPDATE TABLEA SET A.FIELD1 = B.FIELD1, A.FIELD2 = B.FIELD2, A.FIELD3 = B.FIELD3, A.FIELD4 = B.FIELD4, A.FIELD5 = B.FIELD5, A.FIELD6 = B.FIELD6, A.FIELD7 = B.FIELD7, A.FIELD8 = B.FIELD8, A.FIELD9 = B.FIELD9, A.FIELD10 = B.FIELD10, A.FIELD11 = B.FIELD11, A.FIELD12 = B.FIELD12, A.FIELD13 = B.FIELD13, A.FIELD14 = B.FIELD14, A.FIELD15 = B.FIELD15, A.FIELD16 = B.FIELD16, FROM TABLEA AS A INNER JOIN TABLEB AS B ON A.ID = B.ID On Fri, Sep 17, 2010 at 11:28 AM, jwcolby <jwcolby at colbyconsulting.com> wrote: > UPDATE TABLEA SET > FIELD1 = (SELECT TABLEB.FIELD1 FROM TABLEB WHERE TABLEB.ID = TABLEA.ID), > FIELD2= (SELECT TABLEB.FIELD2 FROM TABLEB WHERE TABLEB.ID = TABLEA.ID) > > It seems darned ugly but I can see the logic. However I have about 16 fields to update for (in some > cases) a subset of 2 million records in a 65 million record table. > > It seems like this is going to be very inefficient. > > -- > John W. Colby > www.ColbyConsulting.com > -- > AccessD mailing list > AccessD at databaseadvisors.com > http://databaseadvisors.com/mailman/listinfo/accessd > Website: http://www.databaseadvisors.com >