I thought the same thing until I tried SELECT NBRA, NBRB, NBRC, NBRD, nbra/nbrb, nbrc/nbrd from rob/davesmith ....+....1....+....2....+....3....+....4....+....5....+....6....+....7....+....8....+....9....+. NBRA NBRB NBRC NBRD NBRA / NBRB NBRC / NBRD 1 2 3.00 4.00 .5000000000000000000000000000 .7500000000000000000000000000 But this gives me: SELECT NBRA, (sum(nbra)/sum(nbrb)) from rob/davesmith group by nbra order by nbra ....+....1....+....2....+....3....+....4....+.... NBRA Numeric Expression 1 0 However your casting was a good idea: SELECT NBRA, (float(sum(nbra))/float(sum(nbrb))) from rob/davesmith group by nbra order by nbra ....+....1....+....2....+....3. NBRA Numeric Expression 1 5.0000000000000000E-001 Rob Berendt -- "They that can give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety." Benjamin Franklin "Elvis Budimlic" <ebudimlic@xxxxxxxxxxxxxxxxxxxxxxxxx> Sent by: midrange-l-bounces@xxxxxxxxxxxx 09/18/2003 12:44 PM Please respond to Midrange Systems Technical Discussion To: <midrange-l@xxxxxxxxxxxx> cc: Fax to: Subject: RE: SQL dividing the sum of two results on a select Dave, AFAIK, SQL performs integer division, meaning it rounds to the nearest integer value. If your first sum is smaller than the second one, you'll always get a result of zero. So, check your sum values independently and make sure that's not the case. Another thing you can try is casting the values within the sum function to floats, i.e. SUM(FLOAT(field1)). Division should then return the real division (with decimal points and such). Elvis _______________________________________________ This is the Midrange Systems Technical Discussion (MIDRANGE-L) mailing list To post a message email: MIDRANGE-L@xxxxxxxxxxxx To subscribe, unsubscribe, or change list options, visit: http://lists.midrange.com/mailman/listinfo/midrange-l or email: MIDRANGE-L-request@xxxxxxxxxxxx Before posting, please take a moment to review the archives at http://archive.midrange.com/midrange-l.
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