Force the order in which WHERE clauses are applied?

Question

I need to make sure that one where clause filtering is applied before the other. How do I do that?
Several of my where clauses are below.
I need to make sure than the where clause in bold is applied before several where clauses on the bottom. Otherwise, I get error: division by 0.
where ...
and a.surveydate = (select max(surveydate) from survey
where id = a.id
and surveydate < b.surveydate
and b.surveydate between TO_DATE('&lowdate','DD-MON-YYYY')
and TO_DATE('&highdate','DD-MON-YYYY')
and (
( abs( (b.totvacant/totunits - a.totvacant/totunits) /
(b.surveydate - a.surveydate) * 365 ) >= 0.00 )
or ( abs( (b.avgcost/ a.avgcost -1 ) /
(b.surveydate - a.surveydate) * 365 ) >= 0 )
or ( abs( (b.free_rent - a.free_rent ) /
(b.surveydate - a.surveydate) * 365 ) >= 0 )
Thank you. I know very little about query optimization, but I have a sense that I need to use some kind of SQL hint.

Answer 1

Why not rewrite the where clause a bit to intercept »ORA-01476: divisor is equal to zero«?:
SELECT *
  FROM tab
WHERE a.surveydate =
          (SELECT MAX (surveydate)
             FROM survey
            WHERE ID = a.ID
              AND surveydate < b.surveydate
              AND b.surveydate BETWEEN TO_DATE ('&lowdate', 'DD-MON-YYYY')
                                   AND TO_DATE ('&highdate', 'DD-MON-YYYY')
              AND (   (ABS ((b.totvacant / totunits - a.totvacant / totunits
                            / NULLIF (b.surveydate - a.surveydate, 0) * 365
                           ) >= 0.00
                   OR (ABS ((b.avgcost / a.avgcost - 1)
                            / NULLIF (b.surveydate - a.surveydate, 0) * 365
                           ) >= 0
                   OR (ABS ((b.free_rent - a.free_rent)
                            / NULLIF (b.surveydate - a.surveydate, 0) * 365
                           ) >= 0
                  ))