In Oracle 10g I have a table that holds timestamps showing how long certain operations took. It has two timestamp fields: starttime and endtime. I want to find averages of the durations given by these timestamps. I try:
select avg(endtime-starttime) from timings;
But get:
SQL Error: ORA-00932: inconsistent datatypes: expected NUMBER got INTERVAL DAY TO SECOND
This works:
select
avg(extract( second from endtime - starttime) +
extract ( minute from endtime - starttime) * 60 +
extract ( hour from endtime - starttime) * 3600) from timings;
But is really slow.
Any better way to turn intervals into numbers of seconds, or some other way do this?
EDIT: What was really slowing this down was the fact that I had some endtime's before the starttime's. For some reason that made this calculation incredibly slow. My underlying problem was solved by eliminating them from the query set. I also just defined a function to do this conversion easier:
FUNCTION fn_interval_to_sec ( i IN INTERVAL DAY TO SECOND )
RETURN NUMBER
IS
numSecs NUMBER;
BEGIN
numSecs := ((extract(day from i) * 24
+ extract(hour from i) )*60
+ extract(minute from i) )*60
+ extract(second from i);
RETURN numSecs;
END;
There is a shorter, faster and nicer way to get DATETIME difference in seconds in Oracle than that hairy formula with multiple extracts.
Just try this to get response time in seconds:
(sysdate + (endtime - starttime)*24*60*60 - sysdate)
It also preserves fractional part of seconds when subtracting TIMESTAMPs.
See http://kennethxu.blogspot.com/2009/04/converting-oracle-interval-data-type-to.html for some details.
Note that custom pl/sql functions have significant performace overhead that may be not suitable for heavy queries.
If your endtime and starttime aren't within a second of eachother, you can cast your timestamps as dates and do date arithmetic:
select avg(cast(endtime as date)-cast(starttime as date))*24*60*60
from timings;
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