How to subtract 2 dates in oracle to get the resul

Question

I want to subtract 2 dates and represent the result in hour and minute in one decimal figure.

I have the following table and I am doing it in this way but the result is not as desired.

There is some slight variation, I'm sure this is simple arithmetic but I'm not getting it right.

select start_time, end_time, (end_time-start_time)*24 from 
come_leav;    

START_TIME          END_TIME            (END_TIME-START_TIME)*24    
------------------- ------------------- ------------------------      
21-06-2011 14:00:00 21-06-2011 16:55:00  2.9166667      
21-06-2011 07:00:00 21-06-2011 16:50:00  9.8333333      
21-06-2011 07:20:00 21-06-2011 16:30:00  9.1666667      

I want the result (end_time-start_time) as below.

16:55- 14:00 = 2.55      
16:50-07:00 = 9.5      
16:30-7:20 = 9.1 and so on.    

How can I do that?

Answer 1

Try this

round(to_number(end_time - start_time) * 24)

Answer 2

This query is very usefull for me and if any body want diffrence between start_date and end_date with time like HH:MI:SS please use this query.

SELECT
    TRIM(TO_CHAR(TRUNC(((86400*(end_date - start_date))/60)/60)-24*(trunc((((86400(end_date - start_date))/60)/60)/24)),'00')) ||'.'||
    TRIM(TO_CHAR(TRUNC((86400*(actual_completion_date - actual_start_date))/60)-60*(trunc(((86400*(end_date - start_date))/60)/60)),'00')) ||'.'||
    TRIM(TO_CHAR(TRUNC((86400*(end_date - start_date)))-60*(trunc((86400*(end_date - actual_start_date))/60)),'00')) as duration  
FROM fnd_concurrent_requests; 

Answer 3

Edit: if you need a number, then

    trunc(end_date-start_date)*24+
    to_number(to_char(trunc(sysdate)+(end_date-start_date),'HH24.MI'))

For string result, if delta is LESS THAN 24H: I would go with

    to_char(trunc(sysdate)+(end_date-start_date),'HH24.MI')

or ...'HH24:MI:SS', but thats my personal preference.

for longer than 24H terms, I would prefix with

    trunc(end_date-start_date)||"days "||
    to_char(trunc(sysdate)+(end_date-start_date),'HH24.MI')

Yes, as oracle counts in days, with seconds precision, you are dealing with arithmetical problems. Once because you are only handling minutes (so you might round your number to trunc(days*24*60+0.5)/24/60), but the binary arithmetic imprecision on the number 1/24/60 might still cause you troubles.

Edit2.1:

    to_char(24*(trunc(end_date)-trunc(start_date))+to_number(to_char(end_date,'HH24.MI'))-to_number(to_char(start_date,'HH24.MI')),'99999.99')

But The result could be quite confusing for the average, as the decimal 7.50 would suggest seven and a half hour, or at least 7 hour 50 minutes, opposed to the elapsed time of 7 hours 10 minutes.

Answer 4

SQL> edit
Wrote file afiedt.buf

  1  select start_date
  2      , end_date
  3      , (24 * extract(day from (end_date - start_date) day(9) to second))
  4          + extract(hour from (end_date - start_date) day(9) to second)
  5          + ((1/100) * extract(minute from (end_date - start_date) day(9) to second)) as "HOUR.MINUTE"
  6* from t
SQL> /

START_DATE          END_DATE            HOUR.MINUTE
------------------- ------------------- -----------
21-06-2011 14:00:00 21-06-2011 16:55:00        2.55
21-06-2011 07:00:00 21-06-2011 16:50:00         9.5
21-06-2011 07:20:00 21-06-2011 16:30:00         9.1

It should be noted for those coming across this code that the decimal portions are ACTUAL minute differences, and not part of an hour. .5, therefore, represents 50 minutes, not 30 minutes.

Answer 5

try this:

    SELECT
    TRIM(TO_CHAR(TRUNC(((86400*(end_time - start_time))/60)/60)-24*(trunc((((86400*(end_time - start_time))/60)/60)/24)),'00')) ||'.'||
    TRIM(TO_CHAR(TRUNC((86400*(end_time - start_time))/60)-60*(trunc(((86400*(end_time - start_time))/60)/60)),'00')) ||'.'  as duration
FROM come_leav;

Answer 6

you can work with the extract:

SELECT start_time, end_time, extract(HOUR FROM end_time-start_time) || '.' || extract(MINUTE FROM end_time-start_time)  
From come_leav;