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

This is a very ugly way to do it, and this first part doesn't exactly question by the OP, but it gives a way to get results by subtracting 2 date fields -- in my case, the CREATED_DATE and today represented by SYSDATE:

SELECT FLOOR(ABS(MONTHS_BETWEEN(CREATED_DATE, SYSDATE)) / 12) || ' years, '  
|| (FLOOR(ABS(MONTHS_BETWEEN(CREATED_DATE, SYSDATE))) - 
   (FLOOR(ABS(MONTHS_BETWEEN(CREATED_DATE, SYSDATE)) / 12)) * 12) || ' months, '  
-- we take total days - years(as days) - months(as days) to get remaining days
|| FLOOR((SYSDATE - CREATED_DATE) -      -- total days
   (FLOOR((SYSDATE - CREATED_DATE)/365)*12)*(365/12) -      -- years, as days
   -- this is total months - years (as months), to get number of months, 
   -- then multiplied by 30.416667 to get months as days (and remove it from total days)
   FLOOR(FLOOR(((SYSDATE - CREATED_DATE)/365)*12 - (FLOOR((SYSDATE - CREATED_DATE)/365)*12)) * (365/12)))  
|| ' days, '   
-- Here, we can just get the remainder decimal from total days minus 
-- floored total days and multiply by 24       
|| FLOOR(
     ((SYSDATE - CREATED_DATE)-(FLOOR(SYSDATE - CREATED_DATE)))*24
   )
|| ' hours, ' 
-- Minutes just use the unfloored hours equation minus floored hours, 
-- then multiply by 60
|| ROUND(
       (
         (
           ((SYSDATE - CREATED_DATE)-(FLOOR(SYSDATE - CREATED_DATE)))*24
         ) - 
         FLOOR((((SYSDATE - CREATED_DATE)-(FLOOR(SYSDATE - CREATED_DATE)))*24))
       )*60
    )
|| ' minutes'  
AS AGE FROM MyTable`

It delivers the output as x years, x months, x days, x hours, x minutes. It could be reformatted however you like by changing the concatenated strings.

To more directly answer the question, I've gone ahead and written out how to get the total hours with minutes as hours.minutes:

select  
((FLOOR(end_date - start_date))*24)
|| '.' ||
ROUND(
       (
         (
           ((end_date - start_date)-(FLOOR(end_date - start_date)))*24
         ) - 
         FLOOR((((end_date - start_date)-(FLOOR(end_date - start_date)))*24))
       )*60
    )
from 
come_leav;   

Answer 2

Oracle represents dates as a number of days, so (end_time-start_time)*24 gives you hours. Let's assume you have this number (eg. 2.9166667) in h column. Then you can easily convert it to the format you want with: FLOOR(h) + (h-FLOOR(h))/100*60.

Example:

WITH diff AS (
    SELECT (TO_DATE('21-06-2011 16:55:00', 'DD-MM-YYYY HH24:MI:SS') - TO_DATE('21-06-2011 14:00:00', 'DD-MM-YYYY HH24:MI:SS'))*24 h
    FROM dual
) SELECT FLOOR(h) + (h-FLOOR(h))/100*60
FROM diff

In your case:

SELECT start_time, end_time,
    FLOOR((end_time-start_time)*24) + ((end_time-start_time)*24-FLOOR((end_time-start_time)*24))/100*60 AS hours_diff
FROM come_leav