subtract date in SQL

my table structure is like this

Table Name : Receive
Field Name         Field Type
 FileName          Varchar2(50)    
 StartDateTime     Date
 EndDateTime       Date
 TotalRecord       Number

i want to know Total time process in hour:minute:second


the output is more like this:
FileName     : A.GFD
StartDateTime: 20-Nov-1999 10:00:00
EndDateTime  : 20-Nov-1999 15:23:33
Process Time : 03:23:33   <-- i want the process time in this format(hh:mi:ss)

how to do it ?





hdrikAsked:
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sgantaCommented:
Hi

SELECT floor((enddatetime-startdatetime)*24)
     ||':'||MOD((enddatetime-startdatetime)*1440,60)||':'||
     MOD((enddatetime-startdatetime)*86400,60)
"Diff time in HH24:MI:SS" FROMreceive;

Hope this helps you!
0
sgantaCommented:
Hi

SELECT floor((enddatetime-startdatetime)*24)                            ||':'||MOD((enddatetime-startdatetime)*1440,60)||':'||
                           MOD((enddatetime-startdatetime)*86400,60)
                      "Diff time in HH24:MI:SS" FROM receive;
0
Mark GeerlingsDatabase AdministratorCommented:
Sganta's proposed solution looks good as long as the start time and end times are in the same day.  I've developed a PL\SQL function to handle a situation similar to yours where sometimes the start and end times are near midnight, but in different days.

Leave a comment here if the proposed solution does not give you what you wanted.
0
hdrikAuthor Commented:
sganta,
  i agree with markgeer, your proposed answer only work at the same day

markgeer,
  can you help me ?
   
0
mshaikhCommented:
The following will give you the hours minutes and seconds  correctly even is the days are different  in the two dates

SELECT  floor((enddatetime-startdatetime)*24)||':'|| to_char(mod(floor((enddatetime-startdatetime)*1440),60),'09')||':'||to_char(mod((enddatetime-startdatetime)*86400,60),'09')
FROM Receive;

0

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