Formula for subtracting time from a datetime field

I have a field that is formatted as a datetime field. I just need to know how to subtract 3.5 hours from this time. How do I write a formula in crystal reports kindly help me.
Thanks in advance
Ramakrishna Reddy

This should do the trick...
DateAdd("n", -210, {TableName.DateField})
HTH,
Jason

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How to subtract time from dateTime in xquery?

How to subtract time from dateTime in xquery??
In ALDSP i tried to use the function xf:add-days but it is showing an error saying invalid function name. I think this namespace is not valid for ALDSP. Is there is any way to add this function in ALDSP???

xquery functions and operators documentation is here :
http://www.w3.org/TR/xpath-functions/
It sounds like you want to ..
op:subtract-dayTimeDuration-from-dateTime
this is not an xquery function that you can call, it is an xquery operation defined by the specification
I believe you can simply use the '-' operator to use that operation.
<newDateTime>
{ $myDateTime - $myTime }
</newDateTime>

How to subtract Time from Sysdate

Hello,
I want to subtract time from sysdate and get the sysdate with time.
Example : = (Sysdate with time)03-mar-2002 16:40:55 - 8 (hours). The result should be like :03-mar-2002 08:40:55
How to write a query for this?.
Please let me know as soon as possible.
Thanks,
Ravi.

Hi Ravi...
SYSDATE includes time, TRUNC(SYSDATE) does not include time.
If you subtract these two, you'll get a positive number of 1.00 or less which is the fraction of a day. For instance high noon would have a fraction of 0.50000. 12:01 p.m. would be 0.50000 + (1/(24*60)), or 0.50069444....
If you want to set the clock back, by calculating a "date" that is 8 hours less than the SYSDATE, then SYSDATE - (8/24) will do that. This will work even if SYSDATE was something like 2 a.m., because it will then set it to 6 p.m. of the day before the day in SYSDATE.
If you want to set a "date" to a specific time on the day of SYSDATE, then calculate the fraction for that time and add it to the TRUNC(SYSDATE). High noon would be TRUNC(SYSDATE) + 0.5.
SELECT TO_CHAR(TRUNC(SYSDATE)+0.5,'DD-MON-YY HH:MI:SS am') AS NOON FROM DUAL;Hope this helps...
Larry Johnson

Subtract Time from Time

Dear Sir,
I want to subtract Time from time, Please help me
SELECT A.EMP_CODE,
TO_DATE(A.INTIME,'DD/MM/RR')IN_DATE
,MIN(TO_CHAR(A.INTIME,'HH24:MI')) IN_TIME
,MAX(TO_CHAR(A.OUTTIME,'HH24:MI')) OUT_TIME
,TRUNC(SUM(A.TDAYS)*24)||':'||TRUNC((SUM(A.TDAYS)*24-TRUNC(SUM(A.TDAYS)*24))*60) TOTAL_TIME
FROM HCM.EINOUT A, HCM.PERSONNEL B
     WHERE A.EMP_CODE = B.EMP_CODE AND
     TO_DATE(A.INTIME,'DD/MM/RR')BETWEEN
     TO_DATE('01/08/12','DD/MM/RR') AND TO_DATE('15/08/12','DD/MM/RR')
     AND B.DEPT_CODE = 16 and a.emp_Code = '09833'
     GROUP BY A.EMP_CODE, TO_DATE(A.INTIME,'DD/MM/RR')--, a.tdays
     ORDER BY 1, 2 ASC;
I want to subtract 20:00 from (out_time) column from every record
If I subtract it from first record then result in over_time column will be 2:46
(out_time - 20:00)
--------------Out put........................
code date in_time out_time total_time over_time
09833     02-AUG-12     09:18     22:46     13:28
09833     03-AUG-12     09:16     21:50     12:34
09833     04-AUG-12     09:28     21:41     12:13
thanks

this will help
--And if difference can be > 24 hours:
WITH t AS
     (SELECT TO_DATE ('26-aug-2011 10:10:10',
                      'dd-mon-yyyy hh24:mi:ss'
                     ) start_date_time,
             TO_DATE ('26-aug-2011 10:12:10',
                      'dd-mon-yyyy hh24:mi:ss'
                     ) end_date_time
        FROM DUAL)
SELECT    TRUNC (end_date_time - start_date_time)
       || ' days '
       || TO_CHAR (TRUNC (SYSDATE) + (end_date_time - start_date_time),
                   'hh24:mi:ss'
                  ) time_diff
FROM t
--Or interval based solution for < 24 hour diff:
WITH t AS
     (SELECT TO_DATE ('26-aug-2011 10:10:10',
                      'dd-mon-yyyy hh24:mi:ss'
                     ) start_date_time,
             TO_DATE ('26-aug-2011 10:12:10',
                      'dd-mon-yyyy hh24:mi:ss'
                     ) end_date_time
        FROM DUAL)
SELECT REGEXP_SUBSTR (NUMTODSINTERVAL (end_date_time - start_date_time, 'day'),
                      '\d\d:\d\d:\d\d'
                     ) time_diff
FROM t
--interval based solution for >= 24 hour diff:
WITH t AS
     (SELECT TO_DATE ('26-aug-2011 10:10:10',
                      'dd-mon-yyyy hh24:mi:ss'
                     ) start_date_time,
             TO_DATE ('26-aug-2011 10:12:10',
                      'dd-mon-yyyy hh24:mi:ss'
                     ) end_date_time
        FROM DUAL)
SELECT REGEXP_REPLACE (NUMTODSINTERVAL (end_date_time - start_date_time,
                                        'day'),
                       '^.0*(\d+) (.{8}).+$',
                       '\1 days \2'
                      ) time_diff
FROM t
--another method
SELECT    TO_CHAR (TRUNC (nb_sec / 3600), '00')
       || ':'
       || TO_CHAR (nb_min - TRUNC (nb_sec / 3600) * 60, '00')
       || ':'
       || TO_CHAR (MOD (nb_sec, 60), '00')
FROM (SELECT (end_date_time - start_date_time) * 24 * 3600 nb_sec,
               (end_date_time - start_date_time) * 24 * 60 nb_min
          FROM ps_tb_ln_pdc_status_log)
--another method
SELECT SYSDATE + 3.45 / 24 AS "End Date", SYSDATE AS "Start Date",
       CAST (SYSDATE + 3.45 / 24 AS TIMESTAMP) - SYSDATE AS "Duration"
FROM DUAL;Regards,
friend.

SUBTRACT TIME FROM DATE & TIME

Hi.
I need FM which will SUBTRACT TIME from DATE & TIME.
Example:
Date = 05.05.2007
Time = 05:00:00
Time_subtract = 48:00:00
Return of FM:
Date = 03.05.2007
Time = 05:00:00
Do you know any FM?
Thanks!

Hi, Check these function mod's
SD_DATETIME_DIFFERENCE
HRCM_TIME_PERIOD_CALCULATE
HR_ECM_GET_PERIOD_BETW_DATES
After you execute them in SE37 give the required input and you will get a correct output.
<b>REWARD IF USEFUL</b>
thanks n regards
Dinesh

Need help assigning range for an integer for subtracting time.

Ok, so I'm having a little trouble getting this program to work. :x Maybe somebody could give me some help on what to do next. I'm new to Java, and I'm trying to write a program using NetBeans 6.9.1, and I ran into a problem. I'm trying to write a program for a time traveler Marty, who is going back in time. The time is set on a 24 hour clock, so am and pm is not an issue.
I have the program written, but I'm having trouble with the subtraction of the time. Whenever I do the subtract, and the time traveled back (either in hours, minutes or seconds), is bigger than the current time, I just end up with negative numbers. I don't know how to set the value so it will just recycle through the 24 hours, the 60 minutes, or 60 seconds, instead of giving me negative numbers. Any help would be greatly appreciated. Also, I'm not sure how to make the users input italic. Here is what I got so far.
package timetravel;
* @author Jeff
import java.io.*;
import java.util.*;
public class Main {
* @param args the command line arguments
public static void main(String[] args) {
int hours, minutes, seconds, hoursback, minutesback, secondsback;
Scanner keyboard = new Scanner (System.in);
System.out.println("Hi Marty!");
System.out.println("Enter current hour:");
hours = keyboard.nextInt();
System.out.println("Enter current minute:");
minutes = keyboard.nextInt();
System.out.println("Enter current second:");
seconds = keyboard.nextInt();
System.out.println("Enter how many hours you want to travel back");
hoursback = keyboard.nextInt();
System.out.println("Enter how many minutes you want to travel back");
minutesback = keyboard.nextInt();
System.out.println("Enter how many seconds you want to travel back");
secondsback = keyboard.nextInt();
System.out.println("When you arrive, the local time will be " + (hours - hoursback)
+ " hours, " + (minutes - minutesback) + " minutes, and " + (seconds - secondsback)
+ " seconds.");

Well, basically, here's how I want the program to run
Example program run:
Hi Marty!
Enter current hour: 8
Enter current minute: 12
Enter current second: 11
Enter how many hours you want to travel back: 9
Enter how many minutes you want to travel back: 36
Enter how many seconds you want to travel back: 39
When you arrive, the local time will be: 22 hours, 35 minutes, and 32 seconds
(program input is written in italic)
instead of how it runs now
Hi Marty!
Enter current hour: 8
Enter current minute: 12
Enter current second: 11
Enter how many hours you want to travel back: 9
Enter how many minutes you want to travel back: 36
Enter how many seconds you want to travel back: 39
When you arrive, the local time will be: -1 hours, -24 minutes, and -28 seconds
I want it to cycle through the 24 hour clock, so if I travel back in time from 18 (1800 or 6pm) and i want to travel back in time 20 hours, I should get 22 (2200 or 10 pm) instead of -2.
Same with minutes and seconds, except for it to cycle through 60 instead of 24, since there are 60 minutes in an hour, and 60 seconds in a minute. Also, if the minutes end up being negative, I need it to take away one from the hour spot, and so on for the other ones. Hopefully I explained it well enough... Sorry, I'm new to Java.
Edited by: halo2jak on Sep 7, 2010 12:24 AM

BOXI3.0: Formula for subtract/deduct one day or more from date field

Greetings. Its Alias.
I have one question. Actually I already post this question but at designer forum.
I am using SQL.
Example I have Order Date field and from this field I want to subtract the date by minus one day or two days or three days, etc.
I got the solution for designer (example):
DATEADD (dd , -3, EPJ_IPMS.dbo.DO_DATA.DO_DATE_D)
But at webi level it does not work.
I also do not understand why at designer and webi level may have different SQL select statement.
Can someone help me please.
Best regards,
Alias

Hi there Prashant.
In my case here is I am using Microsoft SQL Server Enterprise Manager and the data is save inside it.
The Designer and Webi are using same SQL database.
I also created an Object at Designer and the result is good.
But say if I don't want to create the object at Designer and let user create their own variables at webi.
Meaning to educate user to put the variable and formula themselves.
This is where the SQL statement a I mentioned earlier cannot be used at webi.
Best Regards,
Alias

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I have a DT field that I want to break into 1/2hr designators (so 1-48). Is there a better way to do this than with just a lengthy CASE WHEN (or "old school" IF THEN ELSE)?
So, for example 03/12/2014 12:05:30am becomes 1 (1st half hour after midnight), while 03/12/2014 11:37:33pm would be 48 (last half hour before midnight).
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else
If minute({DateTime_Field}) in 31 to 59 then
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(SELECT sysdate x FROM dual
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Formula for subtracting time from a datetime field

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