Formula of FormCalc to make the difference between two time fields if they are populated - help please

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  I'm so bothered by the fact that I can't understand that.

  Cell B2 - 08:00

  Cell C2 - 10:50

  (How can cell D2 - I get this cell to calculate the difference and say 02:50?)

  I know it's probably one of the most basic operations, but for the life of me I can't understand it. The cells B2 and C2 are formatted for a 24-hour clock. But if I tell the system to just subtract the two, I get "0,118. Everything I find on the forum search goes beyond what I need. Can someone help me?

  Thank you

  Hi sapirs,

  Departure and arrival of the cells are in Date and time with Date format: no and time: 24-hour clock.

  

  Formula in D2 (fill down)

  = C2−B2

  

  The results become a duration format, but under automatic (numbers automatically worth this format)

  

  If you wish, you can change the cells of lasting results.

  What Data Format have your result cells?

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• Difference between two times

  I am building a working method in LiveCycle, and I want to track how long the procedure takes. To do this, I have two time fields, and then a third time field to find the difference between two time fields. Looking at other posts to discussion, I could find a Javascript code that was part of what I wanted, but I can't get the third time field to display in the format I want. I tried to change the habits of the fields and that did not work. Could someone offer some advice on how to change the format or help me with the code?  I couldn't figure to attach the actual file, but I have attached a picture of the PDF file and the Javascript code I used.  If you look at the TimeIn (hh: mm) and time (hh: mm) areas (on the photo, I have attached), you will notice that the TotalTime comes out like HMM, and I want that it out as hh: mm.

  

  

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  Hello to everyone. Today, I came with the new requirement.
  
  
  I need to know the difference between a date and the current date in the formula column application presentation layer.by.
  
  
  
  Thank you and best regards,
  Prates

  Hi Navin,

  TIMESTAMPDIFF function first determines the timestamp component that corresponds to the specified interval setting. For example, SQL_TSI_DAY corresponds to the day component and SQL_TSI_MONTH corresponds to the component "month".

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• What is the difference between nvl (sum (field), 0) & sum (nvl (field, 0))?

  What is the difference between nvl (sum (field), 0) & sum (nvl (field, 0))?
  
  For the data in the table below I don't see that in the results
  
  CREATE TABLE FRUITS
  (
  TYPE VARCHAR2 (10 BYTE),
  VARIETY VARCHAR2 (10 BYTE),
  PRICE NUMBER (18.2)
  )
  
  VARIETY OF TYPE PRICE
  
  Gala Apple 2.79
  Apple fuji 0.24
  Apple limbertwig 2.87
  Orange valencia 3.59
  9.36 navel Orange
  7.77 bradford PEAR
  7.77 bartlett PEAR
  Cherry bing 2.55
  Cherry Chelan 6.33
  Bradford PEAR
  navel 6.39
  
  
  Choose the variety, nvl (sum (price), 0)
  fruit
  Group of variety;
  
  
  VARIETY NVL (SUM (PRICE), 0)
  
  limbertwig 2.87
  Bartlett 7.77
  Bing 2.55
  0 Marathon
  Gala 2.79
  Fuji 0.24
  navel 15.75
  Bradford 7.77
  Chelan 6.33
  Valencia 3.59
  
  
  Choose the variety, sum (nvl(price,0))
  fruit
  Group of variety;
  
  VARIETY SUM (NVL(PRICE,0))
  
  limbertwig 2.87
  Bartlett 7.77
  Bing 2.55
  0 Marathon
  Gala 2.79
  Fuji 0.24
  navel 15.75
  Bradford 7.77
  Chelan 6.33
  Valencia 3.59
  
  
  no difference in the output.
  
  What is the difference?
  
  Thanks in advance

  Do you see the difference now?

  PRAZY@11gR1> create table testing(field number);
  
  Table created.
  
  Elapsed: 00:00:00.10
  PRAZY@11gR1> insert into testing select null from dual connect by level<=5;
  
  5 rows created.
  
  Elapsed: 00:00:00.00
  PRAZY@11gR1> select * from testing;
  
       FIELD
  ----------
  
  Elapsed: 00:00:00.01
  PRAZY@11gR1> select sum(nvl(field,1)) from testing;
  
  SUM(NVL(FIELD,1))
  -----------------
                  5
  
  Elapsed: 00:00:00.00
  PRAZY@11gR1> select nvl(sum(field),1) from testing;
  
  NVL(SUM(FIELD),1)
  -----------------
                  1
  
  Elapsed: 00:00:00.00
  

  In the first case, we are substituting 1 to null and synthesis. as a result, we had 5. in the latest, we are summarizing the value null, which is null and its replacement by 1 if the result is null. We have 1.

  Kind regards
  Prazy

• Find the difference between two date and time

  Hi friends,
  
  I wanted to find the difference between two date and time, but my qury is slightest error "invalid number."
  
  

  select sql_step_num,proc_name,run_seqno,start_date,end_date,(to_char(start_date,'HH24-MI-SS') - to_char(end_date,'HH24-MI-SS') ) as ed  
  from eval.EVAL_RUNTIME_DETAILS
  where trunc(start_date) = trunc(sysdate) 
  order by sql_step_num;

  You try to get the feel between two char strings.
  And more difference between two dates gives a NUMBER of days.
  Try this:

  select sql_step_num,proc_name,run_seqno,start_date,end_date,numtodsinterval(end_date-start_date,'DAY') as ed
  from eval.EVAL_RUNTIME_DETAILS
  where trunc(start_date) = trunc(sysdate)
  order by sql_step_num;
  

• What is the difference between AXIEM and EMSIght? What are the benefits of AXIEM by EMSight if any?

  I wonder what are the differences between AXIEM and EMSight. Are associated with application? One is better than the other? Please explain.

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  Mesh

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  Ports

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  • Ports can be added to most any place in geometry with AXIEM.  For EMSight, the ports are usually only added outside the geometry and have limitations to the aircraft on either side of the reference speaker.

  Problem solvers

  • EMSight does not have a low frequency Solver.  Stop low-frequency and then extrapolate to DC.  This usually works well, but you must take into account if the extrapolation is accurate.  AXIEM presents both a low and high frequency Solver so will do better in a low frequencies generally much better in DC.
  
  
  • Solver of EMSight scales O (n) ^ 2 for memory and O (n) ^ 3 for the moment.   It has a special algorithm to break the question of memory, but the question of time is the killer.  AXIEM has a direct Solver which scales in the same way and an iterative Solver which will scale O (nlogn).   In other words, AXIEM can solve a lot of bigger problems than EMSight.
  
  
  • Several AXIEM processes are multithreaded as EMSight.

  Speaker

  • EMSight is inside a perfect driver side wall.   This works well if you the circuit is in the same cavity.  If not, you must know the rules of them on to the forms of space of the flanks and how that much to extend the port could reference planes.   AXIEM is not here.   AXIEM does not currently have to simulate a perfect driver side wall.

  I don't know that I missed a few aspects of the tools.   From the point of view of the AWR that AXIEM is the tool of choice, except in a few cases have been EMSight may be required.

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  Hi all

  Is there a function in labview other than the subtraction function that can find the difference between the two numbers given?

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• How to store the difference between two timestamps as a number?

  Hello
  I need to find the difference between 2 variables of timestamp and the result will be stored as a number in minutes?
  as
  declare
  timestamp (7) of the fir_timestamp: = 1 January 2012 13:30 ';
  timestamp (7) of the sec_timestamp: = 1 January 2012 14:00 ';
  c number;
  Start
  c:=((sec_timestamp-fir_timestamp)/(24*60));
  dbms_output.put_line (c);
  end;

  962813 wrote:
  declare

  -fir_timestamp timestamp (7): ='01 - jan - 2012 13:30 ';
  timestamp (7) of the sec_timestamp: = 1 January 2012 14:00 ';  Incorrect type declaration

  fir_timestamp timestamp: = 1st January 12 01.30.00.000000 PM';
  sec_timestamp timestamp: = 1st January 12 02.00.00.000000 PM';
  c number;
  Start
  c: = ABS ((extrait (HEURE de sec_timestamp) - extract (fir_timestamp TIME)) * 60) +.
  ABS ((extrait (MINUTE de sec_timestamp) - extract (fir_timestamp MINUTE)));
  dbms_output.put_line (' interval Min Total:-' | c);
  end;
  /
  declare
  *
  ERROR on line 1:
  ORA-01830: date format picture ends before converting all of the input string
  ORA-06512: at line 6

  Follow what Paul has done...

  In fact, you do not need timestamp, date data type is enough

  SQL> declare
    2     fir_timestamp date :=
    3        to_date('01-jan-2012 13:30:00','dd-mon-yyyy hh24:mi:ss');
    4     sec_timestamp date :=
    5        to_date('01-jan-2012 14:00:00','dd-mon-yyyy hh24:mi:ss');
    6     c number;
    7  begin
    8     c:= round((sec_timestamp-fir_timestamp)*24*60);
    9     dbms_output.put_line(c);
   10  end;
   11  /
  30
  
  PL/SQL procedure successfully completed.
  

  If you want to place on timestamp

  SQL> declare
    2     fir_timestamp timestamp :=
    3        to_timestamp('01-jan-2012 13:30:00','dd-mon-yyyy hh24:mi:ss');
    4     sec_timestamp timestamp :=
    5        to_timestamp('01-jan-2012 14:00:00','dd-mon-yyyy hh24:mi:ss');
    6     c number;
    7  begin
    8     c:= round((
    9     to_date(to_char(sec_timestamp,'ddmmyyyyhh24miss'),
   10             'ddmmyyyyhh24miss')-
   11          to_date(to_char(fir_timestamp,'ddmmyyyyhh24miss'),
   12             'ddmmyyyyhh24miss')
   13           )*24*60);
   14     dbms_output.put_line(c);
   15  end;
   16  /
  30
  

• How to find the difference between two dates in time except Sunday

  Hi all
  
  I have a table, as shown below.

  SQL> select * from test;
  
  TR_ID                                              CREATE_TIME                                                                       CODE
  -------------------------------------------------- --------------------------------------------------------------------------- ----------
  S12341                                             05-JUN-12 12.20.52.403000 AM                                                      1003
  S12342                                             11-JUN-12 11.15.33.182000 AM                                                      1003
  S12342                                             07-JUN-12 12.00.36.573000 PM                                                      1002
  S12343                                             20-JUN-12 12.34.37.102000 AM                                                      1003
  S12343                                             15-JUN-12 11.34.27.141000 PM                                                      1002
  S12344                                             01-JUL-12 10.01.06.657000 PM                                                      1002
  S12344                                             06-JUL-12 12.01.04.188000 AM                                                      1003
  S12341                                             31-MAY-12 11.20.38.529000 PM                                                      1002

  I would like to know the difference between same tr_ids create_time, which should give out in hours except Sunday.
  
  For example:
  
  TR_ID: S12344
  1002_Create_time: July 1, 12 PM 10.01.06.657000 (i.e. Sunday)
  1003_Create_time: 12.01.04.188000 AM 6 July 12
  
  1002 create time is 22:00 Sunday.
  
  If the query must exclude only the hours of Sunday which is 10 p.m. to Monday 00 h which is 2 Hrs.
  
  I tried the sub query after doing a search on this forum but I am not getting the desired output.
  

  SELECT count(*) FROM (SELECT ROWNUM RNUM,tr_id,create_time CT_1002 FROM test c WHERE c.tr_id='S12344' and 
  ROWNUM <= (select (cast(a.create_time as date)-cast((select create_time from test b where a.tr_id=b.tr_id and code=1002) as date)) 
  from test a where a.code=1003 and a.tr_id=c.tr_id) + 1) d 
  WHERE to_char(cast((select create_time from test e where e.tr_id=d.tr_id and code=1002) as date) + RNUM - 1, 'DY') NOT IN('SUN');

  Need help to get the desired o/p

  Hello

  If I unederstand the problem correctly, that's what you want:

  WITH       got_extrema     AS
  (
       SELECT       tr_id
       ,       CAST (MIN (create_time) AS DATE)     AS start_date
       ,       CAST (MAX (create_time) AS DATE)     AS end_date
       FROM       test
       GROUP BY  tr_id
  )
  SELECT       tr_id
  ,       start_date, end_date          -- If wanted
  ,       24 * ( ( ( TRUNC (end_date,   'IW')          -- Count -1 day for every full week
                      - TRUNC (start_date, 'IW')
               )
             / -7
                )
              + LEAST ( end_date               -- If end_date is a Sunday
                          , TRUNC (end_date, 'IW') + 6     -- consider it 00:00:00 on Sunday
                   )
              - CASE
                        WHEN  start_date >= TRUNC (start_date, 'IW') + 6     -- If start_date is a Sunday
                 THEN  TRUNC (start_date, 'IW') + 6               -- consider it 00:00:00 Sunday
                 ELSE  start_date
                    END
              )     AS total_hours
  FROM      got_extrema
  ;
  

  I guess that you don't need to worry about fractions of a second. As you did in the code you have posted, I am to convert the TIMESTAMP to date values, because of DATE arithmetic and functions are so much better than what is available for timestamps.

  Basically, it's just to find the number of days between start_date and end_date and multiplying by 24, with these twists:
  (a) 1 day is deducted for each week between start_date and end_date
  (b) if End_date is a Sunday, none of the end_date himself hours are counted
  (c) If start_date is a Sunday, then all the start_date himself hours are counted. Why these hours should be counted? Because 1 day is already being deducted for the week which includes start_date, which contains only this Sunday.

  TRUNC (dt, 'IW') is the beginning of the ISO week containing dt; in other words, 00:00:00 the or before the dt last Monday. This is not the NLS parameters.

  Of course, I can't test without some sample data and the exact results you want from these data. You may need a little something more If start_date and end_date are both on the same Sunday.
  Whenever you have a problem, please post a small example of data (CREATE TABLE and only relevant columns, INSERT statements) of all of the tables involved.
  Also post the results you want from this data, as well as an explanation of how you get these results from these data, with specific examples.
  Always tell what version of Oracle you are using.
  See the FAQ forum {message identifier: = 9360002}

• The most elegant way to get the difference between two tables - not least!

  Hello
  
  
  Simplified example of what I'm trying to achieve - I have two tables ORIGINAL and REVISED.
  
  
  My aim is to compare the two, such as; -
  
  
  When there is data in the two tables I get the difference between the Budget column, and if there is no difference, so I don't want no lines.
  
  When data exists in the ORIGINAL, but not in review, I want to the inverse of the current value of the Budget column.
  
  Where the data exist in REVISED I want the REVISED value.
  
  I can see how I can do this, see below, but is there a more elegant solution?
  
  
  
  
  Data for the ORIGINAL table
  

  select '801040' entity, '2186' expense_type, 234000 budget
  from dual
  union all
  select '801040' entity, '3001' expense_type, 1000 budget
  from dual
  union all
  select '801040' entity, 'P132' expense_type, 34000 budget
  from dual
  union all
  select '801040' entity, 'P135' expense_type, 43000 budget
  from dual

  Data for the REVISED table
  

  select '801040' entity, '2186' expense_type, 235000 budget
  from dual
  union all
  select '801040' entity, 'P132' expense_type, 34000 budget
  from dual
  union all
  select '801040' entity, 'P139' expense_type, 56000 budget
  from dual

  Desired output
  
  
  
  ENTITY EXPENSE_TYPE DIFFERENCE
  ------ ------------ ----------
  801040 2186 1000
  801040 3001-1000
  801040 P135-43000
  801040 P139 56000
  
  5 selected lines.
  
  
  
  Code current to achieve this, is there a better way?
  

  select original.entity
  ,      original.expense_type
  ,       (nvl(revised.budget,0) - original.budget) as difference
  from   original
  ,      revised
  where  original.entity = revised.entity(+)
  and    original.expense_type = revised.expense_type(+)
  and   (nvl(revised.budget,0) - original.budget) != 0
  union all
  select  revised.entity
  ,       revised.expense_type
  ,       revised.budget as difference
  from   revised
  where  not exists
  (select 'x'
  from   original
  where  original.entity = revised.entity
  and    original.expense_type = revised.expense_type)
  and    revised.budget != 0

  Thanks a lot for your comments,.
  
  
  Robert.
  
  Published by: Robert Angel on January 17, 2012 03:31 to change is not equal to! = - Thanks for the heads up

  SQL> with original
    2  as
    3  (
    4    select '801040' entity, '2186' expense_type, 234000 budget
    5    from dual
    6    union all
    7    select '801040' entity, '3001' expense_type, 1000 budget
    8    from dual
    9    union all
   10    select '801040' entity, 'P132' expense_type, 34000 budget
   11    from dual
   12    union all
   13    select '801040' entity, 'P135' expense_type, 43000 budget
   14    from dual
   15  )
   16  , revised
   17  as
   18  (
   19    select '801040' entity, '2186' expense_type, 235000 budget
   20    from dual
   21    union all
   22    select '801040' entity, 'P132' expense_type, 34000 budget
   23    from dual
   24    union all
   25    select '801040' entity, 'P139' expense_type, 56000 budget
   26    from dual
   27  )
   28  select *
   29    from (
   30          select nvl(o.entity, r.entity) entity,
   31                 nvl(o.expense_type, r.expense_type) expense_type,
   32                 nvl(r.budget,0) - nvl(o.budget,0) budget
   33            from original o
   34            full join revised r
   35              on o.entity = r.entity
   36             and o.expense_type = r.expense_type
   37         )
   38   where budget <> 0
   39  /
  
  ENTITY EXPE     BUDGET
  ------ ---- ----------
  801040 2186       1000
  801040 P135     -43000
  801040 3001      -1000
  801040 P139      56000
  
  SQL>
  

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  You posted this in the two forums, its been answered in the mac forum.