outer join: difference between two queries

Similar Questions

• 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.

  

  

  Any help would be greatly appreciated!

  Thank you

  this.rawValue = (totalHrs > 9? String (totalHrs): '0' + String (totalHrs)) + ':' + (totalMins > 9? String (totalMins): '0' + String (totalMins));

• Difference between two data aquistion

  Hello

  I got a data aquistion system that is supposed to keep data acquisition selection until the difference between two points a lot. For example, I get voltage (mv) as 2.31, 2.32, 2.34. 2.33 2.32, 2.33 2.32. the data up to a point, say 0.1V higher or lower than the first. Thewhole procss stop. Thank you for your suggestions.

  Try this... (Labview 8.2)

• using outer joins if the two column is null? Use only (+)

  Hi all

  create the table xxc_tr_num (tl_number number, tr_no number tl_no_id);

  insert into xxc_tr_num values (123,100,222);

  insert into xxc_tr_num values (124,100,333);

  create the table xxc_od_tab (tl_number number, tl_id number);

  insert into xxc_od_tab values (123,001);

  insert into xxc_od_tab values (null, null);

  create table xxc_oth_tab (name varchar2 (10), number of tl_id);

  insert into xxc_oth_tab values('abc',,001);

  insert into xxc_oth_tab values (null, null);

  Wait it out put

  tr_no tl_no_id name

  100 222 abc

  100 333

  using outer joins if the two column is null? use only please of outer joins

  And I tried to use outer joins on both tl_id column but not get values and I use have County (tr_no ) > 1

  Rajesh123 wrote:

  Thank you Kiss it is not possible to use having clause?

  You need to understand the functioning of the group. If you will not be asked this question.

  Check this box

  SQL> select tr_no,
    2         tl_no_id,
    3         count(*)
    4    from xxc_tr_num a,
    5         xxc_od_tab b,
    6         xxc_oth_tab c
    7   where a.tl_number = b.tl_number(+)
    8     and b.tl_id = c.tl_id(+)
    9   group
   10      by tr_no
   11       , tl_no_id;
  
       TR_NO   TL_NO_ID   COUNT(*)
  ---------- ---------- ----------
         100        333          1
         100        222          1
  

  See what returns the count? You have grouped according to TR_NO and TL_NO_ID. You must take into consideration the TL_NO_ID just put COUNT (TR_NO) does not increase the NUMBER of the whole group. To get the NUMBER on the whole group, I used the analytical function and did. Like this, see the number of the analytical function here

  SQL> select tr_no,
    2         tl_no_id,
    3         count(*),
    4         count(*) over(partition by tr_no)
    5    from xxc_tr_num a,
    6         xxc_od_tab b,
    7         xxc_oth_tab c
    8   where a.tl_number = b.tl_number(+)
    9     and b.tl_id = c.tl_id(+)
   10   group
   11      by tr_no
   12       , tl_no_id;
  
       TR_NO   TL_NO_ID   COUNT(*) COUNT(*)OVER(PARTITIONBYTR_NO)
  ---------- ---------- ---------- ------------------------------
         100        222          1                              2
         100        333          1                              2
  

  So to answer your question, yes you can't do in the HAVING clause...

• difference between two dates in the year/month/day in Obiee

  Hi gurus,

  I know this question have been asked and answered several times but I have a requirement that is a little different, then the previous ones.

  I want to calculate the difference between two dates in OBIEE10g in year/month/day format similar to the below SQL output

  SQL > select end_date, start_date,

  trunc (months_between (end_date, start_date) / 12) years.

  months of mod (trunc (months_between (end_date, start_date)), 12).

  End_date - add_months (start_date, trunc (months_between (end_date, start_date))) days

  t

  Thanks in advance

  SK

  Search for this

  "TIMESTAMPDIFF IN THE FORM OF MTHS # YEARS."

  or else

  To get the current mandate of employees since the date of hiring in the form of # years # mths or # year (s) # month (s)

  Concat (concat (cast (TIMESTAMPDIFF (SQL_TSI_Month, "Employee attributes". (((' ' Hiring last Date employee ", CURRENT_DATE) / 12 as char),"Year (s)"), concat (cast (MOD (TIMESTAMPDIFF (SQL_TSI_Month,"Employee attributes". (((((' ' Hiring last Date employee ", CURRENT_DATE), 12) as char), 'Month (s)'))

  Concat (concat (CAST (TIMESTAMPDIFF (SQL_TSI_YEAR, "Employee attributes". (((' ' Hiring last Date employee ", CURRENT_DATE) as CHAR),"Year (s)"), concat (cast (MOD (TIMESTAMPDIFF (SQL_TSI_Month,"Employee attributes". (((((' ' Hiring last Date employee ", CURRENT_DATE), 12) as char), 'Mth (s)'))

• difference between two timestamps

  Hi gurus,

  I have a procedure which has 2 input parameters of type timestamp. I need to find the difference between these two stamps and be converted to a timestamp and return. Help, please

  Diifernce between two timestamps is interval. You can not convert the difference between two timestamp timestamp and it makes no sense. What you expect:

  25/06/2014 11:25:37.524000 - 13:07:22.987654 20/06/2014

  SY.

• Exact differences between two images

  I need to be able to tell the difference between two images of a pastel artwork.

  
  

  The first image is of the piece, to be hung on a wall or placed on a table.

  
  

  The second is a photo of the piece after it has been moved, shaken or vibrated during which time particles were moved.

  
  

  I tried superimposing the two images and see the difference between the layers.  The images never seem to perfectly align along the lines of reference drawn in ink on the piece, and I do not understand the presentation of grayscale of the layer mode of difference.  I would like to know if anyone has any photographic photoshopic, or insightful ideas on how to capture this specific detail of change.  Another big challenge is now the images themselves to the exact distance and exposure to the room at the before and after photos, especially if they were taken in two different places on different dates.

  Best I can think of is to load into Photoshop layers >, select the two layers and go Edit > auto alignment of layers.  Then to the blending mode of the layer from the top to the difference.

  You can test so that you know what to expect by opening an image and copy the layer of BG and setting the blending mode of the layer copied the difference.  It will show as a pure black.  Now make tiny changes to the upper layer.

• How do the time difference between two dates?

  Hi all
  
  I use this query to get the time difference between two dates.
  
  Select to_timestamp ('2012-10-03 12:00 ',' YYYY-MM-DD hh)-to_timestamp ('2012-10-03 11:00 ',' YYYY-MM-DD hh) as double diff;
  
  but do not get the correct result.
  
  Thank you

  Left KEY... Left Padding of tanks.

• How to find the difference between two dates in the presentation layer

  Hi gurus,
  
  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".

  If you want to display the difference between two dates in days using SQL_TSI_DAY, unlike butterflies SQL_TSI_MONTH and so on...

  hope you understand...

  Award points and to close the debate, if your question is answered.

  See you soon,.
  Aravind

• 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;
  

• 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>
  

• Find the difference between two numbers (DBL)

  Hi all

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

  Thank you

  Davidson

  
  

• Difference between two numbers of laptop models

  Can someone tell me if there is a difference between these models of computers two laptops? As far as I can tell they are the same.
  M7-j120dx and M7-J010D. Thank you.

  Hello

  To the second machine:

  http://support.HP.com/us-en/document/c03770540

  You can check yourself, BUT I found at least 2 things that tell me that the first one is the best: it has screen FULL HD and better card WiFi (b/g/n a/c).

  Kind regards.

• Difference between two dates

  Hello

  I have two columns both of them date data types.

  I would like to know the difference between these two dates.

  SELECTION INTERVAL ' + 2342 23:23:23 ' DAY (4) TO THE SECOND - INTERVAL ' + 102 13:13:13 ' DAY (4) SECOND FROM DUAL;

  But, I get this:

  '+ 2240 10:10:10.000000'

  I'd like to see the result in a year-month and jour-deuxieme groups. I can't do anything with + 2240 days...

  Thanks in advance!

  Hello Adme12

  You can use it.

  SET LINESIZE 110 PAGESIZE 20

  COLUMN today FORMAT A20

  COLUMN TOMORROW FORMAT A20

  COLUMN YEAR_MONTHS FORMAT A15

  COLUMN DAY_SEC FORMAT A30

  SET ECHO ON

  SELECT TO_CHAR (today, ' JJ.) MR. HH24:MI:SS' YYYY) LIKE TODAY

  , TO_CHAR (tomorrow, ' JJ.) MR. HH24:MI:SS' YYYY) TOMORROW

  NUMTOYMINTERVAL (MONTHS_BETWEEN (TOMORROW, TODAY), 'MONTHS') AS YEAR_MONTHS

  , NUMTODSINTERVAL (TOMORROW - ADD_MONTHS (TODAY, TRUNC (MONTHS_BETWEEN (TOMORROW, TODAY))), 'DAY') AS DAY_SEC

  Of

  (SELECT SYSDATE + INTERVAL ' + 102 13:13:13 ' DAY (4) ON THE SECOND THAT TODAY ' HUI)

  , SYSDATE + INTERVAL ' + 2342 23:23:23 ' DAY (4) ON THE SECOND TOMORROW

  OF THE DOUBLE

  )

  ;

  TODAY TOMORROW YEAR_MONTHS DAY_SEC

  -------------------- -------------------- --------------- ------------------------------

  23.04.2014 04:49 10.06.2020 14:59:10 + 000000006-02 + 000000018 10:10:10.000000000

  Of course, it is not so good that you have so many 0, but you can convert the result to a char (TO_CHAR) and you can enter.

  I hope this helps you!

  Kind regards

  David

• Difference between two statement.

  I use 11.2.0.3 oracle version, and when I am running below pieces of code and expect similar results.

  Below is the variable declare package level as below.

  TYPE mycur IS REF CURSOR;

  Now mentioned below is my procedure inside the package. My question is, I want to understand the difference between the following two statement in the same procedure. as as statement: 1 give me several records to pk1, where as statement: 2 gives me the good result which is recorded under the outlet of the slider back.

  Statement: 1

  PROCEDURE p1 (pk1 in NUMBER,
  ret_cursor ON mycur)
  AS
  
  BEGIN
  Open Select ret_cursor *.
  A,.
  b,
  WHERE a.id = b.id
  AND (a.c1 = b.c1
  GOLD a.c2 = b.c2)
  AND a.c3 = pk1;
  
  END;
  /
  
  Statement: 2
  
  PROCEDURE p1 (pk1 in NUMBER,
  ret_cursor ON mycur)
  AS
  V_QUERY VARCHAR2 (4000);
  
  BEGIN
  V_QUERY: =.
  "SELECT."
  A,.
  b,
  WHERE a.id = b.id
  AND (a.c1 = b.c1
  GOLD a.c2 = b.c2)
  AND a.c3 = ' | PK1;
  
  Ret_cursor OPEN for V_QUERY;
  END;
  
  

  actually my mistake, I just the reason behind the question. I had published the name of the table variable / by their and are different than real.
  So the question was naming convention given by the developers, the name of the input variable looked exactly like that of the name of the column, it's why he estimated at something like 'and (1 = 1)'.

  The input variable name has been 'c3' as column name 'a' causing problem.

  a.C3 = c3