Results of the difference of two queries with a result zero

Similar Questions

• Help to combine the two queries with dependencies

  Combining two queries with dependencies:
  First ask: Returns USER_ID that has a type of access to a table with the OWNER (I also shoot the TLA of the USER_ID standard: ' of TLA #)

  Select distinct SUBSTR (DBA_USERS. Username, 2, 3) AS TLA

  DBA_USERS. Username USED_ID

  DBA_TAB_PRIVS. Owner OWNER

  from DBA_USERS

  DBA_ROLE_PRIVS

  DBA_TAB_PRIVS

  where DBA_USERS. Username = DBA_ROLE_PRIVS. Dealer

  and DBA_USERS.default_tablespace not in ('SYSTEM', "SYSAUX")

  and DBA_USERS. Username like would be %'

  and DBA_ROLE_PRIVS. Granted_role = DBA_TAB_PRIVS. Dealer

  and SUBSTR (DBA_USERS. Username, 2, 3) <>DBA_TAB_PRIVS. Owner

  ;

  Example of results (red font used in the last citation):

  TLA USED_ID OWNER

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

  DGTX999 GTX GRR

  DGTX999 GTX ABG

  DGTX999 GTX HTC

  FWS DFWS999 GRR

  OCN DOCN999 GRR

  RHR DRHR999 DAS

  ETM DETM999 DAS

  FWS DFWS995 DAS

  CD DCDS999 DAS

  ABE DABE999 DAS

  
  
  Second request: matches the USER name (same standards as above) and MACHINES (filter PC connections) and the number of connections made of

  Select DBA_USERS. Username

  DBA_HIST_ACTIVE_SESS_HISTORY. Machine

  COUNT (*) AS CONN_COUNT

  of DBA_HIST_ACTIVE_SESS_HISTORY

  DBA_USERS

  where DBA_HIST_ACTIVE_SESS_HISTORY. User_id = DBA_USERS. User_id

  and DBA_USERS. Username like would be %'

  and DBA_HIST_ACTIVE_SESS_HISTORY. Machine not as "WINAD%\%-%".

  Group of DBA_USERS. Username

  DBA_HIST_ACTIVE_SESS_HISTORY. Machine

  order by 1

  3 desc;

  Example of results (red font used in the last citation):

  MACHINE CONN_COUNT USERNAME

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

  DFWS999 home13e0 7557

  DGAM800 vu1246 37

  DGAM999 vu1246 2558

  DABE999 back18cb 4639

  DINL888 home162f 84

  DFWS999 WINDAD\OMHQ58BF 178

  DCDS999 back18cb 788

  DGTX999 home18c8 7

  DGTX999 home13d3 62

  DOCN999 vu1261 44

  DFWS999 back1976 3

  DCDS999 home18c8 173

  DGTX999 home19c9 13

  What I'd like to have (just made two first USER_IDs from the first query) (COL duplicated for each separate owner 1-4).  So 3 machines and 3 owners would result in 9 lines (I could reorganize the collar and use a 'pause'):

  TLA USED_ID MACHINE CONN_COUNT OWNER

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

  GTX DGTX999 home13d3 62 GRR

  Home18c8 GTX DGTX999 7 GRR

  DGTX999 GTX home19c9 GRR 6

  GTX DGTX999 home13d3 62 ABG

  DGTX999 GTX home18c8 ABG 7

  DGTX999 GTX home19c9 6 GBS

  GTX DGTX999 home13d3 62 CTH

  DGTX999 GTX HTC 7 home18c8

  DGTX999 GTX home19c9 6 HTC

  FWS DFWS999 home13e0 7557 GRR

  FWS DFWS999 WINDAD\OMHQ58BF 178 GRR

  FWS DFWS999 back1976 GRR 3

  
  So it would be logically read: TLA as USER_ID of MACHINE, CONN_COUNT consultation times: OWNER information.

  Thought as only those with the necessary values. Just try this and let me know in the case of all conflicts in the output.

  WITH T1 AS)

  Select distinct SUBSTR (DBA_USERS. Username, 2, 3) AS TLA

  DBA_USERS. Username USED_ID

  DBA_TAB_PRIVS. Owner OWNER

  ROWNUM RN

  from DBA_USERS

  DBA_ROLE_PRIVS

  DBA_TAB_PRIVS

  where DBA_USERS. Username = DBA_ROLE_PRIVS. Dealer

  and DBA_USERS.default_tablespace not in ('SYSTEM', "SYSAUX")

  and DBA_USERS. Username like would be %'

  and DBA_ROLE_PRIVS. Granted_role = DBA_TAB_PRIVS. Dealer

  and SUBSTR (DBA_USERS. Username, 2, 3) <> DBA_TAB_PRIVS. Owner),

  () AS T2

  Select DBA_USERS. Username

  DBA_HIST_ACTIVE_SESS_HISTORY. Machine

  COUNT (*) AS CONN_COUNT

  of DBA_HIST_ACTIVE_SESS_HISTORY

  DBA_USERS

  where DBA_HIST_ACTIVE_SESS_HISTORY. User_id = DBA_USERS. User_id

  and DBA_USERS. Username like would be %'

  and DBA_HIST_ACTIVE_SESS_HISTORY. Machine not as "WINAD%\%-%".

  Group of DBA_USERS. Username

  DBA_HIST_ACTIVE_SESS_HISTORY. Machine

  order by 1

  3 desc)

  SELECT T1. TLA

  ,         T1. USED_ID

  ,         T2. MACHINE

  ,         T2. CONN_COUNT

  ,         T1. OWNER

  FROM T1

  T2

  WHERE T1. USED_ID = T2. USER NAME;

• 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

• Dynamic action - Get the difference between two dates + times

  I have problems a little dynamic to work action. I'm trying to get the time between two dates with the time difference.
  
  Here is what I got (this is apex 4.0):
  
  Two date pickers + two numbers fields (date/start/end times)
  
  I created a dynamic action on the page who fires on the point lose focus (above points).
  
  The real action for the DA is the body of the PL/SQL function:

  declare
    end_date DATE;
    start_date DATE;
  Begin
    start_date := to_char(:P1_START_DATE || ' ' || :P1_START_TIME, 'DD-MON-YYYY HH:MIAM');
    end_date := to_char(:P1_END_DATE || ' ' || :P1_END_TIME, 'DD-MON-YYYY HH:MIAM');
    :P1_HOURS := end_date-start_date;
  End;

  When I change the values on the page, I get the following error:
  
  AJAX call back Server error ORA-06502: PL/SQL: digital or value error: character number conversion error to set the value.
  
  I'm guessing that there is a problem with the date formatting, but I can't make it work. Thanks in advance!

  Hi djston,

  because you chose the dynamic action of 'Set value' with the "Body of the PL/SQL function" type you need to return the value. Try the following code

  declare
    end_date DATE;
    start_date DATE;
  Begin
    start_date := to_date(:P1_START_DATE || ' ' || :P1_START_TIME, 'DD-MM-YYYY HH:MIAM');
    end_date := to_date(:P1_END_DATE || ' ' || :P1_END_TIME, 'DD-MM-YYYY HH:MIAM');
    RETURN (end_date-start_date)*24;
  End;
  

  and P1_REQUESTED_HOURS like 'item affected. "

  Concerning
  Patrick
  -----------
  My Blog: http://www.inside-oracle-apex.com
  APEX 4.0 Plug-Ins: http://apex.oracle.com/plugins
  Twitter: http://www.twitter.com/patrickwolf

  Published by: Patrick Wolf on January 17, 2011 10:54

• compare the fields of two queries

  I need to compare the fields of two queries.
  
  I went through many articles to asktom help operator less. but I'm not convinced by this method.
  
  Comaprisons are for about 10 fields.
  
  So I'm planing to do using sliders and nested loops. performance can be a problem.
  
  Please give your suggestions or solutions.

  OK, using the data you provided, I get this:

  with map_test_fac as (select 100 idid, 789 ifid, 300 edid, 741 efid from dual union all
                        select 100 idid, 123 ifid, 300 edid, 963 efid from dual union all
                        select 100 idid, 456 ifid, 300 edid, 258 efid from dual),
           test_fac as (select 741 facid, 'ETY' type, 'EUR' ccy, 'Ext' src, 300 did from dual union all
                        select 789 facid, 'WE' type, 'EUR' ccy, 'Int' src, 100 did from dual union all
                        select 123 facid, 'VPR' type, 'EUR' ccy, 'Int' src, 100 did from dual union all
                        select 456 facid, 'ETY' type, 'USD' ccy, 'Int' src, 100 did from dual union all
                        select 963 facid, 'WE' type, 'EUR' ccy, 'Ext' src, 300 did from dual union all
                        select 258 facid, 'VPR' type, 'EUR' ccy, 'Ext' src, 300 did from dual)
  -- end of mimicking your data: USE SQL below:
  select tfi.facid int_facid,
         tfi.type int_type,
         tfi.ccy int_ccy,
         tfi.src int_src,
         tfi.did int_did,
         tfe.facid ext_facid,
         tfe.type ext_type,
         tfe.ccy ext_ccy,
         tfe.src ext_src,
         tfe.did ext_did
  from   test_fac tfi,
         test_fac tfe,
         map_test_fac mtf
  where  mtf.ifid = tfi.facid
  and    mtf.efid = tfe.facid
  and    mtf.idid = tfi.did
  and    mtf.edid = tfe.did
  and    (tfi.type != tfe.type
          or tfi.ccy != tfe.ccy)
  
   INT_FACID INT INT INT    INT_DID  EXT_FACID EXT EXT EXT    EXT_DID
  ---------- --- --- --- ---------- ---------- --- --- --- ----------
         789 WE  EUR Int        100        741 ETY EUR Ext        300
         123 VPR EUR Int        100        963 WE  EUR Ext        300
         456 ETY USD Int        100        258 VPR EUR Ext        300
  

  However, I think that your mapping data is incorrect, so with keys of information:

  with map_test_fac as (select 100 idid, 789 ifid, 300 edid, 963 efid from dual union all
                        select 100 idid, 123 ifid, 300 edid, 258 efid from dual union all
                        select 100 idid, 456 ifid, 300 edid, 741 efid from dual),
           test_fac as (select 741 facid, 'ETY' type, 'EUR' ccy, 'Ext' src, 300 did from dual union all
                        select 789 facid, 'WE' type, 'EUR' ccy, 'Int' src, 100 did from dual union all
                        select 123 facid, 'VPR' type, 'EUR' ccy, 'Int' src, 100 did from dual union all
                        select 456 facid, 'ETY' type, 'USD' ccy, 'Int' src, 100 did from dual union all
                        select 963 facid, 'WE' type, 'EUR' ccy, 'Ext' src, 300 did from dual union all
                        select 258 facid, 'VPR' type, 'EUR' ccy, 'Ext' src, 300 did from dual)
  -- end of mimicking your data: USE SQL below:
  select tfi.facid int_facid,
         tfi.type int_type,
         tfi.ccy int_ccy,
         tfi.src int_src,
         tfi.did int_did,
         tfe.facid ext_facid,
         tfe.type ext_type,
         tfe.ccy ext_ccy,
         tfe.src ext_src,
         tfe.did ext_did
  from   test_fac tfi,
         test_fac tfe,
         map_test_fac mtf
  where  mtf.ifid = tfi.facid
  and    mtf.efid = tfe.facid
  and    mtf.idid = tfi.did
  and    mtf.edid = tfe.did
  and    (tfi.type != tfe.type
          or tfi.ccy != tfe.ccy)
  
   INT_FACID INT INT INT    INT_DID  EXT_FACID EXT EXT EXT    EXT_DID
  ---------- --- --- --- ---------- ---------- --- --- --- ----------
         456 ETY USD Int        100        741 ETY EUR Ext        300
  

  That's more like what you're looking for?

• How to quickly find the difference of two version of RPD?

  Hi all
  
  Someone has a good idea to quickly find the difference of two versions of the RPD in the project?
  
  
  Kind regards
  Anne

  This could be a help for you
  http://varanasisaichand.blogspot.com/2010/03/how-to-merge-two-repositories.html
  http://www.rittmanmead.com/2010/07/OBIEE-11gr1-incremental-patches-to-the-RPD/

  See you soon,.
  KK

  Published by: Kranthi.K on December 7, 2011 21:41

• 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;
  

• How to calculate the difference between two times by the NUMBERS

  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?

  Kind regards

  Ian.

• outer join: difference between two queries

  Below two queries which should give the same results in my opinion. I want that all records of u_protocol and the pval.u_protocol_variable_value value if necessary.
  
  Why the outer join in query2 does not work as in query1?
  
  
  Query1:

  select p.u_protocol_id, i.u_protocol_variable_value
  from lims_sys.u_protocol p,
  (
     select pval.u_protocol_id, pval.u_protocol_variable_value
     from lims_sys.u_protocol_variable pvar, lims_sys.u_protocol_value_user pval
     where pvar.u_protocol_variable_id = pval.u_protocol_variable_id
     and pvar.name = 'VALUE_Protocol_Group'
  ) i   
  where p.u_protocol_id  = i.u_protocol_id (+)

  Query2:

  select prt.u_protocol_id, pval.u_protocol_variable_value
  from lims_sys.u_protocol prt, lims_sys.u_protocol_variable pvar, lims_sys.u_protocol_value_user pval
  where pvar.u_protocol_variable_id = pval.u_protocol_variable_id
  and prt.u_protocol_id = pval.u_protocol_id (+)
  and pvar.name = 'VALUE_Protocol_Group'

  Hello

  When you perform an outer join, all conditions on the tables in option must be outer join conditions. Otherwise, the effect is an inner join.
  There is one exception: cascade of outer joins, external-junction b of a and c is outside attached to b. In this case, in the b - c join only columns of c are marked with the sign +.

  Re - write to obtain the same results as query1, query2 like this:

  select  prt.u_protocol_id
  ,      pval.u_protocol_variable_value
  from     lims_sys.u_protocol          prt
  ,      lims_sys.u_protocol_variable     pvar
  ,      lims_sys.u_protocol_value_user pval
  where     pvar.u_protocol_variable_id (+)     = pval.u_protocol_variable_id     -- cascading outer join
  and     prt.u_protocol_id           = pval.u_protocol_id (+)
  and     pvar.name (+)               = 'VALUE_Protocol_Group'     -- (+) added
  ;
  

  I find the syntax ANSI easier to use for all the joins and much easier for outer joins.

• Computer with Vista unable to connect to the internet through two computers with Windows 7.

  Dear engineers,

  In our House, we have two new computers Windows 7 connect to Wired internet access.  Our friend came in hoping to play some MMO games with us on our connection (it has windows vista).  When we hung everything upward, by linking to our modem, his computer was visible on my husbands computer, but was visible on mine as a media device.  He could see the two computers, but could not get an internet connection and no viable ping.  We checked the network map and he showed my computer with one? related to my husband then connected to the internet.  Inside, there was an announcement indicating that the following devices are not displayed in the network map.  under of course was listed both our friends media device and the computer device.  I searched Control Panel and gave access to anyone on the login to my husbands system as long as me.  What happens if nothing else can I try to have its system you can connect to our internet?

  Thank you for your attention,

  Cerisea

  Hello

  If a computer is not properly visible it generally means that sharing is not set correctly to match your system.

  You must also set your computers with the computer account of the friend as a user on your computer.

  Everyone does not mean everyone wishing to connect.

  If the accounts are presented only for John, Jill and Jack, setting in the folder with all permissions saves the effort of definition permission separately for John, Jill and Jack.

  Everyone isn't everyone group, only John, Jill and Jack.

  In general.

  Win7 when configured on the peer-to-peer network has three types of configurations of sharing.

  Group residential network = only works between Win 7 computers. This type of configuration, it is very easy to entry level users to start sharing network.

  Working network = fundamentally similar to previous methods of sharing that allow you to control what, how and to whom the records would be shared with.
  
  Public share = network Public (as Internet Café) in order to reduce security risks.

  For the best newspaper of the results of each computer screen system and together all computers on a network of the same name, while each computer has its own unique name.

  http://www.ezlan.NET/Win7/net_name.jpg

  Make sure that the software firewall on each computer allows free local traffic. If you use 3rd party Firewall on, Vista/XP Firewall Native should be disabled, and the active firewall has adjusted to your network numbers IP on what is sometimes called the Zone of confidence (see part 3 firewall instructions

  General example, http://www.ezlan.net/faq#trusted
  Please note that some 3rd party software firewall continue to block the same aspects it traffic Local, they are turned Off (disabled). If possible, configure the firewall correctly or completely uninstall to allow a clean flow of local network traffic. If the 3rd party software is uninstalled, or disables, make sure Windows native firewall is active .

  Win 7 networking with other version of Windows as a work network.

  In the center of the network, by clicking on the type of network opens the window to the right.

  Choose your network type. Note the check box at the bottom and check/uncheck depending on your needs.

  http://www.ezlan.NET/Win7/net_type.jpg

  Win 7 - http://windows.microsoft.com/en-us/windows7/Networking-home-computers-running-different-versions-of-Windows

  Win 7 network sharing folder specific work - http://www.onecomputerguy.com/windows7/windows7_sharing.htm

  Vista file and printer sharing - http://technet.microsoft.com/en-us/library/bb727037.aspx

  When you have finished the configuration of the system, it is recommended to restart everything the router and all computers involved.

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

  If you have authorization and security problems, check the following settings.

  Point to a folder that wants to share do right click and choose Properties.

  In the properties

  Click on the Security tab shown in the bellows of the photo on the right) and verify that users and their permissions (see photo below Centre and left) are configured correctly. Then do the same for the authorization tab.

  This screen shot is to Win 7, Vista menus are similar.

  http://www.ezlan.NET/Win7/permission-security.jpg

  The Security Panel and the authorization Panel, you need to highlight each user/group and consider that the authorization controls are verified correctly.

  When everything is OK, restart the network (router and computer).

  * Note . The groups and users listed in the screen-shoot are just an example. Your list will focus on how your system is configured.

  * Note . There must be specific users. All means all users who already have an account now as users. This does not mean everyone who feel they would like to connect.

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

  *** Note. Some of the processes described above are made sake not for Windows, but to compensate for different routers and how their firmware works and stores information about computers that are networked.

  Jack-MVP Windows Networking. WWW.EZLAN.NET

• Take the difference of two tables

  Hello

  I have two tables where in I need to take the difference between the two columns of different tables and show as decimal below:

  Table A

  Name, salary

  x 10.5

  y - 10.32

  z 5

  Table B

  Name, salary

  x 10.5

  y - 10.32

  z 3

  a 12

  Output:

  Name_A, Name_B, salary, remarks

  x, x, 0, no diff

  y, y, 0, no diff

  z, z, 2 Diff is here

  , one, no data, no data from the previous

  Hello

  Try ythis:

  with table_a as
  (
  SELECT 'X' NAME, SALARY, 10.5 DUAL UNION ALL
  SELECT THE NAME OF 'Y',-10.32 SALARY OF DOUBLE UNION ALL
  SELECT THE NAME OF 'Z', 5 DOUBLE SALARY
  )
  table_B as
  (
  SELECT 'X' NAME, SALARY, 10.5 DUAL UNION ALL
  SELECT THE NAME OF 'Y',-10.32 SALARY OF DOUBLE UNION ALL
  SELECT THE NAME OF 'Z', 3 DOUBLE UNION ALL SALARIES
  SELECT 'A' NAME, SALARY 12 DUAL FROM
  )
  SELECT
  COALESCE(A.NAME,B.NAME) NAME
  , NVL (TO_CHAR (A.SALARY - B.SALARY), 'NO DATA') SALARY
  , CASE WHEN A.SALARY - B.SALARY = 0 THEN 'NO DIFFERENCE '.
  WHEN B.SID IS NULL, "NO DATA OF THE PREVIOUS.
  WHERE B.NAME IS NULL, "NO DATA FROM NEXT.
  ANOTHER 'DIFFERENCE IS THERE.
  END NOTE

  Of
  TABLE_A HAS
  TABLE_B FULL OUTER JOIN B ON (B.SID = B.NAME)
  ;

  NAME SALARY NOTE
  ---- ---------------------------------------- ---------------------
  X 0 NO DIFFERENCE
  Y 0 NO DIFFERENCE
  Z 2 THE DIFFERENCE IS THERE
  A NUMBER OF DATA NO DATA FROM THE PREVIOUS

  Kind regards

  Peter

• 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>
  

• How can I compare the differences between two files After Effects?

  I have a major problem.  I was working on special effects for a film.  In After Effects, when I change siggificant, I often have to register under and create a new name so I can go back to the old work if necessary.  It comes in a case, I have come back from earlier works.  If I use these two files After Effects allows you to create multiple clips of effects.  And I do not remember just where you look at it, it is better.  Does anyone know a good way to compare the differences between the files?

  One thing to keep this conversation on the right way: I know that I did a stupid thing.  I've never done this before so can we please not spend a lot of time on how to avoid this problem and stick to how we solve this problem?

  If you encounter difficulties to identify differences in the old and new versions, you can use this procedure:

  To avoid confusion, I'll call your former company and your most recent compositions CompB comps.

  CompA drag "New Comp" icon in the project window.  This will create a new temporary layout that contains the nested CompA.

  Drag in your new COMP CompB temporary ensure it aligns image for image with CompA.

  Place the top layer (CompB) difference.

  Now, when you play through the comp, you'll see differences in color at a time where the comparison and CompB are not identical.

• How to tell the difference between two dates

  Hello:
  
  I have a question on how to get the difference in quantities for two dates in the replies and dashboard.
  
  This is the scenario. I have two guests date and three columns. The first column must be the sum of the amounts for the first quick date. The second column should be the sum of the amounts for the second fast date. And the third column is the difference between the first two columns.
  
  Basically, I'm trying to do something similar to a function of sum of two lines but trying to get the difference between the two rows. Any help to solve this problem would be appreciated.

  Use the COLUMN FILTERS level...

  the first column with the first variable of date quick presentation of the filter

  Similarly, the second column in the second variable of quick overview of date filter

  Kind regards
  Rambeau