Compare the min and max columns

Similar Questions

• the min and MAX values maintain their good values tia sal22

  the min and MAX values maintain their good values tia sal22

  Hi all

  I have problems a little value to stay in their correct min and max values min and max.
  Continue to change their values (up and down) trying to get the max value to only display the maximum value at a given time and the same for the min value.  I looked at the examples max and min and it seems that it should be in a while loop
  http://forums.NI.com/NI/board/message?board.ID=170&thread.ID=359822

  but I would add another while loop and if so put it in the loop for? or is there a better way to do it.

  TIA sal22

  Hi sal22,

  I thought it was just one example you found somewhere, because it is called red rooster. See the attached example.

  Mike

• Delete the min and max labels on chart axis markers

  Hello

  Is it possible to remove the min and max labels on chart axis markers? For example, see the screenshot below. Can I remove the numbers circled in green without changing the range of the axis? In some cases (like this one), the presence of the min and max number of cases other marker values to not be displayed (red line). I am aware that "10" is displayed if I enlarge the graph, but this isn't an option for my application. FYI - the screenshot is from the side before the attached VI.

  Thanks for any help,

  Kind regards

  Stu

  

  You can't delete them because they are the main features to be able to set the max and min on the graph of the Panel before if you use no-AutoScaling.

• Grouping and then find the min and max

  DROP TABLE purge of y;

  CREATE TABLE y
  (
  ID NUMBER (10),
  START_DATE_TIME NUMBER (15),
  END_DATE_TIME NUMBER (15)
  );

  INSERT INTO y VALUES (1, 20140602164819, 20140606140851);
  INSERT INTO y VALUES (1, 20140827141743, 20140827142131);
  INSERT INTO y VALUES (1, 20140827141744, 20140827141835);
  INSERT INTO y VALUES (1, 20140827141744, 20140827142131);
  INSERT INTO y VALUES (1, 20140827141944, 20140827142131);
  INSERT INTO y VALUES (1, 20140827141944, 20140924142131);
  INSERT INTO y VALUES (1, 20140927141944, 20141027142131);
  INSERT INTO y VALUES (2, 20140602164819, 20140606140851);
  INSERT INTO y VALUES (2, 20140827141743, 20140827142131);
  INSERT INTO y VALUES (2, 20140827141744, 20140827141835);
  INSERT INTO y VALUES (2, 20140827141744, 20140827142131);
  INSERT INTO y VALUES (2, 20140827141944, 20140827142131);
  INSERT INTO y VALUES (2, 20140827141944, 20140924142131);
  INSERT INTO y VALUES (2, 20140927141944, 20141027142131);

  COMMIT;

  Select
  *
  Of
  There
  order by 2, 3;

  out necessary: when the output of the table are sorted by start_date_time and end_date_time in ascending order.
  and when the value of for a particular id start_date_time is located between the
  another set of values start_date_time and end_date_time for the same id then min (start_date_time) and max (end_date_time) is the desired output.

  and if there is no overlap then (start_date_time) min and max (end_date_time) is the output desired.

  and for each line of the output in ascending order of line number order addd;

  Example output is in the output table named

  drop table output is serving;

  create an output table
  (
  ID NUMBER (10),
  START_DATE_TIME NUMBER (15),
  END_DATE_TIME NUMBER (15),
  ROW_NUMBER NUMBER (10)
  );

  
  Values to INSERT OUTPUT (1, 20140602164819, 20140606140851, 1);
  Values to INSERT OUTPUT (1, 20140827141743, 20140924142131, 2);
  Values to INSERT OUTPUT (1, 20140927141944, 20141027142131, 3);
  Values to INSERT OUTPUT (2, 20140602164819, 20140606140851, 1);
  Values to INSERT OUTPUT (2, 20140827141743, 20140924142131, 2);
  Values to INSERT OUTPUT (2, 20140927141944, 20141027142131, 3);

  COMMIT;

  SELECT
  *
  Of
  output
  ORDER BY 1,2,3,4;

  Thank you

  SQL_Novice

  Hello

  Your postal code.  It is difficult to tell what you're doing wrong without knowing what you are doing.

  user6166680 wrote:

  Hi Frank:

  Max works for ID 3 values only because there is only one group here, as all other values compared, they overlap with max (end_date_time);

  but when I use MAX for ID 1 and 2, the result set is not the same thing as LAG works because there is a GAP...

  Exactly, you have to find the gaps.  If start_date_time is less than or equal to the MAXIMUM of all the end_date_times earlier, then it isn't a gap; Otherwise, there is a gap.

  What is the problem with MAX?

  WITH got_new_group AS

  (

  SELECT id, start_date_time and end_date_time

  CASE

  WHEN start_date_time <= > MAX (end_date_time) OVER (PARTITION BY id )

  ORDER BY start_date_time

  end_date_time

  ROWS BETWEEN UNBOUNDED PRECEDING

  AND 1 PRECEDING

  )

  THEN 0

  1. OTHER

  END AS new_group

  OF y

  )

  got_group_number AS

  (

  SELECT id, start_date_time and end_date_time

  SUM (new_group) OVER (PARTITION BY ID.

  ORDER BY start_date_time

  end_date_time

  ) AS group_number

  OF got_new_group

  )

  SELECT id

  MIN (start_date_time) AS group_start_date_time

  MAX (end_date_time) AS group_end_date_time

  group_number

  OF got_group_number

  GROUP BY id, group_number

  ORDER BY id, group_number

  ;

  You will notice that it is exactly what I posted earlier, in response #5, except that instead of using the OFFSET it uses MAX.

  Output:

  ID GROUP_START_DATE_TIME GROUP_END_DATE_TIME GROUP_NUMBER

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

  1 20140602164819 20140606140851 1

  1 20140827141743 20140924142131 2

  1 20140927141944 20141027142131 3

  2 20140602164819 20140606140851 1

  2 20140827141743 20140924142131 2

  2 20140927141944 20141027142131 3

  3 20140330041350 47121231000000 1

• Using the MIN and MAX functions

  Hello
  
  Guide to please the following:
  
  I have a table with records like below:
  
  
  Invoice ID RegistrationNo InvoiceDate
  1 AB-1 JANUARY 1, 10
  2 CB-4 MARCH 23, 10
  3 AB-1 1ST JULY 10
  4 FG-2 AUG 1, 10
  AB - 5 1 7 FEBRUARY 11
  CB - 6 4 FEBRUARY 8, 11
  7 CB-4 12 AUG 11
  
  I want to pass the result of the SQL query, as below
  
  For RegistrationNo = AB-1
  
  Reffacture RegistrationNo InvoiceDate
  3 AB-1 1ST JULY 10
  AB - 5 1 7 FEBRUARY 11
  
  For RegistrationNo = CB-4
  
  Reffacture RegistrationNo InvoiceDate
  CB - 6 4 FEBRUARY 8, 11
  7 CB-4 12 AUG 11
  
  I want to get that two records by RegistrationNo with the previous date of min, whose date max more forward.
  
  
  
  Kind regards

  Something like that?

  WITH A AS (SELECT 1 INV_ID, 'AB-1' REG_NO, TO_DATE ('01-JAN-10', 'dd-mon-rr') INV_DATE FROM DUAL
              UNION
             SELECT 2 INV_ID, 'CB-4' REG_NO, TO_DATE ('23-mrz-10', 'dd-mon-rr') INV_DATE FROM DUAL
              UNION
             SELECT 3 INV_ID, 'AB-1' REG_NO, TO_DATE ('01-JUL-10', 'dd-mon-rr') INV_DATE FROM DUAL
              UNION
             SELECT 4 INV_ID, 'FG-2' REG_NO, TO_DATE ('01-AUG-10', 'dd-mon-rr') INV_DATE FROM DUAL
              UNION
             SELECT 5 INV_ID, 'AB-1' REG_NO, TO_DATE ('07-FEB-11', 'dd-mon-rr') INV_DATE FROM DUAL
              UNION
             SELECT 6 INV_ID, 'CB-4' REG_NO, TO_DATE ('08-FEB-11', 'dd-mon-rr') INV_DATE FROM DUAL
              UNION
             SELECT 7 INV_ID, 'CB-4' REG_NO, TO_DATE ('12-AUG-11', 'dd-mon-rr') INV_DATE FROM DUAL)
  SELECT INV_ID,
         REG_NO,
         INV_DATE
    FROM (SELECT INV_ID,
                 REG_NO,
                 INV_DATE,
                 ROW_NUMBER() OVER (PARTITION BY REG_NO ORDER BY INV_DATE DESC) RN
            FROM A
         )
   WHERE RN<3
  ORDER BY REG_NO,
           INV_DATE DESC
  

• I've updated to 8.0 and now firefox will not open and I get an error message indicating that 7.0.1 is not compatible with version 8.0 of the Min and max version 8.0. What can I do now?

  I tried to uninstall 7.0.1 and reinstall only and still it would not to date and problems such as the loss of my favorites and have different toolbars and Add ons whenever I opened it. I lost so many bookmarks and it became so frustrating that I decided to try to upgrade to see if that could overcome the problems. And now I have this problem. When I did a search on my hard drive to see the components of firefox how was I found firefox files and documents in all directions and I was wondering if I should just delete everything related to firefox to clean all residual files which could be causing problems and then start a clean download , but I didn't want to delete files in windows and didn't know what I should delete and I have to leave. Can anyone help? I miss my firefox and the wish that I had never upgraded to 7.0.1.

  If you use ZoneAlarm Extreme Security then try turning off virtualization.

  • http://KB.mozillazine.org/Browser_will_not_start_up#XULRunner_error_after_an_update

  See also:

  • [880050/questions/880050]

• Taskbar and min and max do not show on the homepage and e-mail

  When cleaning key board, the taskbar, and the top part of the page containing the min and max symbols disapperared.  How can I restore them?

  Hello

  I suggest you to try the steps below and check if it helps.

  (a) open Internet Explorer by pressing the button of the window of your keyboard, then clicking on Internet Explorer.

  (b) move your cursor to the top right and click the Tools button, and then click file.

  (c) uncheck the option full screen .

  You can also press F11 on your keyboard, which would help you to get to the normal screen.

  Hope this information is useful.

• Min and Max on a table

  Hello. I want to make a program where I want to generate a sine and take only the monotonous positive part.

  So I thought to find the min and max of the function index and make a program like the one attached.

  As you can see that if you set the frequency to 10 Hz, it works fine... the problem is when you change the frequency such as 20 or 50 H.. t seems that it is not find the min and max correctly.

  Can you suggest me some tips?

  Thank you

  s.turino84 wrote:

  and take only positive part monotonous.

  This means that the derivative is positive.  So, make a derivative.  If the derivative is greater than 0 keep the present value, or the value 0.

  

• Min and Max in a row!

  Hi guys,.

  Do you know how I can get the MIN and MAX in a row?

  For example:

  C1 | C2 | C3 | C4 | C5 | C6 | C7 | C8

  10. 50. 70.  5. 90. 135. 4   | 60

  I return 4 and 135.

  I can return it without using the service?

  Thank you

  Eric

  create table t as
  with d as (
    select 1 n from dual
    union all
    select null from dual
  )
  select a.n c1, b.n*2 c2, c.n*3 c3
  from d a, d b, d c;
  
  select c1,c2,c3,
  greatest(
    coalesce(c1,c2,c3),
    coalesce(c2,c3,c1),
    coalesce(c3,c1,c2)
  ) greatest_c,
  least(
    coalesce(c1,c2,c3),
    coalesce(c2,c3,c1),
    coalesce(c3,c1,c2)
  ) least_c
  from t;
  

  C1	C2	C3	GREATEST_C	LEAST_C
  1	2	3	3	1
  1	2		2	1
  1		3	3	1
  1			1	1
  	2	3	3	2
  	2		2	2
  		3	3	3

• Help the query to list all the days between MIN and MAX date in a table

  Hello

  Sorry, this may have already responded earlier, but I really struggled to find a previous response with the new provision of the RTO.

  In a DB 11 g, I date MIN and MAX in a table:

  {code}

  SELECT MIN (process_date) start_date, MAX (process_date) end_date FROM my_table;

  {code}

  I would get every day between these 2 dates. I mean even is there is no record in the table for a date.

  Is this possible?

  Thanks in advance,

  Olivier

  Hello

  Do you mean something like this?

  ranit@XE11GR2>> ed
  Wrote file c:\rb\1.sql
  
    1  with xx as
    2  (
    3      select
    4     to_date('05-06-2013','dd-mm-yyyy') min_d,
    5     to_date('20-06-2013','dd-mm-yyyy') max_d
    6      from dual
    7  )
    8  --
    9  -- end of test data
   10  --
   11  select
   12     min_d + level date_x
   13  from xx
   14* connect by level<=(max_d-min_d)
  ranit@XE11GR2>> /
  
  DATE_X
  -------------------
  06-06-2013 00:00:00
  07-06-2013 00:00:00
  08-06-2013 00:00:00
  09-06-2013 00:00:00
  10-06-2013 00:00:00
  11-06-2013 00:00:00
  12-06-2013 00:00:00
  13-06-2013 00:00:00
  14-06-2013 00:00:00
  15-06-2013 00:00:00
  16-06-2013 00:00:00
  17-06-2013 00:00:00
  18-06-2013 00:00:00
  19-06-2013 00:00:00
  20-06-2013 00:00:00
  
  15 rows selected.
  

• button conrol min and max range option during execution

  Is it possible to change the window button conrol min and max Beach setting during execution?

  Thank you very much Nick.

• Find dates MIN and MAX

  Please, advise how correctly find MIN and MAX dates.
  The XML code:

  <DATES>
  <DT>2011-02-24</DT>
  <DT>2011-02-25</DT>
  <DT>2011-02-26</DT>
  <DT>2011-02-27</DT>
  <DT>2011-02-28</DT>
  </DATES>

  Model:

  <?xdoxslt:minimum(DT)?>
  <?xdoxslt:maximum(DT)?>

  The output:

  0
  20110224

  Expected results:

  2011-02-24
  2011-02-28

  Hello

  If your data is:
  
  
  
  

  2008-10-04
  
  1
  
  
  
  2009-10-21
  
  3
  
  
  
  2009-11-05
  
  4
  
  
  
  2008-11-10
  
  2
  
  

  The maximum is given by: / ROWSET/ROW/DT [.. / DT_RANK = count (//ROW)]
  and the minimum is: / ROWSET/ROW/DT [.. / DT_RANK = 1]

  There may be more effective, but it works.

  concerning

  Jorge

• How can I go into a used computer and change the username to the mine and to make a new password or delete the old user of my computer completely? (preferably get rid of the former user)?

  I need to know how to solve this problem in detail please

  I bought a computer with windows xp service pack 3 and when I start it the old user keeps popping up asking for a password. I want to know how to enter and change the username to the mine and to make a new password or delete the old user of my computer completely (preferably get rid of the former user)?

  See the following link to delete the account: http://support.microsoft.com/kb/279783

• Gap between the left and right columns. Both floated and checked the meansurements. Help!

  Gap between the left and right columns. Both floated and checked the meansurements. Help!

  Nancy - I think I might have fixed. I'll get back to you. Thank you very much for answering. Sometimes I feel so lonely there.

• in comparing the index and a table between instances PK

  Aloha!
  
  I'll compare the indices and the PK's (Constraint) of 3 instances.
  
  This table of s/n/s best use? The comparison is made on a specific schema only.
  
  Please be kind enough to write the script that I need or tables.
  
  PS
  Which is better dba_ind_columns or dba_objects?
  
  BR,
  Hades
  
  Published by: TheHades0210 on December 2, 2012 21:58

  TheHades0210 wrote:
  Hello

  Thank you for these elaborate explanation. I just want to know why under dba_ind_Columns INDEX_NAME, some of the names of the characters $ and #. Is it system generated index?

  very probably are generated, the system but not necessarily.

  If so, what application generates this index?

  I don't know what applications are installed in your DB.
  No objects appear spontaneously.
  Someone has published the DDL that created.