Find the gap and overlap

Similar Questions

• Calculations of dates for the gaps and overlaps

  Oracle version: 11.2.0.3

  Platform: Linux

  EmpNo position start_dt end_dt

  1 X 01/01/2008-01/02/2010

  1 Y 01/02/2009-12/31/2012

  1 X 02/01/2012-04/01/2013

  1 Z 2012 12/31/5 / 01 / 2012

  1 O 01/05/2013 presents

  I need to find the exact period the employee has worked for the company in a particular situation. The query must be able to calculate the overlap/gaps as a result

  For example in this case, the result must be

  Days of position empno Total_Days

  1 x 1156 2943

  1 1543 2943 y

  1             z               244        2943

  Thank you

  Kevin

  SQL > select empno, position, hard

  2, sum (hard) earlier (partition by empno)

  3 of

  (4)

  5. Select empno, position, sum (end_dt - start_dt) hard

  6 t

  Group 7 by empno, position

  8)

  9;

  EMPNO P HARD TOT

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

  1 X 1157 2944

  1 1543 2944 Y

  1 Z                  244                 2944

  Don't know exactly what you mean by cross-checking.  Can an employee cause two different positions at the same time?  Your sample data does not display it.

• 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

• find the gaps between the sequences

  Hi all

  I have a request to find the gap between the sequences.

  Here's my table structure

  user_number | seq_number

  1            |   101

  1            |      102

  1            |    109

  1            |    110

  1            |    111

  2            |    201

  2            |    202

  2            |    203

  2            |    207

  2            |    208

  .            |    .

  .            |    .

  .            |    .

  .            |    .

  Now I have to find the gap between sequences, what I mean by gap's user_number 1 ' 6 between seq_number gap ' 102 'and 109'.

  Here's the query that I'm writing, but not giving the correct answer, he always gives the gap of 2:

  Select gap_start - gap_end gap

  de)

  Select a.seq_number + 1 as gap_start, b.seq_number - 1 as gap_end

  from my_table a, b from my_table

  where a.seq_number = b.seq_number

  and a.seq_number - (select c.seq_number

  from my_table c

  where c.seq_number = a.seq_number - 1) is set to null

  and a.user_number = '1'

  order of a.seq_number

  );

  Thank you.

  Hello

  Whenever you have a question, please post a small example of data (CREATE TABLE and only relevant columns, INSERT statements) for all of the tables involved and the accurate results you want from this data, so that people who want to help you can recreate the problem and test their ideas.

  Explain, using specific examples, how you get these results from these data.

  Always say what version of Oracle you are using (for example, 11.2.0.2.0).

  See the FAQ forum: Re: 2. How can I ask a question on the forums?

  Maybe you want something like this:

  WITH got_prev_seq_number AS

  (

  SELECT user_number, seq_number

  LAG (seq_numebr) OVER (PARTITION BY user_number

  ORDER BY seq_number

  ) AS prev_seq_number

  FROM my_table

  )

  SELECT user_number, seq_number, prev_seq_number

  OF got_prev_seq_number

  WHERE seq_number > prev_seq_number + 1

  ORDER BY user_number, seq_number

  ;

  .

• Toshiba 40TL938G - find the date and time settings

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  Can someone help me?

  SSY
  I read the manual and it does ' t say anything about these settings.

  THX

  In the settings of digital television, you should see the option called local time settings

  To display the time on the TV you press the sign of the small clock on the remote control while watching a normal TV show (analog only)

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  Hello

  I use laptop Vaio VGN-CR36G with vista ultimate. Now, I want to install windows 8 enterprise edition.

  But I can't find the drivers and utilities on the site of sony vaio.

  But my product is compatible for windows 8 OS.

  Here is the link below:

  http://www.Sony-Asia.com/support/product/VGN-CR36G

  Can someone please help me to download all drivers and software?

  Thank you

  Hi Whiteeasy.

  This model does not support Windows 8 operating system which you can confirm from the link below.

  http://www.Sony-Asia.com/microsite/VAIO/support/GetWindows8/GetWindows8_EN_NA.html?referer=http%3A%2...

  If my post answered your question, please mark it as an "accepted Solution".

• How to find the Maxima and Minima for each column of a 2D array?

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  I have a 2D chart and I would find the maxima and minima of each column of the 2-D table. Even though I know how to get maxima and minima for the whole picture but don't know how columnwise? Any ideas please?

  Thank you

  Rohit

  Hello

  @Smercurio-What you said is true, I should have shown using automatic indexing enabled which is really excellent choice. I just tried to show in a very simple way.

  Anyway, here's the best way

  

• How to find the maximum and minimum of a waveform between cursors

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  Andreas

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  attached, you get an example of it in MeasurementStudio.

• Cocomputer cannot find the printer and I do not have a disk for it

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  For example, if we see the role of manager accounts payable, it conatins Pages and the list of permissions.

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• To find the months and days between 2 dates

  Hello

  I want to find the months and days between 2 dates.

  For example.

  1 - Date: August 25, 2013

  2 - Date: October 23, 2013

  If we consider each month 30 days, it should give

  August 25, 2013 to August 30, 2013 = 6 days

  01-Sep-2013-30-Sep-2013 = 1 month

  October 23, 2013 to October 30, 2013 = 8 days

  Total = 1 month and 14 days.

  Kindly help as soon as possible.

  Thanks and greetings

  Suresh

  Assuming that d2 > d1,.

  where d)

  Select sysdate d1, sysdate + 56 double d2

  Union all select to_date (March 1, 2013 ',' dd-mon-yyyy "") d1, to_date (March 31, 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date (5 February 2013 ',' dd-mon-yyyy ') d1, to_date (March 31, 2013 ',' dd-mon-yyyy "") double d2

  Union all select to_date (February 25, 2013 ',' dd-mon-yyyy "") d1, to_date (March 23, 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date (February 25, 2013 ',' dd-mon-yyyy ') d1, to_date (March 31, 2013 ',' dd-mon-yyyy "") double d2

  Union all select to_date (August 2, 2013 ',' dd-mon-yyyy "") d1, to_date (29 October 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date (February 1, 2013 ',' dd-mon-yyyy "") d1, to_date (May 31, 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date (25 August 2013 ',' dd-mon-yyyy "") d1, to_date ('03-Sep-2013', 'Mon-dd-yyyy') d2 double

  Union all select to_date (July 30, 2013 ',' dd-mon-yyyy "") d1, to_date (August 31, 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date (July 31, 2013 ',' dd-mon-yyyy ') d1, to_date (August 30, 2013 ',' dd-mon-yyyy "") double d2

  Union all select to_date (July 31, 2013 ',' dd-mon-yyyy ') d1, to_date (3 August 2013 ',' dd-mon-yyyy "") double d2

  Union all select to_date (3 July 2013 ',' dd-mon-yyyy "") d1, to_date (August 31, 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date ('31-08-2013', ' dd-mm-yyyy'), to_date('05-10-2013','dd-mm-yyyy') of the double

  Union all select to_date ('05-02-2013', ' dd-mm-yyyy'), to_date('31-03-2013','dd-mm-yyyy') of the double

  Union all select to_date ('05-02-2013', ' dd-mm-yyyy'), to_date('05-03-2013','dd-mm-yyyy') of the double

  Union all select to_date ('05-02-2013', ' dd-mm-yyyy'), to_date('05-02-2013','dd-mm-yyyy') of the double

  )

  Select d1, d2,

  1 + 30 * trunc (months_between (d2, d1)) + LESS (extract (day of d2), 30)-LESS (excerpt (d1 day), 30)

  + CASE when extracted (d2 day)< extract(day="" from="" d1)="" then="" 30="" else="" 0="" end ="">

  d

  D1 D2 DAYSBETWEEN

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

  October 10, 2013 5 December 2013 56

  March 1, 2013 30 March 31, 2013

  5 February 2013 March 31, 2013 56

  February 25, 2013 March 23, 2013 29

  February 25, 2013 March 31, 2013 36

  August 2, 2013 29 October 2013 88

  February 1, 2013 may 31, 2013 120

  August 25, 2013 03 - Sep-2013 9

  July 30, 2013 31 August 31, 2013

  July 31, 2013 August 30, 2013 31

  July 31, 2013 3 August 2013 4

  July 3, 2013 August 31, 2013 58

  31 August 2013 5 October 2013 36

  5 February 2013 March 31, 2013 56

  5 February 2013 March 5, 2013 31

  February 5, 2013 February 5, 2013 1

  In my view, which corresponds to your rules.

• I posted a question can not find the question and or response. Help, please

  I posted a question can not find the question and or response. Help, please

  AFAIK, you can't share a catalog like this.

  The only recommended method is to store your images and catalogue on an external drive, you can alternatively plug in both computers.

  With today the external USB drives, the speed is still correct, and many users are satisfied with this solution.