without the use of order by and connect by clause

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• without the use of union or union of all the

  Hello experts

  I have the following data below

  create the table t_me

  (

  first_one varchar (1000).

  second_one varchar (1000).

  third_one varchar (1000).

  week_needed varchar (1000)

  );

  insert into t_me values ('A', 'B', 'C', ' week 1 "');

  insert into t_me values ('A', 'B', 'C', ' week 1 "');

  insert into t_me values ('B', 'B', 'C', ' week 1 "');

  insert into t_me values ('C', 'A', 'C', ' week 2 "");

  insert into t_me values ('C', 'A', 'C', "3 week");

  insert into values of t_me ('B', 'C', 'B', "3 week");

  output desired without the use of union

  the characters first_one, second_one, third_one week_needed

  Week 1             A:                2               0              0

  Week 1             B:               1               3              0

  Week 1             C:               0                0              3

  Week 2             A:               0                1              0

  Week 2             C:               1                0              1

  Week 3             A:               0                 1             0

  Week 3             B:               1                 1              0

  Week 3             C:               1                 0              2

  Any help is appreciated. Thank you

  Hello

  If you want something that works in your version, you should tell what version it is, especially if this version is 12 years old.

  Here's a way to make the unpivot operator and rotate in Oracle 9:

  WITH cntr AS

  (

  SELECT LEVEL AS col_num

  OF the double

  CONNECT BY LEVEL<=>

  )

  unpivoted_data AS

  (

  SELECT week_needed

  col_num

  CASE col_num

  WHEN 1 THEN first_one

  WHEN 2 THEN second_one

  WHEN 3 THEN third_one

  AS END characters

  OF t_me

  CROSS JOIN cntr

  )

  SELECT week_needed, characters

  , COUNT (CASE WHEN col_num = 1, 1 END) AS first_one

  , COUNT (CASE WHEN col_num = END 2 THEN 1) AS second_one

  , COUNT (CASE WHEN col_num = END of the 3 THEN 1) AS third_one

  Of unpivoted_data

  GROUP BY week_needed, characters

  ORDER BY week_needed, characters

  ;

• without the use of grouping sets

  Dear expert;

  
  

  I need help in the creation of this report without the use of the grouping sets. Any help is appreciated

  Week                                                          hours

  John                                                            24

  Tim                                                             1

  Mark                                                           16

  Adam                                                          10

  Subtotal (30/06/2013-06/07/2013) 66

  Adam                                                           7

  Subtotal (07/07/2013-08/07/2013) 7

  Month Total                                                   73

  Any help is appreciated. Thank you.

  see examples of data below. Thank you

  create table info_mech
  (
  
  req_name varchar(1000),
  beginwork_date date,
  endwork_date date
  );
  
  insert into info_mech values ('John', to_date('07/01/2013 08:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/01/2013 15:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  insert into info_mech values ('John', to_date('07/02/2013 10:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/02/2013 19:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  insert into info_mech values ('John', to_date('07/05/2013 11:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/05/2013 21:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  insert into info_mech values ('Tim', to_date('07/02/2013 08:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/02/2013 15:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  insert into info_mech values ('Tim', to_date('07/04/2013 09:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/04/2013 18:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  
  insert into info_mech values ('Mark', to_date('07/01/2013 08:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/01/2013 15:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  insert into info_mech values ('Mark', to_date('07/02/2013 10:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/02/2013 19:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  insert into info_mech values ('Adam', to_date('07/05/2013 11:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/05/2013 21:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  insert into info_mech values ('Adam', to_date('07/08/2013 08:00:00', 'MM-DD-YYYY HH24:MI:SS'), to_date('07/05/2013 15:00:00', 'MM-DD-YYYY HH24:MI:SS'));
  
  
  
  
  

  Hello

  It's a good start.  Is this really what you want to GROUP BY?

  Group of (next_day (beginwork_date, 'Sunday') - 7 - next_day (beginwork_date, "Saturday))

  Looks like always, which evaluates to-6 or - 13, according to was looking for beginwork_date is a Saturday or not.  If you want something which returns a different value for each week?  I would like to use

  TRUNC (beginwork_date + 1, 'IW')

  which returns the Monday of the week even as beginwork_date.  (Always ISO weeks start on Mondays.  The 'magic number' 1 in the above expression reflects the fact that your startw week 1 day earlier, on Sunday.)

  UNION, you try will work.  Finally, you want 3 different levels of rollup, so eventually you will need a Union of 3 tracks, each with a GROUP BY clause to different.  Here's a different approach, which gets 3 copies of the same data and totals in a single GROUP BY:

  WITH cntr AS

  (

  SELECT LEVEL AS grouping_level

  OF the double

  CONNECT BY LEVEL<=>

  )

  unaggregated_data AS

  (

  SELECT THE CHECK BOX

  WHEN c.grouping_level<=>

  THEN TRUNC (i.beginwork_date + 1, 'IW')

  Week END AS

  CASE

  WHEN c.grouping_level = 1

  THEN i.req_name

  END AS req_name

  24 * (i.endwork_date - i.beginwork_date)

  As schedules

  c.grouping_level

  Info_mech I have

  CROSS JOIN cntr c

  WHERE i.beginwork_date > = DATE ' 01-07-2013'

  AND i.endwork_date<  date="">

  )

  SELECT CASE grouping_level

  WHEN 1

  THEN req_name

  WHEN 2

  THEN "subtotal (' |)" To_char (week - 1, ' MM/DD/YYYY' -)

  || To_char (week + 6, MM/DD/YYYY")')

  ANOTHER "Grand Total".

  END AS name_or_week

  The amount (in hours) AS total_hours

  Of unaggregated_data

  GROUP BY week

  grouping_level

  req_name

  ORDER a week

  grouping_level

  req_name

  ;

  This assumes that, on a given line of the table, beginwork_date is in the same week as endwork_date.  The sample data you posted include 1 row when this was not true:

  insert into info_mech values ('Adam', to_date (July 8, 2013 08:00 ',' ' DD-MM-YYYY HH24:MI:SS), to_date (July 5, 2013 15:00 ',' ' DD-MM-YYYY HH24:MI:SS));))

  Not only are the two dates in different weeks, but beginwork_date is later than endwork_date.  I assumed that was a typo and changed endwork_date to July 8.  With this change, the result I got is:

  NAME_OR_WEEK TOTAL_HOURS

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

  Adam                                          10

  John                                          26

  Mark                                          16

  Tim                                           16

  Subtotal (30/06/2013-07/07/2013) 68

  Adam                                           7

  Subtotal (07/07/2013-14/07/2013) 7

  Grand Total 75

  The last line is the total of all hours for all dates included.  In the query above, this means that the month of July 2013, because I've hardcoded the dates in the WHERE clause.  You can climatiqueLes WHERE clause for what you want, or omitted completely.

  Note how much more complicated the query above is that of the solution of GROUPING SETS below:

  WITH got_week AS

  (

  SELECT req_name

  TRUNC (beginwork_date + 1, 'IW') AS the week

  24 * (endwork_date - beginwork_date) AS hours

  Of info_mech

  WHERE beginwork_date > = DATE ' 01-07-2013'

  AND endwork_date<  date="">

  )

  SELECT THE CHECK BOX

  WHEN you GROUP (req_name) = 0

  THEN req_name

  WHEN you GROUP (week) = 0

  THEN "subtotal (' |)" To_char (week - 1, ' MM/DD/YYYY' -)

  || To_char (week + 6, MM/DD/YYYY")')

  ANOTHER "Grand Total".

  END AS name_or_week

  The amount (in hours) AS total_hours

  OF got_week

  GROUP BY GROUPING SETS ((semaine, req_name))

  (week)

  , ()

  )

  ORDER a week

  req_name

  ;

  GROUPING SETS will also be more effective.

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