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.

Tags: Database

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