Given a date, select all the week numbers that month contains - sql

Question

I have an ApEx application for tracking working hours.
I have a view that looks like this:
CREATE OR REPLACE VIEW HOURSDAYS
AS SELECT
(MAX("TO_X") - MIN("FROM_X"))*24 -
(max(case when PROJECT_ID = 999 then to_x else to_date('01012000','DDMMYYYY') end) -
max(case when PROJECT_ID = 999 then from_x else to_date('01012000','DDMMYYYY') end))*24 AS TIME_SPENT,
DAY,
PERSON_ID
FROM ATTENDANCE_HOURS
GROUP BY PERSON_ID, DAY
ORDER BY DAY DESC
I need the sum of hours per week. So I have this:
SELECT TO_CHAR(DAY,'IW'), MIN(DAY), MAX(DAY), SUM(TIME_SPENT)
FROM HOURSDAYS
WHERE PERSON_ID = (SELECT ID FROM ATTENDANCE_PEOPLE WHERE MAIL_SSO = V('APP_USER') AND ROLE = 0) AND
EXTRACT (YEAR FROM DAY) = EXTRACT (YEAR FROM TO_DATE(:P100_DATE_PICKER,'DD-MON-RR'))
GROUP BY TO_CHAR(DAY,'IW')
ORDER BY TO_CHAR(DAY,'IW') ASC
And now the fun begins: I have a page item P100_DATE_PICKER and I need to display only those weeks that have > 0 days that belong to the month to which belongs the day I've picked using the date picker.
For example, for 1.1.2015 I want only weeks 1, 2, 3, 4, 5 displayed. 24.3.2015: 9, 10, 11, 12, 13, 14.
If anyone is interested in why would I do that, it is for validation - number of work hours per week cannot exceed 20/32, depends on what type of contract do you have.

Answer 1

You can do like this:
where day >= trunc(trunc(TO_DATE(:P100_DATE_PICKER,'DD-MON-RR'), 'MM'),'IW')
and day < trunc(last_day(TO_DATE(:P100_DATE_PICKER,'DD-MON-RR')),'IW') + 7
The first line starts by trunc to MM, which gives you the first day of the month of your date. Then trunc again to IW gives you the date of the monday in the week that contains the first day of month.
The second line uses last_day to get the last day of the month of your date. Then trunc to IW gets the monday of that week, and adding 7 days get the monday after. Then by using < rather than <= you get the desired result.
Also I suggest you do not group and order by TO_CHAR(DAY,'IW') but instead use TRUNC(DAY,'IW') for your grouping and ordering. Otherwise you can get into problems around new year, because your code for example when showing the weeks for December 2014 would have to show 49 to 52 of year 2014 as well as week 1 of year 2015. If you use TO_CHAR that week 1 would be wrongly sorted before weeks 49 to 52.
Similarly do not use EXTRACT(YEAR FROM ...) or even TO_CHAR(...,'YYYY') when you are working with ISO weeks. Because the date '2014-12-31' belongs to ISO week 1 year 2015, there is a special date format string 'IYYY' which gives the correct year for the ISO week. Try it out and see the difference between 'YYYY' and 'IYYY'.

Answer 2

This will not answer all your questions. This is the quick example of week calculation from a given date according to the subject line of your question. Pls copy/paste to see results. Also, here's the link to check the week numbers:
http://www.epochconverter.com/date-and-time/weeknumbers-by-year.php
SELECT given_date
, end_date
, TRUNC(calc_start_date, 'iw') wk_starts
, TRUNC(calc_start_date, 'iw') + 7 - 1/86400 wk_ends
, TO_CHAR(calc_start_date, 'iw') wk_number
, calc_start_date
FROM
(
SELECT trunc(sysdate, 'mm') given_date
, trunc(sysdate, 'mm')-7 + LEVEL*7 AS calc_start_date
, Last_Day(trunc(SYSDATE, 'mm')) end_date
FROM dual
CONNECT BY LEVEL <= ROUND((trunc(last_day(sysdate)) - trunc(sysdate, 'mm')+7)/7) -- number of weeks --
)
/
GIVEN DATE END DATE WK_STARTS WK_ENDS WK NUMBER
-------------------------------------------------------------------------
1/1/2015 1/31/2015 12/29/2014 1/4/2015 11:59:59 PM 01
1/1/2015 1/31/2015 1/5/2015 1/11/2015 11:59:59 PM 02
1/1/2015 1/31/2015 1/12/2015 1/18/2015 11:59:59 PM 03
1/1/2015 1/31/2015 1/19/2015 1/25/2015 11:59:59 PM 04
1/1/2015 1/31/2015 1/26/2015 2/1/2015 11:59:59 PM 05

Answer 3

The best way to achieve the last part is to have a calendar table that has the following columns
Date, Day eg Mon, Month, Quarter, Year, WeekNo
Then for a given date you can simply select all the weeks that belong to the month.
This allows you to define weekNo so that it fits in with company circumstances. Eg occasionally you may get a 53 week year, especially if they operate 4-4-5 accounting periods

Answer 4

with days as (
select
to_number(to_char(trunc(to_date('2015-03-24', 'yyyy-MM-dd'), 'month'), 'WW')) w1,
to_number(to_char(last_day(to_date('2015-03-24', 'yyyy-MM-dd')), 'WW')) w2
from dual)
select w1+level-1 week from days
connect by level <= w2-w1+1
Subquery days gives minimum and maximum week, query with connect by returns numbers between these values. Please replace example date with :P100_DATE_PICKER.