November 27, 2012 at 3:55 am
Hello,
I have a script problem.
I calculate data per week (group by week)
However, for indicators, I have business rules which apply in the month.
Here is the problem because I have weeks that are mounted on 2 different months.
I 'd like so:
1 - Identify the weeks that straddle two months
2 - For the week, identify the last day of the month is the day of the week.
3 - If the date is before Wednesday when the week belongs to month
Otherwise, the week belongs to the next month.
I hope I have clearly expressed my need.
And sorry for my English.:))
November 27, 2012 at 5:47 am
Kindly post some sample SQL that would definitely help!
Do read about best-practices link underneath my signature. That will always help you to get response to your queries in timely manner.
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~ Lokesh Vij
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November 27, 2012 at 8:40 pm
Lidou123 (11/27/2012)
Hello,I have a script problem.
I calculate data per week (group by week)
However, for indicators, I have business rules which apply in the month.
Here is the problem because I have weeks that are mounted on 2 different months.
I 'd like so:
1 - Identify the weeks that straddle two months
2 - For the week, identify the last day of the month is the day of the week.
3 - If the date is before Wednesday when the week belongs to month
Otherwise, the week belongs to the next month.
I hope I have clearly expressed my need.
And sorry for my English.:))
Hi, Lidou123,
What day of the week is the first day of the week for you?
Also, how do you want the weeks numbered? From the start of the year or the start of the month?
--Jeff Moden
Change is inevitable... Change for the better is not.
November 28, 2012 at 2:48 am
Hi Jeff,
Thank you for your answer.
For my company, the first day of the week is Monday.
And the weeks are numbered by the start of the year.
We use the ISO Calendar.
All I want is to determine when a week is between two months, how many days are in the month M and how many are in the Month M+1.
I found the sql script in a Oracle forum but I don't know to translate it into T-SQL.
This is the URL where i found the Oracle Script. It s a french forum :)) : http://www.developpez.net/forums/d908298/bases-donnees/oracle/sql/nb-jours-semaine-cheval-2-mois/
And this the Oracle script:
WITH cal AS
(
SELECT date '2010-01-01' + level -1 AS dt
FROM dual
connect BY level <= 365
)
, sr AS
(
SELECT to_char(dt, 'yyyy-mm') AS mois ,
to_char(dt, 'iyyy"W"iw') AS iweek,
count(*) AS nb_jours,
count(*) over(partition BY to_char(dt, 'iyyy"W"iw')) AS nb_mois
FROM cal
GROUP BY to_char(dt, 'yyyy-mm'),
to_char(dt, 'iyyy"W"iw')
)
SELECT *
FROM sr
WHERE nb_mois = 2
ORDER BY mois ASC, iweek ASC;
MOIS IWEEK NB_DAY NB_MONTH
------- ------- ---------- ----------
2010-03 2010W13 3 2
2010-04 2010W13 4 2
2010-04 2010W17 5 2
2010-05 2010W17 2 2
2010-05 2010W22 1 2
2010-06 2010W22 6 2
2010-06 2010W26 3 2
2010-07 2010W26 4 2
2010-07 2010W30 6 2
2010-08 2010W30 1 2
2010-08 2010W35 2 2
2010-09 2010W35 5 2
2010-09 2010W39 4 2
2010-10 2010W39 3 2
2010-11 2010W48 2 2
2010-12 2010W48 5 2
I am waiting your answer.
November 28, 2012 at 11:09 am
Lidou123 (11/28/2012)
Hi Jeff,Thank you for your answer.
For my company, the first day of the week is Monday.
And the weeks are numbered by the start of the year.
We use the ISO Calendar.
All I want is to determine when a week is between two months, how many days are in the month M and how many are in the Month M+1.
I found the sql script in a Oracle forum but I don't know to translate it into T-SQL.
This is the URL where i found the Oracle Script. It s a french forum :)) : http://www.developpez.net/forums/d908298/bases-donnees/oracle/sql/nb-jours-semaine-cheval-2-mois/
And this the Oracle script:
WITH cal AS
(
SELECT date '2010-01-01' + level -1 AS dt
FROM dual
connect BY level <= 365
)
, sr AS
(
SELECT to_char(dt, 'yyyy-mm') AS mois ,
to_char(dt, 'iyyy"W"iw') AS iweek,
count(*) AS nb_jours,
count(*) over(partition BY to_char(dt, 'iyyy"W"iw')) AS nb_mois
FROM cal
GROUP BY to_char(dt, 'yyyy-mm'),
to_char(dt, 'iyyy"W"iw')
)
SELECT *
FROM sr
WHERE nb_mois = 2
ORDER BY mois ASC, iweek ASC;
MOIS IWEEK NB_DAY NB_MONTH
------- ------- ---------- ----------
2010-03 2010W13 3 2
2010-04 2010W13 4 2
2010-04 2010W17 5 2
2010-05 2010W17 2 2
2010-05 2010W22 1 2
2010-06 2010W22 6 2
2010-06 2010W26 3 2
2010-07 2010W26 4 2
2010-07 2010W30 6 2
2010-08 2010W30 1 2
2010-08 2010W35 2 2
2010-09 2010W35 5 2
2010-09 2010W39 4 2
2010-10 2010W39 3 2
2010-11 2010W48 2 2
2010-12 2010W48 5 2
I am waiting your answer.
There are a couple of fairly simple ways to start this. Please see "Example A" of "Create Function" in Books Online. I won't be able to write any code for this (the example is RBAR, which I recommend avoiding) until after work today. My recommendation is to search for "ISO Calendar Table) on Goolge for a quick legup on this problem, until then.
--Jeff Moden
Change is inevitable... Change for the better is not.
November 29, 2012 at 3:37 am
Thank You Jeff.
November 29, 2012 at 9:14 pm
I'm so used to working in 2005 that I didn't think of a possible 2008 solution. Since you're posting in a 2008 forum, can I safely assume that you're using 2008?
--Jeff Moden
Change is inevitable... Change for the better is not.
December 2, 2012 at 3:17 pm
In SQLServer 2008, there is a DATEPART called ISOWK. I've not used it because I don't work with week numbers but that should fill the bill for you.
--Jeff Moden
Change is inevitable... Change for the better is not.
December 3, 2012 at 8:51 am
Hi Jeff.
The iso_wk function works well
Thank You Jeff
December 3, 2012 at 5:13 pm
You're welcome. Are you all set on this problem or do you have additional questions on it?
--Jeff Moden
Change is inevitable... Change for the better is not.
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