I need to change the week start date from Monday to Saturday in PostgreSQL. I have tried SET DATEFIRST 6; but it doesn't work in PostgreSQL. Please suggest solution for this.
It appears that DATEFIRST is a thing in Microsoft Transact-SQL.
I don't believe Postgres has any exact equivalent, but you should be able to approximate it.
Postgres supports extracting various parts of a TIMESTAMP via the EXTRACT function. For your purposes, you would want to use either DOW or ISODOW.
DOW numbers Sunday (0) through Saturday (6), while ISODOW, which adheres to the ISO 8601 standard, numbers Monday (1) through Sunday (7).
From the Postgres doc:
This:
SELECT EXTRACT(DOW FROM TIMESTAMP '2001-02-18 20:38:40');
Returns 0, while this:
SELECT EXTRACT(ISODOW FROM TIMESTAMP '2001-02-18 20:38:40');
Returns 7.
So you would use a version of EXTRACT in your queries to get the day of the week number. If you're going to use it in many queries, I would recommend creating a function which would centralize the query in one spot, and return the number transposed as you would like such that it started on Saturday (the transposition would vary depending on which numbering method you used in EXTRACT). Then you could simply call that function in any SELECT and it would return the transposed number.
You can just add +x, where x is the offset between postgres dow and what you want to achieve.
For example:
You want Sunday to be the first day of the week. 2018-06-03 is a Sunday and you want to extract the dow of it:
SELECT EXTRACT(DOW FROM DATE '2018-06-03')+1;
returns 1
That is probably the reason why postgres dow yields 0 for Sunday. Would it be 7, then
SELECT EXTRACT(DOW FROM DATE '2018-06-03')+1;
would return 8.
Related
Using TSQL I need to get the ISO Week Number in a Month for a give ISO Year Week Number.
For example: The following code will give me Week #1 for 12/31/2001, which is correct. It is the first Monday in 2002 and the first day of the ISO Year 2002.
select DATEPART(ISO_WEEK, '12-31-2001'); --Week 1 January 2002
My question is how do I...
(1) Take the ISO Week Number Example: ISO Year Week Number: 14 for April 4, 2016 (April Week #1).
(2) Now Take ISO Year Week Number 14 and return April Month Week Number = 1 for the example above.
There seems to be nothing in SQL Server to get the ISO Month Week# from the ISO Year Week Number. I have a function I wrote but it is has some hacks to get it to work, but not 100%.
I think you want something like this... but am not sure why you need the ISO_WEEK. Just replace getdate() with your column.
select datepart(wk, getdate()) - datepart(wk,dateadd(m, DATEDIFF(M, 0, getdate()), 0)) + 1
After attempting to handle getting ISO Month Week Number in functions I decided an easier solution was to create an ISO_Calendar table in SQL Sever.
We know in TSQL you can get the ISO Year Week Number and some a bit of work the ISO Year Number. The ISO Month Week Number is another story.
I build the table shown below with data from 2000 to 2040 by populating all the columns other than the ISO Month number. Then on a second pass I looped through the table and set the ISO Month number based on the Month# in the Monday Date.
Now if I want to get the ISO Month number for a date I check #Date between Monday and Sunday. In my case I am only concerned with dates between Monday and Friday since this is for a stock analysis site.
select #ISOMonthWeekNo = c.ISOMonthWeekNo
from ISO_Calendar c
where #TransDate between c.Monday and c.Sunday;
The advantage is the table is a one time build and easier to verify the accuracy of the data.
I have read in this online PostgreSQL documentation... http://www.postgresql.org/docs/9.4/static/datatype-datetime.html#INTERVAL-STYLE-OUTPUT-TABLE
in the point 8.5.5 something about how to tweak the default Interval Output. . I mean the default interval is shown like this...
00:00:00.000 (if the timedifference is lower than a day or month or year)
1 day 00:00:00.000 (if the timedifference reaches days, but is lower than a month or a year)
1 month 1 day 00:00:00.000 (if the timediffence reaches months, but is lower than a year)
1 year 1 month 1 day 00:00:00.000 (if it reaches years, months, days)
it evens uses plurarl cases (years, mons, days) when their values are greater than one.
All these variations make difficult to any other app when SELECTing (query) this interval values (as text) to convert it to a proper time. So I would like postgresql to always show year, month n days, even if their value are 0 (it could be even better if it could show the date part of the interval like this... 01-11-30, adding zeros to the left side when values are less than ten)
I know I can change the interval to text, using to_char() but I really would like to avoid that, I would like some good fellow postgresql programmer to tell me if it is true that there is a way to tweak the Interval Output as is said in the postgresql documentation.
Thanks Advanced.
PD: two more links about the subject
https://my.vertica.com/docs/7.1.x/HTML/Content/Authoring/SQLReferenceManual/Statements/SET/SETDATESTYLE.htm
http://my.vertica.com/docs/6.1.x/HTML/index.htm#13874.htm
You can set the interval output style, but only to one of a few pre-defined formats that are unambigious on input, and that PostgreSQL knows how to parse back into intervals. Per the documentation these are the SQL standard interval format, two variants of PostgreSQL specific syntax, and iso_8601 intervals.
If you want something familiar and easy to parse, consider using:
SET intervalstyle = 'iso_8601'
and using an off-the-shelf parser for ISO1601 intervals.
NEWBIE at work! I am trying to create a simple summary that counts the number of customer visits and groups by 1) date and 2) hour, BUT outputs this:
Date Day of Wk Hour #visits
8/12/2013 Monday 0 5
8/12/2013 Monday 1 7
8/12/2013 Monday 6 10
8/13/2013 Tuesday 14 25
8/13/2013 Tuesday 16 4
We are on military time, so 14 = 2:00 pm
Select
TPM300_PAT_VISIT.adm_ts as [Date]
,TPM300_PAT_VISIT.adm_ts as [Day of Week]
,TPM300_PAT_VISIT.adm_ts as [Hour]
,count(TPM300_PAT_VISIT.vst_ext_id) as [Total Visits]
From
TPM300_PAT_VISIT
Where
TPM300_PAT_VISIT.adm_srv_cd='22126'
and TPM300_PAT_VISIT.adm_ts between '07-01-2013' and '08-01-2013'
Group by
cast(TPM300_PAT_VISIT.adm_ts as DATE)
,datepart(weekday,TPM300_PAT_VISIT.adm_ts)
,datepart(hour,TPM300_PAT_VISIT.adm_ts)
Order by
CAST(TPM300_PAT_VISIT.adm_ts as DATE)
,DATEPART(hour,TPM300_PAT_VISIT.adm_ts)
This should solve the problem:
; With Streamlined as (
SELECT
DATEADD(hour,DATEDIFF(hour,'20010101',adm_ts),'20010101') as RoundedTime,
vst_ext_id
from
TPM300_PAT_VISIT
where
adm_srv_cd='22126' and
adm_ts >= '20130701' and
adm_ts < '20130801'
)
Select
CONVERT(date,RoundedTime) as [Date],
DATEPART(weekday,RoundedTime) as [Day of Week],
DATEPART(hour,RoundedTime) as [Hour],
count(vst_ext_id) as [Total Visits]
From
Streamlined
Group by
RoundedTime
Order by
CONVERT(date,RoundedTime),
DATEPART(hour,RoundedTime)
In the CTE (Streamlined)'s select list, we floor each adm_ts value down to the nearest hour using DATEADD/DATEDIFF. This makes the subsequent grouping easier to specify.
We also specify a semi-open interval for the datetime comparisons, which makes sure we include everything in July (including stuff that happened at 23:59:59.997) whilst excluding events that happened at midnight on 1st August. This is frequently the correct type of comparison to use when working with continuous data (floats, datetimes, etc), but means you have to abandon BETWEEN.
I'm also specifying the dates as YYYYMMDD which is a safe, unambiguous format. Your original query could have been interpreted as either January 7th - January 8th or 1st July - 1st August, depending on the settings of whatever account you use to connect to SQL Server. Better yet, if these dates are being supplied by some other (non-SQL) code, would be for them to be passed as datetimes in the first place, to avoid any formatting issues.
In MySql I can use YEARWEEK() to receive the week and the related year of this week in one string. (E.g. SELECT YEARWEEK('1987-01-01'); which leads to "198653").
Is there anything like that in Oracle10g?
I only know about the TO_CHAR function. But if I use TO_CHAR(sysdate, 'YYYYIW'); I receive 198753 and not 198653. So, how I am able to calculate this correctly?
Does using IYYYIW format with TO_CHAR() make any difference? Note the "I" in the beginning instead of first "Y", it is for 4-digit year based on the ISO standard.
I can't reproduce your example that Oracle returns 198753.
select TO_CHAR(DATE '1987-01-01', 'YYYYIW') from dual returns 198701 for me which is correct according to the ISO definition of week numbers.
Oracle has another format mask WW (instead of IW) that uses the week where the first day of the year is in as week #1 - which again would return week number 1 for the January 1st.
Have a look here: http://en.wikipedia.org/wiki/Week_number#Week_numbering
I find MySQL's week number a bit strange actually, because no week numbering scheme I know would return week 53 for January 1st, 1987 (but that doesn't mean very much though...)
The MS SQL DateDiff function counts the number of boundaries crossed when calculating the difference between two dates.
Unfortunately for me, that's not what I'm after. For instance, 1 June 2012 -> 30 June 2012 crosses 4 boundaries, but covers 5 weeks.
Is there an alternative query that I can run which will give me the number of weeks that a month intersects?
UPDATE
To try and clarify exactly what I'm after:
For any given month I need the number of weeks that intersect with that month.
Also, for the suggestion of just taking the datediff and adding one, that won't work. For instance February 2010 only intersects with 4 weeks. And the DateDiff calls returns 4, meaning that simply adding 1 would leave me the wrong number of weeks.
Beware: Proper Week calculation is generally trickier than you think!
If you use Datepart(week, aDate) you make a lot of assumptions about the concept 'week'.
Does the week start on Sunday or Monday? How do you deal with the transition between week 1 and week 5x. The actual number of weeks in a year is different depending on which week calculation rule you use (first4dayweek, weekOfJan1 etc.)
if you simply want to deal with differences you could use
DATEDIFF('s', firstDateTime, secondDateTime) > (7 * 86400 * numberOfWeeks)
if the first dateTime is at 2011-01-01 15:43:22 then the difference is 5 weeks after 2011-02-05 15:43:22
EDIT: Actually, according to this post: Wrong week number using DATEPART in SQL Server
You can now use Datepart(isoww, aDate) to get ISO 8601 week number. I knew that week was broken but not that there was now a fix. Cool!
THIS WORKS if you are using monday as the first day of the week
set language = british
select datepart(ww, #endofMonthDate) -
datepart(ww, #startofMonthDate) + 1
Datepart is language sensistive. By setting language to british you make monday the first day of the week.
This returns the correct values for feburary 2010 and june 2012! (because of monday as opposed to sunday is the first day of the week).
It also seems to return correct number of weeks for january and december (regardless of year). The isoww parameter uses monday as the first day of the week, but it causes january to sometimes start in week 52/53 and december to sometimes end in week 1 (which would make your select statement more complex)
SET DATEFIRST is important when counting weeks. To check what you have you can use select ##datefirst. ##datefirst=7 means that first day of week is sunday.
set datefirst 7
declare #FromDate datetime = '20100201'
declare #ToDate datetime = '20100228'
select datepart(week, #ToDate) - datepart(week, #FromDate) + 1
Result is 5 because Sunday 28/2 - 2010 is the first day of the fifth week.
If you want to base your week calculations on first day of week is Monday you need to do this instead.
set datefirst 1
declare #FromDate datetime = '20100201'
declare #ToDate datetime = '20100228'
select datepart(week, #ToDate) - datepart(week, #FromDate) + 1
Result is 4.