I was wondering if it is possible in SQL to read in a date column and based on that date create a new column and automatically have the Week number as well. For example today is 4/7/2020 , so the query would have Week 15 populated for that?
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In the picture the week column would ideally be populated beside 'datestr'.
Thank you]2
In redshift, you can use date_part() with the w specifier to get the week number of a date or timestamp:
select t.*, date_part(w, datestr) week_number from mytable t
Note that weeks starts on Monday in Redshift. If you want the week to start on Sunday:
select t.*, date_part(w, datestr + interval '1' day) week_number from mytable t
You could use extract. I am not 100% sure if weeks in Redshift start from Sunday or Monday, but you can adjust the interval to test the edge cases.
select datestr, extract(week from datestr + interval '1 day') as weeknum
from your_table
Related
I google data studio, I have a field called week that contains the week number the value is string 'Week 2' for example. How can I extract the last day of the week based on the week number. In this case I want to get 2023-01-14 which is the last day of week 2?
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Have you tried the last_day function?
I would build from the using the date using last_day, because from the week number it would require a bit more code.
SELECT *,
LAST_DAY(today, WEEK) AS last_day_of_week,
EXTRACT(WEEK FROM today) AS week_number
FROM (SELECT CURRENT_DATE() AS today)
But if you need to go with just the week number, I recommend taking a look into this other question How to create date based on year, week number and day in bigquery .
The following code works:
DATEtime_ADD(DATEtime_TRUNC(DATEtime_ADD(DATEtime_TRUNC(date, WEEK), INTERVAL (week-1) * 1 DAY), WEEK), INTERVAL 6 DAY)
I'm trying to extract the week number from a date, and I want the week to be counted from Sunday to Saturday. This is what I currently have, but I can't seem to find any solution for this is SQL Presto.
SELECT WEEK(date) AS weeknum
Can this be solved?
Thank you!
One method is:
select week(date + interval '1 day') - interval '1 day'
Note: This may not work on the last day of the year.
Alternatively you can use the MySQL-like functions:
select date_format(date, '%V')
This has the week starting on Sunday.
I am trying to run a query in AWS Athena to get the week number from a date. I need the week to start on Sunday.
I have tried the below query and it works but it considers the week starts from Monday.
Extract(week from date) as week_number
Google bigquery has an option where you can specify the week start. Eg:
Extract(week(Sunday) from date) as week_number
Is there something similar in Athena?
Request assistance.
Thank you.
If there isn't, you can add one day:
extract(week from date + interval '1' day) as week_number
Need help figuring out how to determine if the date is the same 'day' as today in teradata. IE, today 12/1/15 Tuesday, same day last year was actually 12/2/2014 Tuesday.
I tried using current_date - INTERVAL'1'Year but it returns 12/1/2014.
You can do this with a bit of math if you can convert your current date's "Day of the week" to a number, and the previous year's "Day of the week" to a number.
In order to do this in Teradata your best bet is to utilize the sys_calendar.calendar table. Specifically the day_of_week column. Although there are other ways to do it.
Furthermore, instead of using CURRENT_DATE - INTERVAL '1' YEAR, it's a good idea to use ADD_MONTHS(CURRENT_DATE, -12) since INTERVAL arithmetic will fail on 2012-02-29 and other Feb 29th leap year dates.
So, putting it together you get what you need with:
SELECT
ADD_MONTHS(CURRENT_DATE, -12)
+
(
(SELECT day_of_week FROM sys_calendar.calendar WHERE calendar_date = CURRENT_DATE)
-
(SELECT day_of_week FROM sys_calendar.calendar WHERE calendar_date = ADD_MONTHS(CURRENT_DATE, -12))
)
This is basically saying: Take the current dates day of week number (3) and subtract from it last years day of week number (2) to get 1. Add that to last year's date and you'll have the same day of the week as current date.
I tested this for all dates between 01/01/2010 and CURRENT_DATE and it worked as expected.
Why don't you simply subtract 52 weeks?
current_date - 364
The SQL below will get you to the abbreviated name for the day of week, it's cumbersome but it works across versions of Teradata.
SELECT CAST(CAST(ADD_MONTHS(CURRENT_DATE, -12) AS DATE FORMAT 'E3') AS CHAR(3)) AS LY_DayOfWeek
, CAST(CAST(CURRENT_DATE) AS DATE FORMAT 'E3') AS CHAR(3)) AS CY_DayOfWeek
Dates are internally represented at integers in Teradata as (Year-1900) * 100000 + (MONTH * 100) + DAY. You may be able to do some creative arithmetic to figure out that 12/1/2015 Tuesday was 12/2/2014 Tuesday last year.
I'm trying to select data grouped by week, which I have working, but I need to be able to specify a different day as the last day of the week. I think something needs to go near INTERVAL (6-weekday('datetime')) but not sure. This kind of SQL is above my pay-grade ($0) :P
SELECT
sum(`value`) AS `sum`,
DATE(adddate(`datetime`, INTERVAL (6-weekday(`datetime`)) DAY)) AS `dt`
FROM `values`
WHERE id = '123' AND DATETIME BETWEEN '2010-04-22' AND '2010-10-22'
GROUP BY `dt`
ORDER BY `datetime`
Thanks!
select
sum(value) as sum,
CASE WHEN (weekday(datetime)<=3) THEN date(datetime + INTERVAL (3-weekday(datetime)) DAY)
ELSE date(datetime + INTERVAL (3+7-weekday(datetime)) DAY)
END as dt
FROM values
WHERE id = '123' and DATETIME between '2010-04-22' AND '2010-10-22'
GROUP BY dt
ORDER BY datetime
This does look pretty evil but, this query will provide you with a sum of value grouped by a week ending on a Thursday (weekday() return of 3).
If you wish to change what day the end of the week is you just need to replace the 3's in the case statement, ie if you wanted Tuesday you would have it say
CASE WHEN (weekday(datetime)<=1) THEN date(datetime + INTERVAL (1-weekday(datetime)) DAY)
ELSE date(datetime + INTERVAL (1+7-weekday(datetime)) DAY)
I hope this helps.
Simple solution that I like. This will return the date for the start of the week assuming the week ends Sunday and starts Monday.
DATE(`datetime`) - INTERVAL WEEKDAY(`datetime`) AS `dt`
This can easily be adjusted to have a week ending on Thursday because Thursday is 3 days earlier than Sunday
DATE(`datetime`) - INTERVAL WEEKDAY(`datetime` + INTERVAL 3 DAY) AS `dt`
this returns for the start of the week that starts on Friday and ends on Thursday.
You can group on this no problem. If you want to use get the end of the week based on the start you do this
DATE(`datetime`) - INTERVAL -6 + WEEKDAY(`datetime` + INTERVAL 3 DAY) AS `dt`
I think you must choose between Sunday and Monday? When you can use DATE_FORMAT for grouping by string format of date, and use %v for grouping by Mondays and %v for grouping by Sundays.
SELECT
sum(`value`) AS `sum`,
DATE_FORMAT(`datetime`,'%v.%m.%Y') AS `dt`
FROM `values`
WHERE id = '123' AND DATETIME BETWEEN '2010-04-22' AND '2010-10-22'
GROUP BY DATE_FORMAT(`datetime`,'%v.%m.%Y')
ORDER BY `datetime`
How to use DATE_FORMAT
I don't remember the exact math, but you can get WEEKDAY to wrap around on different days of the week by adding or subtracting days to its argument. You'll need to tinker with different values of x and y in the expression:
x-weekday(adddate(`datetime`, INTERVAL y DAY))