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))
Related
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?
]1
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
Is there a way to change the showing date to the end of each week instead of the beginning of the week.
Here's my code:
SELECT date_trunc('week', day + '1 day'::interval)::date - '1 day'::interval AS anchor, AVG(value) AS average
FROM daily_metrics
WHERE metric = 'daily-active-users'
GROUP BY anchor
ORDER BY anchor
And the result is as below:
What I want to achieve is to make it 2018-03-03 (the end of the self defined week) instead of 2018-02-25 (the beginning of the self defined week), 2018-03-10 instead of 2018-03-04...
Your trick with shifting back and forth by one day works just fine to get the start of your custom week. You get the end or your custom week (Saturday) by adding 5 instead of subtracting 1:
SELECT date_trunc('week', day + interval '1 day')::date + 5 AS anchor ...
Adding an integer to the date, signifying days.
Related:
PostgreSQL custom week number - first week containing Feb 1st
Simply try
SELECT date_trunc('week', day::DATE + 1)::date + 5 AS anchor, AVG(value) AS average
FROM daily_metrics
WHERE metric = 'daily-active-users'
GROUP BY anchor
ORDER BY anchor
When a date is the start date of a week adding 6 (1 + 5) days will move the date to the last date of the week. The the addition of one is to move sundays to the following week and the 5 to get the end of the week from the start date.
Note, PostgreSQL allows the addition of integers (= days) to dates.
I am trying to count 6am-12:30am(next day) as one date.
For some reason I cannot pull this data from the next day in for the previous day.
Is this possible?
(CASE WHEN TO_CHAR(ITD.TRAN_DATE,'HH24MI')>='0600' THEN TO_CHAR(ITD.TRAN_DATE,'HHAM')
WHEN TO_CHAR(TRUNC(ITD.TRAN_DATE+1),'HH24MI')<='0030' THEN TO_CHAR(ITD.TRAN_DATE,'HHAM')
END)
I am using this case statement to have everything until 12:30am the next day to count as the previous day. It will not work when I set a date parameter.
The answer to your question is yes.
The pseudo-code is:
IF (TRAN_TIME <= 00:30 AND TRAN_DATE = TODAY + 1) OR
(TRAN_TIME >= 06:00 AND TRAN_DATE = TODAY)
THEN ...
What you are currently doing, is taking the existing date, and adding one to it, before comparing the times, and that won't return what you are expecting.
Use an INTERVAL data type to add an offset to a date:
SELECT *
FROM itd
WHERE ITD.TRAN_DATE
BETWEEN TRUNC( :date_to_match ) + INTERVAL '00 06:00' DAY TO MINUTE
AND TRUNC( :date_to_match ) + INTERVAL '01 12:30' DAY TO MINUTE;
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 keep track of recurring weekly events in a table using just a DATETIME. I only care about the TIME and the day of the week it falls on.
I need to be able to convert the set DATETIME into the current or upcoming future one.
IE How can I convert a date stored as 2013-02-22 12:00:00 using the current date to the next occurrence? Ie this next Friday at 12:00:00 or 2013-03-01 12:00:00 so that I can then order events by date?
Or I could store the TIME and day of the week separately as a number 0-6.
UPDATE:
From Erwin I got something like:
Event.order("date_trunc('week', now()::timestamp) + (start_at - date_trunc('week', start_at))")
Which seems order them except that the first dates I get are Monday skipping over events I know exist for Sunday which it puts as last.
Your best choice is to store a timestamp or timestamptz (timestamop with time zone). If you have or ever will have to deal with more than one time zone, make that timestamptz and define whether you want to operate with local time or UTC or whatever. More details in this related answer:
Ignoring timezones altogether in Rails and PostgreSQL
Demo how to transpose a timestamp into the current week efficiently (same day of week and time). Assuming timestamp here:
SELECT date_trunc('week', now()::timestamp) + (t - date_trunc('week', t))
FROM (SELECT '2013-02-15 12:00:00'::timestamp AS t) x;
The trick is to compute the interval between the start of the corresponding week and the given timestamp and add that to the start of the current week with the help of date_trunc().
The ISO week starts with Monday, putting Sunday last.
Or, to just add a week to a given timestamp:
SELECT t + interval '1 week';
If You just want to ORDER BY, you only need the interval:
ORDER BY (t - date_trunc('week', t))
If you want to put Sunday first (shifting days):
ORDER BY ((t + interval '1d') - date_trunc('week', (t + interval '1d'))
Or simpler:
ORDER BY EXTRACT(dow FROM t), t::time
Quoting the manual on EXTRACT():
dow
The day of the week as Sunday(0) to Saturday(6)
isodow
The day of the week as Monday(1) to Sunday(7)
Answer to question in comment
I'm only interested in ordering them relative to the current date. Ie
if it's tuesday, I want tuesday first, monday last.
Wrapping at midnight of "today":
ORDER BY (EXTRACT(dow FROM t)::int + 7 - EXTRACT(dow FROM now())::int) % 7
,t::time
Using the modulo operator % to shift the day according to "today".
Using dowinstead of isodow, because starting with 0 makes % simpler.
Keep using the datetime. It's simple and gives you flexibility. You can use the extract function to get your time of day and day of week results. This page will help you. http://www.postgresql.org/docs/9.3/static/functions-datetime.html