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
Related
I know of date_part('days', age(release_date)), which will show you days of the age of the release_date col (timestamp). For example, if the release date is 1994-05-30, the date_part calculation would yield 5 days.
However, how can I find out from a countdown perspective, ie, for another release. There are 2 days left till the anniversary of this release hits? Is it just a matter of 365 - date_part('days', age(release_date)), for example? Or is there a better way?
Transpose the release date to the current year and subtract the current date from it. Assuming release_date is an actual date:
SELECT *
, (release_date + (date_trunc('year', LOCALTIMESTAMP)
- date_trunc('year', release_date)))::date
- CURRENT_DATE AS days_till_aniversary
FROM release;
db<>fiddle here
There is a reason for LOCALTIMESTAMP in one spot and CURRENT_DATE in the other. This way, the calculation is done without involving time zones. (Except that either depends on the time zone setting of your session to begin with.) And subtracting dates yields an integer, signifying the difference in days.
Produces negative numbers past anniversaries this year.
I have a below query that I run to extract material movements from the last 7 days.
Purpose is to get the data for the last calender week for certain reports.
select
*
From
redshift
where
posting_date between CURRENT_DATE - 7 and CURRENT_DATE - 1
That means I need to run the query on every Monday to get the data for the former week.
Sometimes I am too busy on Monday or its vacation/bank holiday. In that case I would need to change the query or pull the data via SAP.
Question:
Is there a function for redshift that pulls out the data for the last calender week regardless when I run the query?
I already found following solution
SELECT id FROM table1
WHERE YEARWEEK(date) = YEARWEEK(NOW() - INTERVAL 1 WEEK)
But this doesnt seem to be working for redshift sql
Thanks a lot for your help.
Redshift offers a DATE_TRUNC('week', datestamp) function. Given any datestamp value, either a date or datetime, it gives back the date of the preceding Sunday.
So this might work for you. It filters rows from the Sunday before last, up until but not including, the last Sunday, and so gets a full week.
SELECT id
FROM table1
WHERE date >= DATE_TRUNC('week', NOW()) - INTERVAL 1 WEEK
AND date < DATE_TRUNC('week', NOW())
Pro tip: Every minute you spend learning your DBMS's date/time functions will save you an hour in programming.
I have spent days trying to figure this out to no avail, so hopefully someone can help me. I have a queried date set which contains several fields including a column of dates. What I want to do is create a new field in my query that tells what the Monday and Friday is for the week of that row's particular date.
So for example; if the date in one of my rows is "1/16/18",
the new field should indicate "1/15/18 - 1/19/18".
So basically I need to be able to extract the Monday date (1/15/18) and the Friday date (1/19/18) of the week of 1/16/18 and then concatenate the two with a dash ( - ) in between. I need to do this for every row.
How on earth do I do this? I've been struggling just to figure out how to find the Monday or Friday of the given date...
Assuming that your column is of type date, you can use trunc to get the first day of the week (monday) and then add 4 days to get the friday.
For example:
with yourTable(d) as (select sysdate from dual)
select trunc(d, 'iw'), trunc(d, 'iw') + 4
from yourTable
To format the date as a string in the needed format, you can use to_char; for example:
with yourTable(d) as (select sysdate from dual)
select to_char(trunc(d, 'iw'), 'dd/mm/yy') ||'-'|| to_char(trunc(d, 'iw') + 4, 'dd/mm/yy')
from yourTable
gives
15/01/2018-19/01/18
There may be a simpler, canonical Oracle method to this but you can still reduce it to a simple calculation on your own either way. I'm going to assume you're dealing with only dates falling Monday through Friday. If you do need to deal with weekend dates then you might have to be more explicit about which logical week they should be attached to.
<date> - (to_char(<date>, 'D') - 2) -- Monday
<date> + (6 - to_char(<date>, 'D')) -- Friday
In principle all you need to do is add/subtract the appropriate number of days based on the current day of week (from 1 - 7). There are some implicit casts going on in there and it would probably be wise to handle those better. You might also want to check into NLS settings to make sure you can rely on to_char() using Sunday as the first day of week.
https://docs.oracle.com/cd/B19306_01/server.102/b14200/sql_elements004.htm
You can also use the NEXT_DAY function, as in:
SELECT TRUNC(NEXT_DAY(SYSDATE, 'MON')) - INTERVAL '7' DAY AS PREV_MONDAY,
TRUNC(NEXT_DAY(SYSDATE, 'FRI')) AS NEXT_FRIDAY
FROM DUAL;
Note that using the above, on weekends the Monday will be the Monday preceding the current date, and the Friday will be the Friday following the current date, i.e. there will be 11 days between the two days.
You can also use
SELECT TRUNC(NEXT_DAY(SYSDATE, 'MON')) - INTERVAL '7' DAY AS PREV_MONDAY,
TRUNC(NEXT_DAY(SYSDATE, 'MON')) - INTERVAL '3' DAY AS NEXT_FRIDAY
FROM DUAL;
in which case the Monday and Friday will always be from the same week, but if SYSDATE is on a weekend the Monday and Friday returned will be from the PREVIOUS week.
Pretty new to SQL and have hit a roadblock.
I have this query, which works fine:
SELECT
(COUNT(*)::float / (current_date - '2017-05-17'::date)) AS "avg_per_day"
FROM "table" tb;
I now want it to include only data from the last month, not all time.
I've tried doing something along the lines of:
SELECT
(COUNT(*)::float / (current_date - (current_date - '1 month' ::date)) AS "avg_per_day"
FROM "table" tb;
The syntax is clearly wrong, but I am not sure what the right answer is. Have googled around and tried various options to no avail.
I can't use a simple AVG because the number I require is an AVG per day for the last month of data. Thus I've done a count of rows divided by the number of days since the first occurrence to get my AVG per day.
I have a column which tells me the date of the occurrence, however there are multiple rows with the same date in the dataset. e.g.
created_at
----------------------------
Monday 27th June 2017 12:00
Monday 27th June 2017 13:00
Tuesday 28th June 2017 12:00
and so on.
I am counting the number of occurrences per day and then need to work out an average from that, for the last month of results only (they date back to May).
The answer depends on the exact definition of "last month" and the exact definition of "average count".
Assuming:
Your column is defined created_at timestamptz NOT NULL
You want the average number of rows per day - days without any rows count as 0.
Cover 30 days exactly, excluding today.
SELECT round(count(*)::numeric / 30, 2) -- simple now with a fixed number of days
FROM tbl
WHERE created_at >= (now()::date - 30)
AND created_at < now()::date -- excl. today
Rounding is optional, but you need numeric instead of float to use round() this way.
Not including the current day ("today"), which is ongoing and may result in a lower, misleading average.
If "last month" is supposed to mean something else, you need to define it exactly. Months have between 28 and 31 days, this can mean various things. And since you obviously operate with timestamp or timestamptz, not date, you also need to be aware of possible implications of the time of day and the current time zone. The cast to date (or the definition of "day" in general) depends on your current timezone setting while operating with timestamptz.
Related:
Ignoring timezones altogether in Rails and PostgreSQL
Select today's (since midnight) timestamps only
Subtract hours from the now() function
I think you just need a where clause:
SELECT
(COUNT(*)::float / (current_date - (current_date - '1 month' ::date)) AS "avg_per_day"
FROM "table" tb
WHERE created_at > (current_date - '1 month' ::date)
I believe Postgresql and other RDBMS has AVG() to calculate average.
SELECT AVG(tb.columnName) AS avg_per_month
FROM someTable tb
WHERE
tb.createdDate >= [start date of month] AND
tb.createdDate <= [end date of month]
Edit: I subtract current date with INTERVAL. I am on mobile phone so I cannot test.
SELECT
(COUNT(*)::float / (current_date - ( current_date - INTERVAL '1 month')) AS "avg_per_day"
FROM "table" tb;
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))