Make date_trunc() start on Sunday instead of Monday - postgresql

Select date_trunc('week',dateTime) Date_week, Max(Ranking) Runing_Total_ID
from (select datetime, id , dense_rank () over (order by datetime) as Ranking
from Table1)
group by 1
This query is working for me to give me the running total of total IDs by week. But the week starts on Monday in Postgres by default. Is there any way to change the week start to SUNDAY?

Shift the timestamp back and forth:
Add a day before feeding the timestamp to date_trunc(), then subtract again:
SELECT date_trunc('week', datetime + interval '1 day') - interval '1 day' AS date_week
, max(ranking) AS runing_total_id
FROM (
SELECT datetime, dense_rank() OVER (ORDER BY datetime) AS ranking
FROM table1
) sub
GROUP BY 1;
See:
PostgreSQL custom week number - first week containing Feb 1st

Related

Dynamic value passing in Postgres

Here is a complex query where i need to pass some dates as dynamic to this, As of now i have hardcoded this '2021-08-01' AND '2022-07-31' these 2 dates.
But i have to pass this dates dynamically in such a way that next dates ie, 2022-06 month , thew dates passed will be '2021-07-01' and '2022-06-30' , basically 12 months behind data.
if we take 2022-05 then the passed date should be '2021-06-01' and '2022-05-31'.
How can we achieve this ? Any suggestions or help will be much appreciated.
below is the query for reference
WITH base as
(
SELECT created_at as period ,order_number, TRIM(email) as email ,is_first_order
FROM orders
WHERE created_at::DATE BETWEEN '2021-08-01' AND '2022-07-31'
)
,base_agg as
(
select TO_CHAR(period,'YYYY-MM') as period
,COUNT(DISTINCT email)FILTER(WHERE is_first_order IS TRUE) as new_users
,COUNT(DISTINCT order_number)FILTER(WHERE is_first_order IS FALSE) as returning_orders
FROM base
GROUP BY 1
)
,base_cumulative as
(
SELECT ROW_NUMBER() OVER(ORDER BY PERIOD DESC ) as rno
,period
,new_users
,returning_orders
,sum("new_users")over (order by "period" asc rows between unbounded preceding and current row) as "cumulative_total"
from base_agg
)
SELECT
(SELECT period FROM base_cumulative WHERE rno=1) period
,(SELECT cumulative_total FROM base_cumulative WHERE rno=1) as cumulated_customers
,SUM(returning_orders) as returning_orders
,SUM(returning_orders)/NULLIF((SELECT cumulative_total FROM base_cumulative WHERE rno=1),0) as rate
FROM base_cumulative
You can calculate the end of current month based on NOW() and some logic, the same can be applied with the rest of the calculation
select date_trunc('month', now())::date + interval '1 month - 1 day' end_of_this_month,
date_trunc('month', now())::date + interval '1 month - 1 day'::interval - '1 year'::interval + '1 day'::interval first_day_of_prev_year_month
;
Result
end_of_this_month | first_day_of_prev_year_month
---------------------+------------------------------
2022-08-31 00:00:00 | 2021-09-01 00:00:00
(1 row)

Fetch records of current month using PostgreSQL query

Suppose I have following data in a table
id createdAt
1 2021-02-26T06:29:03.482Z
2 2021-02-27T06:29:03.482Z
3 2021-03-14T06:29:03.482Z
4 2021-03-17T06:29:03.482Z
I want data of current month. ie, if I generate report in march, I need to fetch results of march, so we need only current month data from table.
wanted output is
id createdAt
3 2021-03-14T06:29:03.482Z
4 2021-03-17T06:29:03.482Z
Anyone please help. Thank you.
You can use date_trunc():
select *
from the_table
where date_trunc('month', createdat) = date_trunc('month', current_timestamp);
date_trunc('month', ...) returns the first day of the month.
However, the above is not able to make use of an index on createdat. To improve performance, use a range query:
select *
from the_table
where createdat >= date_trunc('month', current_timestamp)
and createdat < date_trunc('month', current_timestamp) + interval '1 month'
The expression date_trunc('month', current_timestamp) + interval '1 month' returns the start of the next month (that's way this is compared with <)
You can compare the month and year of a date with the current one. But the index by field will not be used, you can build a separate index by year and month for this.
select *
from your_table
where extract(YEAR FROM createdAt) = extract(YEAR FROM now())
and extract(MONTH FROM createdAt) = extract(MONTH FROM now())

postgresql list of time slots from 'Monday' | 09:00:00 | 11:00:00

I’m building a booking system where a user will set their availability eg: I’m available Monday’s from 9am to 11am, Tuesdays from 9am to 5pm etc… and need to generate a list of time slots 15mins apart from their availability.
I have the following table (but am flexible to changing this):
availabilities(day_of_week text, start_time: time, end_time: time)
which returns records like:
‘Monday’ | 09:00:00 | 11:00:00
‘Monday’ | 13:00:00 | 17:00:00
‘Tuesday’ | 08:00:00 | 17:00:00
So I’m trying to build a stored procedure to generate a list of time slots so far I've got this:
create or replace function timeslots ()
return setof timeslots as $$
declare
rec record;
begin
for rec in select * from availabilities loop
/*
convert 'Monday' | 09:00:00 | 11:00:00 into:
2020-02-03 09:00:00
2020-02-03 09:15:00
2020-02-03 09:30:00
2020-02-03 09:45:00
2020-02-03 10:00:00
and so on...
*/
return next
end loop
$$ language plpgsql stable;
I return a setof instead of a table as I'm using Hasura and it needs to return a setof so I just create a blank table.
I think I'm on the right track but am currently stuck on:
how do I create a timestamp from 'Monday' 09:00:00 for the next monday as I only care about timeslots from today onwards?
how do I convert 'Monday' | 09:00:00 | 11:00:00 into a list of time slots 15 mins apart?
how do I create a timestamp from 'Monday' 09:00:00 for the next monday
as I only care about timeslots from today onwards?
You can use date_trunc for this (see this question for more info):
SELECT date_trunc('week', current_date) + interval '1 week';
From the docs re week:
The number of the ISO 8601 week-numbering week of the year. By
definition, ISO weeks start on Mondays
So taking this value and adding a week gives next Monday (you may need to ammend this behaviour based upon what you want to do if today is monday!).
how do I convert 'Monday' | 09:00:00 | 11:00:00 into a list of time
slots 15 mins apart?
This is a little tricker; generate_series will give you the timeslots but the trick is getting it into a result set. The following should do the job (I have included your sample data; change the values bit to refer to your table) - dbfiddle :
with avail_times as (
select
date_trunc('week', current_date) + interval '1 week' + case day_of_week when 'Monday' then interval '0 day' when 'Tuesday' then interval '1 day' end + start_time as start_time,
date_trunc('week', current_date) + interval '1 week' + case day_of_week when 'Monday' then interval '0 day' when 'Tuesday' then interval '1 day' end + end_time as end_time
from
(
values
('Monday','09:00:00'::time,'11:00:00'::time),
('Monday','13:00:00'::time,'17:00:00'::time),
('Tuesday','08:00:00'::time,'17:00:00'::time)
) as availabilities (day_of_week,
start_time,
end_time) )
select
g.ts
from
(
select
start_time,
end_time
from
avail_times) avail,
generate_series(avail.start_time, avail.end_time - interval '1ms', '15 minutes') g(ts);
A few notes:
The CTE avail_times is used to simplify things; it generates two columns (start_time and end_time) which are the full timestamps (so including the date). In this example the first row is "2020-02-03 09:00:00, 2020-02-03 11:00:00" (I'm running this on 2020-02-02 so 2020-02-03 is next Monday).
The way I'm converting 'monday' etc to a day of the week is a bit of a hack (and I have not bothered to do the full week); there is probably a better way but storing the day of week as an integer would make this simpler.
I subtract 1ms from the end time because I'm assuming you dont want this in the result set.
The main query is using a LATERAL Subquery. See this question for more info.
Aditional Question
how to adjust this so I can pass in a start and end date so I can get
time slots for a particular period
You could do something like the following (just adjust the dates CTE to return whatever days you want to include; you could convert to a function or just pass the dates in as parameters).
Note that as #Belayer mentions my original solution did not cater for shifts over midnight so this addresses that too.
with dates as (
select
day
from
generate_series('2020-02-20'::date, '2020-03-10'::date, '1 day') as day ),
availabilities as (
select
*
from
(
values (1,'09:00:00'::time,'11:00:00'::time),
(1,'13:00:00'::time,'17:00:00'::time),
(2,'08:00:00'::time,'17:00:00'::time),
(3,'23:00:00'::time,'01:00:00'::time)
) as availabilities
(day_of_week, -- 1 = monday
start_time,
end_time) ) ,
avail_times as (
select
d.day + start_time as start_time,
case
end_time > start_time
when true then d.day
else d.day + interval '1 day' end + end_time as end_time
from
availabilities a
inner join dates d on extract(ISODOW from d.day) = a.day_of_week )
select
g.ts
from
(
select
start_time,
end_time
from
avail_times) avail,
generate_series(avail.start_time, avail.end_time - interval '1ms', '15 minutes') g(ts)
order by
g.ts;
The following uses much of the techniques mentioned by #Brits. They present some very good information, so I'll not repeat but suggest you review it (and the links).
I do however take a slightly different approach. First a couple table changes. I use the ISO day of week 1-7 (Monday-Sunday) rather than the day name. The day name is easily extracted for the dater later.
Also I use interval instead to time for start and end times. ( A time data type works for most scenarios but there is one it doesn't (more later).
One thing your description does not make clear is whether the ending time is included it the available time or not. If included the last interval would be 11:00-11:15. If excluded the last interval is 10:45-11:00. I have assumed to excluded it. In the final results the end time is to be read as "up to but not including".
-- setup
create table availabilities (weekday integer, start_time interval, end_time interval);
insert into availabilities (weekday , start_time , end_time )
select wkday
, start_time
, end_time
from (select *
from (values (1, '09:00'::interval, '11:00'::interval)
, (1, '13:00'::interval, '17:00'::interval)
, (2, '08:00'::interval, '17:00'::interval)
, (3, '08:30'::interval, '10:45'::interval)
, (4, '10:30'::interval, '12:45'::interval)
) as v(wkday,start_time,end_time)
) r ;
select * from availabilities;
The Query
It begins with a CTE (next_week) generates a entry for each day of the week beginning Monday and the appropriate ISO day number for it. The main query joins these with the availabilities table to pick up times for matching days. Finally that result is cross joined with a generated timestamp to get the 15 minute intervals.
-- Main
with next_week (wkday,tm) as
(SELECT n+1, date_trunc('week', current_date) + interval '1 week' + n*interval '1 day'
from generate_series (0, 6) n
)
select to_char(gdtm,'Day'), gdtm start_time, gdtm+interval '15 min' end_time
from ( select wkday, tm, start_time, end_time
from next_week nw
join availabilities av
on (av.weekday = nw.wkday)
) s
cross join lateral
generate_series(start_time+tm, end_time+tm- interval '1 sec', interval '15 min') gdtm ;
The outlier
As mentioned there is one scenario where a time data type does not work satisfactory, but you may not nee it. What happens when a shift worker says they available time is 23:00-01:30. Believe me when a shift worker goes to work at 22:00 of Friday, 01:30 is still Friday night, even though the calendar might not agree. (I worked that shift for many years.) The following using interval handles that issue. Loading the same data as prior with an addition for the this case.
insert into availabilities (weekday, start_time, end_time )
select wkday
, start_time
, end_time + case when end_time < start_time
then interval '1 day'
else interval '0 day'
end
from (select *
from (values (1, '09:00'::interval, '11:00'::interval)
, (1, '13:00'::interval, '17:00'::interval)
, (2, '08:00'::interval, '17:00'::interval)
, (3, '08:30'::interval, '10:45'::interval)
, (5, '23:30'::interval, '02:30'::interval) -- Friday Night - Saturday Morning
) as v(wkday,start_time,end_time)
) r
;
select * from availabilities;
Hope this helps.

Grouping data by quarter intervals (or any time interval) with a defined starting basis in postgresql

Let's say I have a table orders with columns amount and order_date.
I want to be able to group this data by quarters and aggregate the amount, the catch however is that the quarters do not start on January 1st but on any given arbitrary date, say July 12th. These quarters are also split in 13 week intervals. From what I see using something like date_trunc such as:
SELECT SUM(orders.amount), DATE_TRUNC('quarter', orders.order_date) AS interval FROM orders WHERE orders.order_date BETWEEN [date_start] AND [date_end] GROUP BY interval
is out of the question as this forces quarters to start on Jan 1st and it has 'hardcoded' quarter starting dates (Apr 1st, Jul 1st, etc).
I have tried using something like:
SELECT SUM(orders.amount),
to_timestamp(floor((extract('epoch' from orders.order_date / 7862400 )) * 7862400 ) AT TIME ZONE 'UTC' AS interval
FROM orders
WHERE orders.order_date BETWEEN [date_start] AND [date_end]
GROUP BY interval
(where 7862400 is the time interval that I want)
But with this method I cannot figure out how to set the offset for the initial grouping date, in my example I would like it to start from July 12th of each year (then count 13 weeks and start the next quarter, and so on). Hope I was clear and I would appreciate any help!
You can use generate_series() to create the first day of each quarter, join it and group by it.
SELECT quarters.first_day,
quarters.first_day + '13 weeks'::interval last_day,
sum(orders.amount) amount
FROM orders
LEFT JOIN generate_series('2019-07-12'::timestamp,
'2020-07-10'::timestamp,
'13 weeks'::interval) quarters (first_day)
ON quarters.first_day <= orders.order_date
AND quarters.first_day + '13 weeks'::interval > orders.order_date
WHERE orders.order_date BETWEEN [date_start]
AND [date_end]
GROUP BY quarters.first_day,
quarters.first_day + '13 weeks'::interval;
You just need to make sure, that the boundary days you give the generate_series() cover the whole period you want to query, so that depends on your [date_start] and [date_end].
You can generate your own 'quarterly calendar' and use that in place of the Postgers 'quarter' date extraction.
create or replace function quarterly_calendar(annual_date text default extract('YEAR' from current_date)::text)
returns table( quarter integer
, quarter_start_date date
, quarter_end_date date
)
language sql immutable strict leakproof
as $$
with RECURSIVE quarters as
(select 1 qtr, qdt::date q_start_dt, (qdt + interval '90 day' )::date q_end_dt, (qdt+interval '1 year' - interval '1 day')::date last_dt
from ( select date_trunc('year',current_date) + interval '6 month 11 day' qdt) q
union all
select qtr+1, (q_end_dt + interval '1 day')::date, least ((q_end_dt + interval '91 day')::date,last_dt), last_dt
from quarters
where qtr+1 <=5
)
select qtr, q_start_dt, q_end_dt
from quarters;
$$;
-- test
select * from quarterly_calender();
It does actually create 5 quarters. But that is because a year is not a multiple of 13 weeks (or 91 days or 7862400 seconds). In your given year from 12-July-2019 through 11-July-2020 is 2 days (366 days total) over 4 times that interval. You'll have to decide how to handle that 5th quarter. It occurs every year, having either 1 or 2 days. Hope this helps .

Compare day in current month to same day previous month PostgreSQL

I'm trying to compare values of current month's data to previous months using PostgreSQL. So if today is 4/23/2018, I want the data for 3/23/2018.
I've tried current_date - interval '1 month' but it is problematic for months with 31 days.
My table is structured as simply as
date, value
Check this example query:
WITH dates AS (SELECT date::date FROM generate_series('2018-01-01'::date, '2018-12-31'::date, INTERVAL '1 day') AS date)
SELECT
start_dates.date AS start_date,
end_dates.date AS end_date
FROM
dates AS start_dates
RIGHT JOIN dates AS end_dates
ON ( start_dates.date + interval '1 month' = end_dates.date AND
end_dates.date - interval '1 month' = start_dates.date);
It will output all end_dates and corresponding start_dates. The corresponding dates are defined by interval '1 month' and checked in both ways:
start_dates.date + interval '1 month' = end_dates.date AND
end_dates.date - interval '1 month' = start_dates.date
The output looks like this:
....
2018-02-26 2018-03-26
2018-02-27 2018-03-27
2018-02-28 2018-03-28
2018-03-29
2018-03-30
2018-03-31
2018-03-01 2018-04-01
2018-03-02 2018-04-02
2018-03-03 2018-04-03
2018-03-04 2018-04-04
....
Note, that there are 'gaps' for days without corresponding dates.
Back to your table, join the table with itself (giving aliases) and use given join condition, so the query would look like this:
SELECT
start_dates.value - end_dates.value AS change,
start_dates.date AS start_date,
end_dates.date AS end_date
FROM
_your_table_name_ AS start_dates
RIGHT JOIN _your_table_name_ AS end_dates
ON ( start_dates.date + interval '1 month' = end_dates.date AND
end_dates.date - interval '1 month' = start_dates.date);
Given the following table structure:
create table t (
d date,
v int
);
After populating with some dates and values, there is a way to find the value of the previous month using simple calculations and the LAG function, without resorting to joins. I am not sure how it compares from a performance perspective, so please run your own tests before selecting which solution to use.
select
*,
lag(v, day_of_month) over (order by d) as v_end_of_last_month,
lag(v, last_day_of_previous_month + day_of_month - cast(extract(day from d - interval '1 month') as int)) over (order by d) as v_same_day_last_month
from (
select
*,
lag(day_of_month, day_of_month) over (order by d) as last_day_of_previous_month
from (
select
*,
cast(extract(day from d) as int) as day_of_month
from
t
) t_dom
) t_dom_ldopm;
You may note that between the 29th and 31st of March, the comparison will be made against the 28th of February, since the same day does not exist in February for those particular dates. The same logic applies to other months with different number of days.