I also have the question how do i get code block to work on stack overflow but that's a side issue.
I have this quasi-code that works:
select
*
from
unnest('{2018-6-1,2018-7-1,2018-8-1,2018-9-1}'::date[],
'{2018-6-30,2018-7-31,2018-8-31,2018-9-30}'::date[]
) zdate(start_date, end_date)
left join lateral pipe_f(zdate...
But now I want it to work from 6/1/2018 until now(). What's the best way to do this.
Oh, postgresql 10. yay!!
Your query gives a list of first and last days of months between "2018-06-01" and now. So I am assuming that you want to this in a more dynamic way:
demo: db<>fiddle
SELECT
start_date,
(start_date + interval '1 month -1 day')::date as end_date
FROM (
SELECT generate_series('2018-6-1', now(), interval '1 month')::date as start_date
)s
Result:
start_date end_date
2018-06-01 2018-06-30
2018-07-01 2018-07-31
2018-08-01 2018-08-31
2018-09-01 2018-09-30
2018-10-01 2018-10-31
generate_series(timestamp, timestamp, interval) generates a list of timestamps. Starting with "2018-06-01" until now() with the 1 month interval gives this:
start_date
2018-06-01 00:00:00+01
2018-07-01 00:00:00+01
2018-08-01 00:00:00+01
2018-09-01 00:00:00+01
2018-10-01 00:00:00+01
These timestamps are converted into dates with ::date cast.
Then I add 1 month to get the next month. But as we are interested in the last day of the previous month I subtract one day again (+ interval '1 month -1 day')
Another option that's more ANSI-compliant is to use a recursive CTE:
WITH RECURSIVE
dates(d) AS
(
SELECT '2018-06-01'::TIMESTAMP
UNION ALL
SELECT d + INTERVAL '1 month'
FROM dates
WHERE d + INTERVAL '1 month' <= '2018-10-01'
)
SELECT
d AS start_date,
-- add 1 month, then subtract 1 day, to get end of current month
(d + interval '1 month') - interval '1 day' AS end_date
FROM dates
Related
I'm working on the Accrual Reversal query in PostgreSQL. The system running doesn't have the reversal flag. So I need to consider all the end of the day of previous month accrued invoices as the reversal amount. And need to union them all with the main query. I can do it for last month but invoice date are dynamic, user may give 2 years as invoice period. For those 2 years, all the previous month data should be considered as accrued reversal. Here is the query
select invoicename, * from accountpay where invoice_date between '2020-01-01' and '2021-12-31'
union all
select concat('Accured Reversal', invoicename) as reference, * from accountpay where accrual = true and invoice_date::date = (select concat(date_part('year',((('2021-12-30'::date) - interval '1 month'))), '-', date_part('month',((('2021-12-30'::date) - interval '1 month'))), '-01')::date + interval '1 month' - interval '1 day')
Please help me to do this.
Thanks in Advance
SELECT (
Date_trunc('MONTH',a) + interval '1 month -1 day ')
as last_day_of_month
FROM generate_series(
'2020-01-01 00:00'::timestamp
- interval '12 months',
'2022-01-01 00:00',
'1 month') as dt(a);
get last_day_of_month from '2020-01-01 00:00' till '2022-01-01 00:00'
Then your sql would be
invoice_date in
(SELECT (Date_trunc('MONTH',a) + interval '1 month -1 day ')
as last_day_of_month
FROM generate_series(
'2020-01-01 00:00'::timestamp
- interval '12 months',
'2021-01-01 00:00',
'1 month') as dt(a))
This will get the last day of last 12 months so the number of months will be place holder (dynamic) and all the last day of the months will be in IN clause.
SQL re-written:
WITH date_cte AS
(
SELECT Date_trunc('MONTH',dt)+ interval '1 month -1 day ' last_day_of_month
FROM generate_series('2021-11-30 00:00:00'::timestamp - interval '12 months','2021-11-30 00:00:00','1 month') t(dt))
select invoicename, * from accountpay where invoice_date between '2020-01-01' and '2021-12-31'
union all
select concat('Accured Reversal', invoicename) as reference, * from accountpay where accrual = true and invoice_date::date in (select * from date_cte);
Basically , the last dates are generated this way for 12 months:
WITH date_cte AS
(
SELECT Date_trunc('MONTH',dt)+ interval '1 month -1 day ' last_day_of_month
FROM generate_series('2021-11-30 00:00:00'::timestamp - interval '12 months','2021-11-30 00:00:00','1 month') t(dt))
SELECT *
FROM date_cte;
last_day_of_month
---------------------
2020-11-30 00:00:00
2020-12-31 00:00:00
2021-01-31 00:00:00
2021-02-28 00:00:00
2021-03-31 00:00:00
2021-04-30 00:00:00
2021-05-31 00:00:00
2021-06-30 00:00:00
2021-07-31 00:00:00
2021-08-31 00:00:00
2021-09-30 00:00:00
2021-10-31 00:00:00
2021-11-30 00:00:00
You can replace 12 months by any number of months or you can make it year too like:
...generate_series('2021-11-30 00:00:00'::timestamp - interval '1 year','2021-11-30 00:00:00','1 month')
i have following query in postgresql for dates between 2 ranges.
select generate_series('2019-04-01'::timestamp, '2020-03-31', '1 month')
as g_date
I need to generate specific date in every month .i.e 15 th of every month. Following is my query to generate series
DO $$
DECLARE
compdate date = '2019-04-15';
BEGIN
CREATE TEMP TABLE tmp_table ON COMMIT DROP AS
select *,
case
when extract('day' from d) <> extract('day' from compdate) then 0
when ( extract('month' from d)::int - extract('month' from compdate)::int ) % 1 = 0 then 1
else 0
end as c
from generate_series('2019-04-01'::timestamp, '2020-03-31', '1 day') d;
END $$;
SELECT * FROM tmp_table
where c=1;
;
But every thing is perfect if input date between (1..29)-04-2019 ..
2019-04-25
2019-05-25
2019-06-25
2019-07-25
2019-08-25
2019-09-25
2019-10-25
2019-11-25
2019-12-25
2020-01-25
2020-02-25
2020-03-25
but if i give compdate: 31-04-2019 or 30-04-2019 giving out put:
2019-05-31
2019-07-31
2019-08-31
2019-10-31
2019-12-31
2020-01-31
2020-03-31
Expected Output:
date flag
2019-04-01 0 ----start_date
2019-04-30 1
2019-05-31 1
2019-06-30 1
2019-07-31 1
2019-08-31 1
2019-09-30 1
2019-10-31 1
2019-11-30 1
2019-12-31 1
2020-01-31 1
2020-02-29 1
2020-03-31 0 ---end_date
If matched day not found in the result it should take last day of that month..i.e if 31 not found in month of feb it
should take 29-02-2019 and also in april month instead of 31 it should take 2019-04-30.
Please suggest.
to generate the last days of the month, just generate first days & subtract a 1 day interval
example: the following generates all last day of month in the year 2010
SELECT x - interval '1 day' FROM
GENERATE_SERIES('2010-02-01', '2011-01-01', interval '1 month') x
You cannot accomplish what you want with generate_series. This results due to that process applying a fixed increment from the previous generated value. Your case 1 month. Now Postgres will successfully compute correct end-of-month date from 1 month to the next. So for example 1month from 31-Jan yields 28-Feb (or 29), because 31-Feb would be an invalid date, Postgres handles it. However, that same interval from 28-Feb gives the valid date 28-Mar so no end-of-month adjustment is needed. Generate_Series will return 28th of the month from then on. The same applies to 30 vs. 31 day months.
But you can achieve what your after with a recursive CTE by employing a varying interval to the same initial start date. If the resulting date is invalid for date the necessary end-of-month adjustment will be made. The following does that:
create or replace function constant_monthly_date
( start_date timestamp
, end_date timestamp
)
returns setof date
language sql strict
as $$
with recursive date_set as
(select start_date ds, start_date sd, end_date ed, 1 cnt
union all
select (sd + cnt*interval '1 month') ds, sd, ed, cnt+1
from date_set
where ds<end_date
)
select ds::date from date_set;
$$;
-- test
select * from constant_monthly_date(date '2020-01-15', date '2020-12-15' );
select * from constant_monthly_date(date '2020-01-31', date '2020-12-31' );
Use the least function to get the least one between the computed day and end of month.
create or replace function test1(day int) returns table (t timestamptz) as $$
select least(date_trunc('day', t) + make_interval(days => day-1), date_trunc('day', t) + interval '1 month' - interval '1 day') from generate_series('2019-04-01', '2020-03-31', interval '1 month') t
$$ language sql;
select test1(31);
Have two dates - '2018-05-01' and '2018-06-01'. I would like to expand this window to the past by day difference of those dates.
SELECT * FROM data
WHERE
start_time > CAST('2018-05-01' AS timestamptz) - INTERVAL '30 DAY'
AND start_time < CAST('2018-06-01' AS timestamptz)
How can I replace INTERVAL '30 DAY' with number of days between given dates without explicitly defining number of days? I know to calculate day difference:
date_part('day',age('2018-05-01', '2018-06-01'))
But not sure how to incorporate into the substraction. Dates and days between them will change.
You can use date_trunc('mon', some_date_expression) to round down to the start of a month:
select date_trunc('mon', now() - '3 mon'::interval) as date_begin
, date_trunc('mon', now() - '1 day'::interval) as date_end
;
Result
date_begin | date_end
------------------------+------------------------
2018-03-01 00:00:00+01 | 2018-06-01 00:00:00+02
(1 row)
You can simply subtract the difference from the start date:
with t (start_date, end_date) as (
values (date '2018-05-01', date '2018-06-01')
)
select start_date - (end_date - start_date) as new_start,
end_date
from t;
returns
new_start | new_end
-----------+-----------
2018-03-31 | 2018-06-01
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.
There is one table:
ID DATE
1 2017-09-16 20:12:48
2 2017-09-16 20:38:54
3 2017-09-16 23:58:01
4 2017-09-17 00:24:48
5 2017-09-17 00:26:42
..
The result I need is the last 7-days of data with hourly aggregated count of rows:
COUNT DATE
2 2017-09-16 21:00:00
0 2017-09-16 22:00:00
0 2017-09-16 23:00:00
1 2017-09-17 00:00:00
2 2017-09-17 01:00:00
..
I tried different stuff with EXTRACT, DISTINCT and also used the generate_series function (most stuff from similar stackoverflow questions)
This try was the best one currently:
SELECT
date_trunc('hour', demotime) as date,
COUNT(demotime) as count
FROM demo
GROUP BY date
How to generate hourly series for 7 days and fill-in the count of rows?
SQL DEMO
SELECT dd, count("demotime")
FROM generate_series
( current_date - interval '7 days'
, current_date
, '1 hour'::interval) dd
LEFT JOIN Table1
ON dd = date_trunc('hour', demotime)
GROUP BY dd;
To work from now and now - 7 days:
SELECT dd, count("demotime")
FROM generate_series
( date_trunc('hour', NOW()) - interval '7 days'
, date_trunc('hour', NOW())
, '1 hour'::interval) dd
LEFT JOIN Table1
ON dd = date_trunc('hour', demotime)
GROUP BY dd;