I need to get all week days in a given time interval.
In postgresql, there are dow and isodow
By mixing them together, may I write a function to retrieve weekdays?
demo:db<>fiddle
SELECT
generated_date,
to_char(generated_date, 'Day'), -- 1
EXTRACT(isodow FROM generated_date), -- 2
EXTRACT(dow FROM generated_date) -- 3
FROM
generate_series('2020-11-01'::date, '2020-11-10'::date, interval '1 day') AS generated_date
Returns the name for the weekday
Returns the number of the weekday (Monday = 1, Sunday = 7)
Returns the number of the weekday (Sunday = 0, Saturday = 6)
Edit:
If you want to get the days without weekend, you can filter by the dow/isodow values, e.g.:
SELECT
generated_date::date
FROM
generate_series('2020-11-01'::date, '2020-11-10'::date, interval '1 day') AS generated_date
WHERE
EXTRACT(isodow FROM generated_date) < 6
As far as I understand you need to extract all Monday..Fridays between two dates. Here is an illustration with 2020-11-30 as the beginning of the interval and 2020-12-12 as the end of it.
select d
from generate_series('2020-11-30'::date, '2020-12-12'::date, '1 day'::interval) t(d)
where extract(isodow from d) between 1 and 5;
Related
My financial year start from 01-Jul to 30-Jun every year.
I want to find out all financial year wise periods for a given date range.
Let's say, The date range is From_Date:16-Jun-2021 To_Date 31-Aug-2022. Then my output should be like
Start_Date, End_date
16-Jun-2021, 30-Jun-2021
01-Jul-2021, 30-Jun-2022
01-jul-2022, 31-Aug-2022
Please help me query. First record Start_Date must start from From_Date and Last record End_Date must end at To_Date
This should work for the current century.
with t(fys, fye) as
(
select (y + interval '6 months')::date,
(y + interval '1 year 6 months - 1 day')::date
from generate_series ('2000-01-01'::date, '2100-01-01', interval '1 year') y
),
periods (period_start, period_end) as
(
select
case when fys < '16-Jun-2021'::date then '16-Jun-2021'::date else fys end,
case when fye > '31-Aug-2022'::date then '31-Aug-2022'::date else fye end
from t
)
select * from periods where period_start < period_end;
period_start
period_end
2021-06-16
2021-06-30
2021-07-01
2022-06-30
2022-07-01
2022-08-31
Looks well as a parameterized query too with '16-Jun-2021' and '31-Aug-2022' replaced by parameter placeholders.
You want to create multiple records from one record (your date range). To accomplish this, you will need some kind of helper table.
In this example I created that helper table using GENERATE_SERIES and use it to join it to your date range, with some logic to get the dates you want.
dbfiddle
--Generate a range of fiscal years
WITH FISCAL_YEARS AS (
SELECT
CONCAT(SEQUENCE.YEAR, '-07-01')::DATE AS FISCAL_START,
CONCAT(SEQUENCE.YEAR + 1, '-06-30')::DATE AS FISCAL_END
FROM GENERATE_SERIES(2000, 2030) AS SEQUENCE (YEAR)
),
--Your date range
DATE_RANGE AS (
SELECT
'2021-06-16'::DATE AS RANGE_START,
'2022-08-31'::DATE AS RANGE_END
)
SELECT
--Case statement in case the range_start is later
--than the start of the fiscal year
CASE
WHEN RANGE_START > FISCAL_START
THEN RANGE_START
ELSE FISCAL_START
END AS START_DATE,
--Case statement in case the range_end is earlier
--than the end of the fiscal year
CASE
WHEN RANGE_END < FISCAL_END
THEN RANGE_END
ELSE FISCAL_END
END AS END_DATE
FROM FISCAL_YEARS
JOIN DATE_RANGE
--Join to get all relevant fiscal years
ON FISCAL_YEARS.FISCAL_START BETWEEN DATE_RANGE.RANGE_START AND DATE_RANGE.RANGE_END
OR FISCAL_YEARS.FISCAL_END BETWEEN DATE_RANGE.RANGE_START AND DATE_RANGE.RANGE_END
I want to define the start of a “month” as the 26th day of the previous calendar month (and of course ending on 25th).
How can I group by this definition of month using date_trunc()?
This expression gives the month you want:
date_trunc(
'month',
date_add(
day,
case
when extract(day from date) > 25 then 7
else 0
end),
my_date_col
)
)
Select it and group by it.
The logic is: If the day of the month is greater than 25, then add some days to bump it into the next month before truncating it to the month.
I would use an INTERVAL to calculate the correct dates. Here an example using generate_series():
SELECT d::date as reference_date
, (d + interval '25 days')::date AS first_day
, (d + interval '1 month' + interval '24 days')::date as last_day
FROM generate_series('2020-01-01'::timestamp, '2021-01-01'::timestamp, '1 month') g(d);
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);
I need to truncate dates to the start of week, which is Sunday in my case. How can I do this in PostgreSQL? This truncates to Monday:
date_trunc('week', mydate)
If you subtract the dow value (0 for Sundays, 6 for Saturdays) from the current date than you get the previous Sunday which is the begin of your Sunday-based week
demo:db<>fiddle
SELECT
my_date - date_part('dow', my_date)::int
FROM
my_table
Further reading, documentation
You could truncate the date to the week's Monday, then subtract 1 day, e.g:
SELECT (date_trunc('week', now() + interval '1 day') - interval '1 day')::DATE;
date
------------
2019-06-16
As per documentation, date_trunc() accepts values of type date and timestamp and returns a timestamp (thus the cast at the end).
How to calculate end of the month in Postgres? I have table with column date datatype. I want to calculate end of the month of every date. For Eg. In the table there values like "2015-07-10 17:52:51","2015-05-30 11:30:19" then end of the month should be like 31 July 2015,31 May 2015.
Please guide me in this.
How about truncating to the beginning of this month, jumping forward one month, then back one day?
=# select (date_trunc('month', now()) + interval '1 month - 1 day')::date;
date
------------
2015-07-31
(1 row)
Change now() to your date variable, which must be a timestamp, per the docs. You can then manipulate this output (with strftime, etc.) to any format you need.
Source
SELECT TO_CHAR(
DATE_TRUNC('month', CURRENT_DATE)
+ INTERVAL '1 month'
- INTERVAL '1 day',
'YYYY-MM-DD HH-MM-SS'
) endOfTheMonth
Hi I tried like this and it worked
Date(to_char(date_trunc('month'::text, msm013.msa011) + '1 mon - 1 day '::interval , 'DD-MON-YYYY') )
Thanks a lot!!