I need to get last of the five days from month like:
1-5 = 5
6-10 = 10
...
26-30/31 = 30/31 (here can be 6 days depending on the month)
I've prepared function like
create or replace function getfirstdayoffive()
returns date
as
$$
select date_trunc('month', current_date - 5)::date
+ (least(ceil(extract(day from current_date - 5) / 5) * 5,
date_part('day', date_trunc('month', startOp) + interval '1 month - 1 day')))::int - 1;
$$
language sql
stable;
and it is working fine to return last day of five. How can I modify it so it would recognize if the last period should have 5 or 6 days?
try following function:
CREATE OR REPLACE FUNCTION public.days_in_month(d date)
RETURNS integer
LANGUAGE sql
AS $function$
SELECT date_trunc('month', $1::timestamp + interval '1 month')::date
- date_trunc('month', $1::timestamp)::date;
$function$
Determine the last date of the month, then extract day. If the day is 31 then return minus 6 days, else return minus five days. That assumes you want the last 5 days for Feb. But then except for Feb you could just return the 25th of the month as that is what minus 5 for months with 30 days and minus 6 for days 31 always returns. Note: rather than hard coding current_date this allows a parameter with default value of current_date.
create or replace
function getfirstdayoffive(parm_date_in date default current_date)
returns date
language sql
immutable strict
as $$
with last_of_mon(eom) as
( select date_trunc('month', parm_date_in) + interval '1 month - 1 day' )
select case when extract(day from eom) = 31
then (eom-interval '6 days')::date
else (eom-interval '5 days')::date
end
from last_of_mon;
$$;
select * from getfirstdayoffive();
select * from getfirstdayoffive(date '2021-08-15');
select * from getfirstdayoffive(date '2020-02-15');
hmm for now I've got something like :
create or replace
function getlastdayoffive(parm_date_in date default current_date)
returns timestamp
language sql
immutable strict
as $$
with last_of_mon(eom) as
( select date_trunc('month', parm_date_in) + interval '1 month - 1 day' )
select case when extract(day from eom) = 31
then (least(ceil(extract(day from parm_date_in - 5) / 5) * 5, date_part('day', date_trunc('month', parm_date_in) + interval '1 month - 1 day')))::timestamp
else (least(ceil(extract(day from parm_date_in - 6) / 6) * 6, date_part('day', date_trunc('month', parm_date_in) + interval '1 month - 1 day')))::timestamp
end
from last_of_mon;
$$;
But I can't cast to timestamp, how can it be done? if I return integer then I got what I wanted, but the poblem is I need to have a full date in format YYYY-MM-dd
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);
This question is asked many times and one of the suggested queries to get months between 2 dates is not working.
SELECT date_part('month',age('2016-06-30', '2018-06-30'))
The result of this query is 0. It should be 24 months. Because the months are 06 in both dates.
This works, but it is a bit clumsy compared to the sql server function:
SELECT date_part ('year', f) * 12 + date_part ('month', f)
FROM age ('2016-06-30', '2018-06-30') f
Like sql server (I think):
DATEDIFF(month, date1, date2)
Is there no simple way (like the above) to calculate the months between 2 dates in Postgresql? I prefer not to use a function if it is possible.
Unfortunately you already have the most elegant solution.
If you look at the documentation for extract (same as date_part):
https://www.postgresql.org/docs/current/static/functions-datetime.html#FUNCTIONS-DATETIME-EXTRACT
month
For timestamp values, the number of the month within the year (1 - 12) ; for interval values, the number of months, modulo 12 (0 - 11)
SELECT EXTRACT(MONTH FROM TIMESTAMP '2001-02-16 20:38:40');
Result: 2
SELECT EXTRACT(MONTH FROM INTERVAL '2 years 3 months');
Result: 3
SELECT EXTRACT(MONTH FROM INTERVAL '2 years 13 months');
Result: 1
For your problem it would be nice if there was a version of month that wasn't modulo 12 but that doesn't exist.
The option you have (extract the year * 12 + month) is the best option there is.
Edit
If you do want to create a function then see the following two functions:
CREATE OR REPLACE FUNCTION get_months(i interval) RETURNS double precision AS $$
SELECT date_part ('year', i) * 12 + date_part ('month', i) ;
$$ LANGUAGE SQL IMMUTABLE;
SELECT get_months(age('2016-06-30', '2018-06-30'));
Or
CREATE OR REPLACE FUNCTION get_months(to_date date, from_date date) RETURNS double precision AS $$
SELECT date_part ('year', f) * 12 + date_part ('month', f)
FROM age (to_date, from_date) f;
$$ LANGUAGE SQL IMMUTABLE;
SELECT get_months('2016-06-30', '2018-06-30');
You can actually create both then just use whichever suits your code.
This will give you the # of months between two dates excluding days.
CREATE OR REPLACE FUNCTION get_months_between(to_date date, from_date date) RETURNS double precision AS $$
SELECT (date_part ('year', to_date) * 12 + date_part ('month', to_date)) - (date_part ('year', from_date) * 12 + date_part ('month', from_date))
$$ LANGUAGE SQL IMMUTABLE;
I work with a Postgres database. This DB has a table with users, who have a birthdate (date field). Now I want to get all users who have their birthday in the upcoming week....
My first attempt: SELECT id FROM public.users WHERE id IN (lange reeks) AND birthdate > NOW() AND birthdate < NOW() + interval '1 week'
But this does not result, obviously because off the year. How can I work around this problem?
And does anyone know what happen to PG would go with the cases at 29-02 birthday?
We can use a postgres function to do this in a really nice way.
Assuming we have a table people, with a date of birth in the column dob, which is a date, we can create a function that will allow us to index this column ignoring the year. (Thanks to Zoltán Böszörményi):
CREATE OR REPLACE FUNCTION indexable_month_day(date) RETURNS TEXT as $BODY$
SELECT to_char($1, 'MM-DD');
$BODY$ language 'sql' IMMUTABLE STRICT;
CREATE INDEX person_birthday_idx ON people (indexable_month_day(dob));
Now, we need to query against the table, and the index. For instance, to get everyone who has a birthday in April of any year:
SELECT * FROM people
WHERE
indexable_month_day(dob) >= '04-01'
AND
indexable_month_day(dob) < '05-01';
There is one gotcha: if our start/finish period crosses over a year boundary, we need to change the query:
SELECT * FROM people
WHERE
indexable_month_day(dob) >= '12-29'
OR
indexable_month_day(dob) < '01-04';
To make sure we match leap-day birthdays, we need to know if we will 'move' them a day forward or backwards. In my case, it was simpler to just match on both days, so my general query looks like:
SELECT * FROM people
WHERE
indexable_month_day(dob) > '%(start)%'
%(AND|OR)%
indexable_month_day(dob) < '%(finish)%';
I have a django queryset method that makes this all much simpler:
def birthday_between(self, start, finish):
"""Return the members of this queryset whose birthdays
lie on or between start and finish."""
start = start - datetime.timedelta(1)
finish = finish + datetime.timedelta(1)
return self.extra(where=["indexable_month_day(dob) < '%(finish)s' %(andor)s indexable_month_day(dob) > %(start)s" % {
'start': start.strftime('%m-%d'),
'finish': finish.strftime('%m-%d'),
'andor': 'and if start.year == finish.year else 'or'
}]
def birthday_on(self, date):
return self.birthday_between(date, date)
Now, I can do things like:
Person.objects.birthday_on(datetime.date.today())
Matching leap-day birthdays only on the day before, or only the day after is also possible: you just need to change the SQL test to a `>=' or '<=', and not adjust the start/finish in the python function.
I'm not overly confident in this, but it seems to work in my testing. The key here is the OVERLAPS operator, and some date arithmetic.
I assume you have a table:
create temporary table birthdays (name varchar, bday date);
Then I put some stuff into it:
insert into birthdays (name, bday) values
('Aug 24', '1981-08-24'), ('Aug 04', '1982-08-04'), ('Oct 10', '1980-10-10');
This query will give me the people with birthdays in the next week:
select * from
(select *, bday + date_trunc('year', age(bday)) + interval '1 year' as anniversary from birthdays) bd
where
(current_date, current_date + interval '1 week') overlaps (anniversary, anniversary)
The date_trunc truncates the date at the year, so it should get you up to the current year. I wound up having to add one year. This suggests to me I have an off-by-one in there for some reason. Perhaps I just need to find a way to get dates to round up. In any case, there are other ways to do this calculation. age gives you the interval from the date or timestamp to today. I'm trying to add the years between the birthday and today to get a date in the current year.
The real key is using overlaps to find records whose dates overlap. I use the anniversary date twice to get a point-in-time.
Finally, to show the upcoming birthdays of the next 14 days I used this:
SELECT
-- 14 days before birthday of 2000
to_char( to_date(to_char(c.birthdate, '2000-MM-dd'), 'YYYY-MM-dd') - interval '14 days' , 'YYYY-MM-dd') as _14b_b2000,
-- birthday of 2000
to_date(to_char(c.birthdate, '2000-MM-dd'), 'YYYY-MM-dd') as date_b2000,
-- current date of 2000
to_date(to_char(current_date, '2000-MM-dd'), 'YYYY-MM-dd') as date_c2000,
-- 14 days after current date of 2000
to_char( to_date(to_char(current_date, '2000-MM-dd'), 'YYYY-MM-dd') + interval '14 days' , 'YYYY-MM-dd') as _14a_c2000,
-- 1 year after birthday of 2000
to_char( to_date(to_char(c.birthdate, '2000-MM-dd'), 'YYYY-MM-dd') + interval '1 year' , 'YYYY-MM-dd') as _1ya_b2000
FROM c
WHERE
-- the condition
-- current date of 2000 between 14 days before birthday of 2000 and birthday of 2000
to_date(to_char(current_date, '2000-MM-dd'), 'YYYY-MM-dd') between
to_date(to_char(c.birthdate, '2000-MM-dd'), 'YYYY-MM-dd') - interval '14 days' and
to_date(to_char(c.birthdate, '2000-MM-dd'), 'YYYY-MM-dd')
or
-- 1 year after birthday of 2000 between current date of 2000 and 14 days after current date of 2000
to_date(to_char(c.birthdate, '2000-MM-dd'), 'YYYY-MM-dd') + interval '1 year' between
to_date(to_char(current_date, '2000-MM-dd'), 'YYYY-MM-dd') and
to_date(to_char(current_date, '2000-MM-dd'), 'YYYY-MM-dd') + interval '14 days'
;
So:
To solve the leap-year issue, I set both birthdate and current date to 2000,
and handle intervals only from this initial correct dates.
To take care of the near end/beginning dates,
I compared first the 2000 current date to the 2000 birthday interval,
and in case current date is at the end of the year, and the birthday is at the beginning,
I compared the 2001 birthday to the 2000 current date interval.
Here's a query that gets the right result, most of the time.
SELECT
(EXTRACT(MONTH FROM DATE '1980-08-05'),
EXTRACT(DAY FROM DATE '1980-08-05'))
IN (
SELECT EXTRACT(MONTH FROM CURRENT_DATE + s.a) AS m,
EXTRACT(DAY FROM CURRENT_DATE + s.a) AS d
FROM GENERATE_SERIES(0, 6) AS s(a)
);
(it doesn't take care of leap years correctly; but you could use extract again to work the subselect in terms of a leap year instead of the current year.
EDIT: Got it working for all cases, and as a useful query rather than a scalar select. I'm using some extra subselects so that I don't have to type the same date or expression twice for month and day, and of course the actual data would be in a table instead of the values expression. You might adapt this differently. It might still stand to improve by making a more intelligent series for weeks containing leap days, since sometimes that interval will only contain 6 days (for non-leap years).
I'll try to explain this from the inside-out; First thing I do is normalize the target date (CURRENT_DATE usually, but explicit in this code) into a year that I know is a leap year, so that February 29th appears among dates. The next step is to generate a relation with all of the month-day pairs that are under consideration; Since there's no easy way to do an interval check in terms of month-day, it's all happening using generate_series,
From there it's a simple matter of extracting the month and day from the target relation (the people alias) and filtering just the rows that are in the subselect.
SELECT *
FROM
(select column1 as birthdate, column2 as name
from (values
(date '1982-08-05', 'Alice'),
(date '1976-02-29', 'Bob'),
(date '1980-06-10', 'Carol'),
(date '1992-06-13', 'David')
) as birthdays) as people
WHERE
((EXTRACT(MONTH FROM people.birthdate),
EXTRACT(DAY FROM people.birthdate)) IN (
SELECT EXTRACT(MONTH FROM thedate.theday + s.a) AS m,
EXTRACT(DAY FROM thedate.theday + s.a) AS d
FROM
(SELECT date (v.column1 -
(extract (YEAR FROM v.column1)-2000) * INTERVAL '1 year'
) as theday
FROM (VALUES (date '2011-06-09')) as v) as thedate,
GENERATE_SERIES(0, 6) AS s(a)
)
)
Operating on days, as I've done here, should work splendidly all the way up until a two month interval (if you wanted to look out that far), since december 31 + two months and change should include the leap day. On the other hand, it's almost certainly more useful to just work on whole months for such a query, in which case you don't really need anything more than extract(month from ....
First find out how old the person currently is using age(), then grab the year from that extract(year from age()). This is how old they are currently in years, so for their age at their next birthday add 1 to the year. Then their next birthday is found by adding an interval of this many years * interval '1 year' to their birthday. Done.
I've used a subselect here to add the next_birth_day column in to the complete table to make the select clause simpler. You can then play with the where conditions to suit your needs.
select *
from (
select *,
(extract(year from age(birth_date)) + 1) * interval '1 year' + birth_date "next_birth_day"
from public.users
) as users_with_upcoming_birth_days
where next_birth_day between now() and now() + '7 days'
This is based on Daniel Lyons's anniversary idea, by calculating the interval between the next birthday and today, with just +/- date arithmetic:
SELECT
today,
birthday,
CASE
WHEN this_year_anniversary >= today
THEN this_year_anniversary
ELSE this_year_anniversary + '1 year'::interval
END - today < '1 week'::interval AS is_upcoming
FROM
(
SELECT
today,
birthday,
birthday + years AS this_year_anniversary
FROM
(
SELECT
today,
birthday,
((
extract(year FROM today) - extract(year from birthday)
) || ' years')::interval AS years
FROM
(VALUES ('2011-02-28'::date)) AS t1 (today),
(VALUES
('1975-02-28'::date),
('1975-03-06'::date),
('1976-02-28'::date),
('1976-02-29'::date),
('1976-03-06'::date)
) AS t2 (birthday)
) AS t
) AS t;
In case you want it to work with leap years:
create or replace function birthdate(date)
returns date
as $$
select (date_trunc('year', now()::date)
+ age($1, 'epoch'::date)
- (extract(year from age($1, 'epoch'::date)) || ' years')::interval
)::date;
$$ language sql stable strict;
Then:
where birthdate(birthdate) between current_date
and current_date + interval '1 week'
See also:
Getting all entries who's Birthday is today in PostgreSQL
Exemple: birthdate between: jan 20 and feb 10
SELECT * FROM users WHERE TO_CHAR(birthdate, '1800-MM-DD') BETWEEN '1800-01-20' AND '1800-02-10'
Why 1800?
No matter may be any year;
In my registration form, I can inform the date of birth (with years) or just the birthday (without year), in which case I saved as 1800 to make it easier to work with the date
Here's my take, which works with leap years too:
CREATE OR REPLACE FUNCTION days_until_birthday(
p_date date
) RETURNS integer AS $$
DECLARE
v_now date;
v_days integer;
v_date_upcoming date;
v_years integer;
BEGIN
v_now = now()::date;
IF (p_date IS NULL OR p_date > v_now) THEN
RETURN NULL;
END IF;
v_years = date_part('year', v_now) - date_part('year', p_date);
v_date_upcoming = p_date + v_years * interval '1 year';
IF (v_date_upcoming < v_now) THEN
v_date_upcoming = v_date_upcoming + interval '1 year';
END IF;
v_days = v_date_upcoming - v_now;
RETURN v_days;
END
$$ LANGUAGE plpgsql IMMUTABLE;
I know this post is old, but I had the same issue and came up with this simple and elegant solution:
It is pretty easy with age() and accounts for lap years... for the people who had their birthdays in the last 20 days:
SELECT * FROM c
WHERE date_trunc('year', age(birthdate)) != date_trunc('year', age(birthdate + interval '20 days'))
I have simply created this year date from original birth date.
( DATE_PART('month', birth_date) || '/' || DATE_PART('day', birth_date) || '/' || DATE_PART('year', now()))::date between :start_date and :end_date
I hope this help.
In PostgreSQL 8.4, given a date, if that date is not a Friday, I would like to find the date of the previous Friday. Can someone tell me if there is an inbuilt function or give the logic behind getting my own function.
Try this, works on other days too, blog about it http://www.ienablemuch.com/2010/12/finding-previous-day-of-week.html
create or replace function previous_date_of_day(the_date date, dow int) returns date
as
$$
select
case when extract(dow from $1) < $2 then
$1 - ( extract(dow from $1) + (7 - $2) )::int
else
$1 - ( extract(dow from $1) - $2)::int
end;
$$ language 'sql';
select to_char(z.ds, 'Mon dd yyyy dy') as source,
to_char( previous_date_of_day(z.ds, 5), 'Mon dd yyyy dy') as dest
from
(
select 'Dec 1 2010'::date + x.n as ds
from generate_series(0,17) as x(n)
) as z
You solve it without using case:
select
the_date
from
(
select
now()::date - num as the_date, -- generate rows of possible dates
extract(dow from (now()::date - num)) -- dow for the where condition
from (select generate_series(0,6) as num) as t
) as days
where date_part = 5;
SELECT
CASE
-- 1. if Friday, return date
WHEN EXTRACT(DOW FROM my_date) = 5
THEN my_date
-- 2. if Saturday, subtract 1
WHEN EXTRACT(DOW FROM my_date) = 6
THEN my_date - INTERVAL '1 day'
-- 3. all other days of the week, subtract `DOW + 2` from my_date
-- should be ELSE for future-proofing ;-) MB
ELSE -- WHEN EXTRACT(DOW FROM my_date) < 5 THEN
my_date - ((EXTRACT(DOW FROM my_date) + 2)::TEXT||'days')::INTERVAL
END AS tgif
FROM
my_table
WHERE
my_date IS NOT NULL
select case when extract(dow from your_date) < 5 then
your_date - (extract(dow from your_date) + integer '2')
else when extract(dow from your_date) > 5 then
your_date - integer '1'
else
your_date
end
Reference http://developer.postgresql.org/pgdocs/postgres/functions-datetime.html