update table with dates with month - sql

There's a table dates_calendar:
id | date
-------------------------
13 | 2016-10-23 00:00:00
14 | 2016-10-24 00:00:00
I need to update this table and insert dates until the next month counting from the last date in the table. E.g. last date is 2016-10-24 00:00:00 - I need to insert dates till 2016-10-31. After that (the last date now is 2016-10-31) next statement call should insert dates till 2016-11-30 and so on.
Example of my SQL code, but it inserts 30 days all the time.
INSERT INTO dates_calendar (date)
VALUES (
generate_series(
(SELECT date FROM dates_calendar ORDER BY date DESC LIMIT 1) + interval '1 day',
(SELECT date FROM dates_calendar ORDER BY date DESC LIMIT 1) + interval '1 month',
'1 day'
)
);
I'm using PostgreSQL. As well would be fine to get rid of a duplicated SELECT statement of the last date.

insert into dates_calendar (date)
select dates::date
from (
select max(date)::date+ 1 next_day, '1day'::interval one_day, '1month'::interval one_month
from dates_calendar
) s,
generate_series(
next_day,
date_trunc('month', next_day)+ one_month- one_day,
one_day) dates;

To calculate the first and last date you need to insert you can use this query:
select max(date) + interval '1' day as first_day,
date_trunc('month', max(date) + interval '1' month) - interval '1' day as last_day
from dates_calendar
The expression date_trunc('month', max(date) + interval '1' month) calculates the start date of the next month. Subtracting one day from that will give you the last day of that month.
This can then be used to generate the list of dates:
with from_to (first_day, last_day) as (
select max(date) + interval '1' day,
date_trunc('month', max(date) + interval '1' month) - interval '1' day
from dates_calendar
)
select dt
from generate_series( (select first_day from from_to), (select last_day from from_to), interval '1' day) as t(dt);
And finally this can be used to insert the generated rows into the table:
with from_to (first_day, last_day) as (
select max(date) + interval '1' day,
date_trunc('month', max(date) + interval '1' month) - interval '1' day
from dates_calendar
)
insert into dates_calendar (date)
select dt
from generate_series( (select first_day from from_to), (select last_day from from_to), interval '1' day) as t(dt);

with max_date (d) as (select max(date)::date from dates_calendar)
insert into dates_calendar (date)
select d
from generate_series (
(select d from max_date) + 1,
(select date_trunc('month', d + interval '1 month')::date - 1 from max_date),
'1 day'
) g(d)

Related

How to get given date + n months in SELECT?

How can I add a given number of month n=14 to a given date like 2008-01 (January 2008) to get 2009-03 ?
My goal is to show the date in YYYY-MM format instead of foo_id (which is the number of months) in the query below:
SELECT
foo_id AS "date",
amount
FROM my_table
GROUP BY foo_id
ORDER BY foo_id;
Currently, I can get given date + n months with in months only:
SELECT date_part('month' , DATE '2010-01-01' + INTERVAL '124 month') AS "date";
=> 4
You can use to_char() to format the result with only year and month:
select to_char(current_date + interval '14 month', 'yyyy-mm');
If the number of months is stored in a column:
select to_char(current_date + make_interval(months => foo_id), 'yyyy-mm');
So
with the_table (foo_id) as (
values (14)
)
select to_char(date '2008-01-01' + make_interval(months => foo_id), 'yyyy-mm')
from the_table
Returns: 2009-03
Online example
Interval could be multiplied:
SELECT DATE '2010-01-01' + foo_id * INTERVAL '1 month' AS "date" FROM my_table
-- start date + int * 1 month
The current date can also be obtained with now():
select now() + interval '1 month' * my_table.number_of_months from my_table where id = 222;

create table with dates - sql

I have a query that can create a table with dates like below:
with digit as (
select 0 as d union all
select 1 union all select 2 union all select 3 union all
select 4 union all select 5 union all select 6 union all
select 7 union all select 8 union all select 9
),
seq as (
select a.d + (10 * b.d) + (100 * c.d) + (1000 * d.d) as num
from digit a
cross join
digit b
cross join
digit c
cross join
digit d
order by 1
)
select (last_day(sysdate)::date - seq.num)::date as "Date"
from seq;
How could this be changed to generate only dates
Thanks
demo:db<>fiddle
WITH dates AS (
SELECT
date_trunc('month', CURRENT_DATE) AS first_day_of_month,
date_trunc('month', CURRENT_DATE) + interval '1 month -1 day' AS last_day_of_month
)
SELECT
generate_series(first_day_of_month, last_day_of_month, interval '1 day')::date
FROM dates
date_trunc() truncates a type date (or timestamp) to a certain date part. date_trunc('month', ...) removes all parts but year and month. All other parts are set to their lowest possible values. So, the day part is set to 1. That's why you get the first day of month with this.
adding a month returns the first of the next month, subtracting a day from this results in the last day of the current month.
Finally you can generate a date series with start and end date using the generate_series() function
Edit: Redshift does not support generate_series() with type date and timestamp but with integer. So, we need to create an integer series instead and adding the results to the first of the month:
db<>fiddle
WITH dates AS (
SELECT
date_trunc('month', CURRENT_DATE) AS first_day_of_month,
date_trunc('month', CURRENT_DATE) + interval '1 month -1 day' AS last_day_of_month
)
SELECT
first_day_of_month::date + gs
FROM
dates,
generate_series(
date_part('day', first_day_of_month)::int - 1,
date_part('day', last_day_of_month)::int - 1
) as gs
This answers the original version of the question.
You would use generate_series():
select gs.dte
from generate_series(date_trunc('month', now()::date),
date_trunc('month', now()::date) + interval '1 month' - interval '1 day',
interval '1 day'
) gs(dte);
Here is a db<>fiddle.

how to get date different in postgres using date_part option

How to get date time difference in PostgreSQL
I am using below syntax
select id, A_column,B_column,
(SELECT count(*) AS count_days_no_weekend
FROM generate_series(B_column ::timestamp , A_column ::timestamp, interval '1 day') the_day
WHERE extract('ISODOW' FROM the_day) < 5) * 24 + DATE_PART('hour', B_column::timestamp-A_column ::timestamp ) as hrs
FROM table req where id='123';
If A_column=2020-05-20 00:00:00 and B_column=2020-05-15 00:00:00 I want to get 72(in hours).
Is there any possibility to skip weekends(Saturday and Sunday) in first one, it means to get the result as 72 hours(exclude weekend hours)
i am getting 0
But i need to get 72 hours
And if If A_column=2020-08-15 12:00:00 and B_column=2020-08-15 00:00:00 I want to get 12(in hours).
One option uses a lateral join and generate_series() to enumerate each and every hour between the two timestamps, while filtering out week-ends:
select t.a_column, t.b_column, h.count_hours_no_weekend
from mytable t
cross join lateral (
select count(*) count_hours_no_weekend
from generate_series(t.b_column::timestamp, t.a_column::timestamp, interval '1 hour') s(col)
where extract('isodow' from s.col) < 5
) h
where id = 123
I would attack this by calculating the weekend hours to let the database deal with daylight savings time. I would then subtract the intervening weekend hours from the difference between the two date values.
with weekend_days as (
select *, date_part('isodow', ddate) as dow
from table1
cross join lateral
generate_series(
date_trunc('day', b_column),
date_trunc('day', a_column),
interval '1 day') as gs(ddate)
where date_part('isodow', ddate) in (6, 7)
), weekend_time as (
select id,
sum(
least(ddate + interval '1 day', a_column) -
greatest(ddate, b_column)
) as we_ival
from weekend_days
group by id
)
select t.id,
a_column - b_column as raw_difference,
coalesce(we_ival, interval '0') as adjustment,
a_column - b_column -
coalesce(we_ival, interval '0') as adj_difference
from weekend_time w
left join table1 t on t.id = w.id;
Working fiddle.

postgreSQL: How Select the nearest date that is not null

I got a date that I want to find the all records in the past that got the same month and day.
The problem accrues when there is no such date in the same year. For example, the 29th February.
My goal is to get the nearest date from below the date that does not exist.
This is my currently query with the date 2012-02-29:
SELECT date, amount
FROM table_name
WHERE
EXTRACT(MONTH FROM date) = EXTRACT(MONTH FROM DATE('2012-02-29') )
AND EXTRACT(DAY FROM date) = EXTRACT(DAY FROM DATE('2012-02-29') )
AND date < '2012-02-29'
ORDER BY date DESC LIMIT 10;
If I understand correctly, you want one date per year with the property that that day is nearest to the given date.
I would suggest using distinct on:
select distinct on (date_trunc('year', date)) t.*
from table_name t
order by date_trunc('year', date),
abs(date_part('day, (date -
(date '2012-02-29' -
(extract(year from date '2012-02-29') - extract(year from date)) * interval '1 year'
)
)
)
)
);
EDIT:
An example of working code:
select distinct on (date_trunc('year', date)) t.*
from table_name t
order by date_trunc('year', date),
abs(date_part('day', date - (date '2012-02-29' -
((extract(year from date '2012-02-29') - extract(year from date)) * interval '1 year')
)
))

Daily average for the month (needs number of days in month)

I have a table as follow:
CREATE TABLE counts
(
T TIMESTAMP NOT NULL,
C INTEGER NOT NULL
);
I create the following views from it:
CREATE VIEW micounts AS
SELECT DATE_TRUNC('minute',t) AS t,SUM(c) AS c FROM counts GROUP BY 1;
CREATE VIEW hrcounts AS
SELECT DATE_TRUNC('hour',t) AS t,SUM(c) AS c,SUM(c)/60 AS a
FROM micounts GROUP BY 1;
CREATE VIEW dycounts AS
SELECT DATE_TRUNC('day',t) AS t,SUM(c) AS c,SUM(c)/24 AS a
FROM hrcounts GROUP BY 1;
The problem now comes in when I want to create the monthly counts to know what to divide the daily sums by to get the average column a i.e. the number of days in the specific month.
I know to get the days in PostgreSQL you can do:
SELECT DATE_PART('days',DATE_TRUNC('month',now())+'1 MONTH'::INTERVAL-DATE_TRUNC('month',now()))
But I can't use now(), I have to somehow let it know what the month is when the grouping gets done. Any suggestions i.e. what should replace ??? in this view:
CREATE VIEW mocounts AS
SELECT DATE_TRUNC('month',t) AS t,SUM(c) AS c,SUM(c)/(???) AS a
FROM dycounts
GROUP BY 1;
A bit shorter and faster and you get the number of days instead of an interval:
SELECT EXTRACT(day FROM date_trunc('month', now()) + interval '1 month'
- interval '1 day')
It's possible to combine multiple units in a single interval value . So we can use '1 mon - 1 day':
SELECT EXTRACT(day FROM date_trunc('month', now()) + interval '1 mon - 1 day')
(mon, month or months work all the same for month units.)
To divide the daily sum by the number of days in the current month (orig. question):
SELECT t::date AS the_date
, SUM(c) AS c
, SUM(c) / EXTRACT(day FROM date_trunc('month', t::date)
+ interval '1 mon - 1 day') AS a
FROM dycounts
GROUP BY 1;
To divide monthly sum by the number of days in the current month (updated question):
SELECT DATE_TRUNC('month', t)::date AS t
,SUM(c) AS c
,SUM(c) / EXTRACT(day FROM date_trunc('month', t)::date
+ interval '1 mon - 1 day') AS a
FROM dycounts
GROUP BY 1;
You have to repeat the GROUP BY expression if you want to use a single query level.
Or use a subquery:
SELECT *, c / EXTRACT(day FROM t + interval '1 mon - 1 day') AS a
FROM (
SELECT date_trunc('month', t)::date AS t, SUM(c) AS c
FROM dycounts
GROUP BY 1
) sub;