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.
Related
I have the following code to pull records from a daterange in PostgreSQL, it works as intended. The "end date" is determined by the "date" column from the last record, and the "start date" is calculated by subtracting a 7-day interval from the "end date".
SELECT date
FROM files
WHERE daterange((
(SELECT date FROM files ORDER BY date DESC LIMIT 1) - interval '7 day')::date, -- "start date"
(SELECT date FROM files ORDER BY date DESC LIMIT 1)::date, -- "end date"
'(]') #> date::date
ORDER BY date ASC
I'm trying to rewrite this query using CTEs, so I can replace those subqueries with values such as end_date and start_date. Is this possible using this method or should I look for other alternatives like variables? I'm still learning SQL.
WITH end_date AS
(
SELECT date FROM files ORDER BY date DESC LIMIT 1
),
start_date AS
(
SELECT date FROM end_date - INTERVAL '7 day'
)
SELECT date
FROM files
WHERE daterange(
start_date::date,
end_date::date,
'(]') #> date::date
ORDER BY date ASC
Right now I'm getting the following error:
ERROR: syntax error at or near "-"
LINE 7: SELECT date FROM end_date - INTERVAL '7 day'
You do not need two CTEs, it's one just fine, which can be joined to filter data.
WITH RECURSIVE files AS (
SELECT CURRENT_DATE date, 1 some_value
UNION ALL
SELECT (date + interval '1 day')::date, some_value + 1 FROM files
WHERE date < (CURRENT_DATE + interval '1 month')::date
),
dates AS (
SELECT
(MAX(date) - interval '7 day')::date from_date,
MAX(date) to_date
FROM files
)
SELECT f.* FROM files f
JOIN dates d ON daterange(d.from_date, d.to_date, '(]') #> f.date
You even can make it to be a daterange initially in CTE and use it later like this
WITH dates AS (
SELECT
daterange((MAX(date) - interval '7 day')::date, MAX(date), '(]') range
FROM files
)
SELECT f.* FROM files f
JOIN dates d ON d.range #> f.date
Here the first CTE is used just to generate some data.
It will get all file lines for dates in the last week, excluding from_date and including to_date.
date
some_value
2022-09-26
25
2022-09-27
26
2022-09-28
27
2022-09-29
28
2022-09-30
29
2022-10-01
30
2022-10-02
31
I think this is what you want:
WITH end_date AS
(
SELECT date FROM files ORDER BY date DESC LIMIT 1
),
start_date AS
(
SELECT date - INTERVAL '7 day' as date
FROM end_date
)
SELECT F.date, S.date startDate, E.date endDate
FROM files F
JOIN start_date S on F.date >= S.date
JOIN end_date E on F.date <= E.date
ORDER BY date ASC;
I hope I'm not repeating anything, but if I understand your problem correctly I think this will work:
with cte as (
select max (date)::date as max_date from files
)
select date
from files
cross join cte
where date >= max_date - 7
Or perhaps even:
select date
from files
where date >= (select max (date)::date - 7 from files)
Since you have already determined that the CTE has the max date, there is really no need to further bound it with a between, <= or range. You can simply say anything after that date minus 7 days.
The error in your code above is because you want this:
SELECT date - INTERVAL '7 day' as date FROM end_date
And not this:
SELECT date FROM end_date - INTERVAL '7 day'
You are subtracting from the table, which doesn't make sense.
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.
I want to generate series of month and year from the next month of current year(say, start_month) to 12 months from start_month along with the corresponding data (if any, else return nulls) from another table in PostgreSQL.
SELECT ( ( DATE '2019-03-01' + ( interval '1' month * generate_series(0, 11) ) )
:: DATE ) dd,
extract(year FROM ( DATE '2019-03-01' + ( interval '1' month *
generate_series(0, 11) )
)),
coalesce(SUM(price), 0)
FROM items
WHERE s.date_added >= '2019-03-01'
AND s.date_added < '2020-03-01'
AND item_type_id = 3
GROUP BY 1,
2
ORDER BY 2;
The problem with the above query is that it is giving me the same value for price for all the months. The requirement is that the price column be filled with nulls or zeros if no price data is available for a given month.
Put the generate_series() in the FROM clause. You are summarizing the data -- i.e. calculating the price over the entire range -- and then projecting this on all months. Instead:
SELECT gs.yyyymm,
coalesce(SUM(i.price), 0)
FROM generate_series('2019-03-01'::date, '2020-02-01', INTERVAL '1 MONTH'
) gs(yyyymm) LEFT JOIN
items i
ON gs.yyyymm = DATE_TRUNC('month', s.date_added) AND
i.item_type_id = 3
GROUP BY gs.yyyymm
ORDER BY gs.yyyymm;
You want generate_series in the FROM clause and join with it, somewhat like
SELECT months.m::date, ...
FROM generate_series(
start_month,
start_month + INTERVAL '11 months',
INTERVAL '1 month'
) AS months(m)
LEFT JOIN items
ON months.m::date = items.date_added
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)
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;