Is there any way to get the first day of previous month other than
date_trunc('month', current_date - interval '1 month') ?
I'm trying to save a query in a Report Designer software (DBxtra) but the software freezes while using the "interval" feature of PostgreSQL.
You can try to calculate the previous month manually.
One approach is extract the month and subtract 1 if it is not 12(in this case you return 1):
SELECT to_timestamp(concat(EXTRACT(YEAR from current_date), '-',CASE (EXTRACT(MONTH from current_date)) WHEN 12 THEN 1 ELSE (EXTRACT(MONTH from current_date)-1) END,'-', 1), 'YYYY-MM-DD');
If you need it without timezone:
SELECT to_timestamp(concat(EXTRACT(YEAR from current_date), '-',CASE (EXTRACT(MONTH from current_date)) WHEN 12 THEN 1 ELSE (EXTRACT(MONTH from current_date)-1) END,'-', 1), 'YYYY-MM-DD')::timestamp without time zone;
Try:
make_interval(month := 1)
'1 month'::interval
cast('1 month' as interval)
instead of interval '1 month'
Related
I'd like to filter a dataset to include everything with a created_date >= current year plus 5 full years prior. If I ran this today, I'd want anything with a created_date >= 01/01/2016 (2021 - 5 = 2016)
where extract(year from created_date) >= extract(year from current_date) - interval '5 years'
of course this doesn't work with double precision datatype. Any thoughts? Thank you.
Use date_trunc() to "round" the dates to the first of January.
where date_trunc('year', created_date) >= date_trunc('year', current_timestamp) - interval '5 years'
But the "I'd want anything with a created_date >= 01/01/2016" seems to indicate you might want:
where created_date >= date_trunc('year', current_timestamp) - interval '5 years'
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'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.
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!!
How to get last 3 months records from the table.
SELECT *
from table
where month > CURRENT_DATE-120
and month < CURRENT_DATE
order by month;
I have used the above query is it correct? shall I use this for get last 3 month record from the table.
You can use built-in INTERVAL instruction
Check how this works:
SELECT CURRENT_DATE - INTERVAL '3 months'
and you can rewrite your SQL to:
SELECT * from table where date > CURRENT_DATE - INTERVAL '3 months'
(not checked but this should give you an idea how to use INTERVAL instruction)
Try that:
SELECT *
FROM table
WHERE month BETWEEN EXTRACT(MONTH FROM NOW() - INTERVAL '3 months')
AND EXTRACT(MONTH FROM NOW())
ORDER BY month
;
This filters the last 3 calendar months
SELECT * from table where date >= to_char(CURRENT_DATE - INTERVAL '3 months', 'YYYY-MM-01')::date
select date::date
from generate_series((current_date - INTERVAL '1 Month')::date, (current_date - INTERVAL '1 DAY')::date,'1
day'::interval) date
WHERE date >= date_trunc('month', current_date - interval '3' month)
and date < date_trunc('month', current_date)
This will give last three months date list, excluding current months date. Example if current month is November. This list will give use all dates of August, Septemeber and October.