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.
Related
I want to query data for last 30 days including today from redshift table. below is my query.
my date_column's type is 'timestamp without timezone'
select *
from mytable
WHERE date_column BETWEEN current_date - INTERVAL '30 day' AND current_date
order by date_column desc;
It gives the result for 30 days. But it doesn't include today's result.
I want to query for 30 days result including today's result also.
If it's a timestamp don't use between as it also compares the time part. Use a range query:
where date_column >= current_date - interval '30 day'
and date_column < current_date + interval '1 day'
Note that the upper bound is using < together with "tomorrow"
With Postgres this could be simplified to
where date_column >= current_date - 30
and date_column < current_date + 1
but Redshift isn't Postgres and I don't know if that would work there.
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 the following records in a table, as image below. The last period is December / 2019.
I would like to list the periods within a range of 2 years (backwards) from the current date.
For example: today 09/10/2019, list periods from 01/01/2017 to 12/12/2019
I have difficulty assembling the query below.
SELECT c_period_id, name, startdate, enddate
FROM adempiere.C_Period
WHERE startdate BETWEEN now() - INTERVAL '2 year' AND now()
order by startdate desc
I am not quite sure what your problem is, but if it is rounding the dates to the year boundary, this might serve:
WHERE startdate >= date_trunc('year', current_timestamp) - INTERVAL '2 years'
AND startdate < date_trunc('year', current_timestamp) + INTERVAL '1 year'
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!!