How to find the last day os the month in postgres?
I have a date columns stored as numeric(18) in the format(YYYYMMDD)
I am trying it to make it date using
to_date("act_dt",'YYYYMMDD') AS "act date"
then find the last day of this date:
like this:
(select (date_trunc('MONTH',to_date("act_dt",'YYYYMMDD')) + INTERVAL '1 MONTH - 1 day')::date)
but it gives me this error:
ERROR: Interval values with month or year parts are not supported
Detail:
-----------------------------------------------
error: Interval values with month or year parts are not supported
code: 8001
context: interval months: "1"
query: 673376
location: cg_constmanager.cpp:145
process: padbmaster [pid=20937]
-----------------------------------------------
Any help?
Postgres version:
PostgreSQL 8.0.2 on i686-pc-linux-gnu, compiled by GCC gcc (GCC) 3.4.2 20041017 (Red Hat 3.4.2-6.fc3), Redshift 1.0.874
For anybody coming to this question looking for the Postgres way to do this (not using Redshift), here's how you'd do it:
SELECT (date_trunc('month', '2017-01-05'::date) + interval '1 month' - interval '1 day')::date
AS end_of_month;
Replacing the '2017-01-05' with whatever date you want to use. You can make this into a function like this:
create function end_of_month(date)
returns date as
$$
select (date_trunc('month', $1) + interval '1 month' - interval '1 day')::date;
$$ language 'sql'
immutable strict;
EDIT Postgres 11+
Pulling this out of the comments from #Gabriel, you can now combine interval expressions in one interval (which makes things a little shorter):
select (date_trunc('month', now()) + interval '1 month - 1 day')::date as end_of_month;
-- +--------------+
-- | end_of_month |
-- +--------------+
-- | 2021-11-30 |
-- +--------------+
-- (1 row)
If you're using Amazon AWS Redshift then you can use Redshift's LAST_DAY function. While Redshift is based on PostgreSQL, the LAST_DAY function is not available in PostgreSQL, for a solution for PostgreSQL see #wspurgin's answer.
https://docs.aws.amazon.com/redshift/latest/dg/r_LAST_DAY.html
LAST_DAY( { date | timestamp } )
LAST_DAY returns the date of the last day of the month that contains date. The return type is always DATE, regardless of the data type of the date argument.
For example:
SELECT LAST_DAY( TO_DATE( act_date, 'YYYYMMDD' ) )
Okay, so you've got a numeric(18) column containing numbers like 20150118. You can convert that to a date like:
to_date(your_date_column::text, 'YYYYMMDD')
From a date, you can grab the last day of the month like:
(date_trunc('month', your_date_column) +
interval '1 month' - interval '1 day')::date;
Combined, you'd get:
select (date_trunc('month', to_date(act_dt::text, 'YYYYMMDD')) +
interval '1 month' - interval '1 day')::date
from YourTable;
Example at SQL Fiddle.
date_trunc('month',current_date) + interval '1 month' - interval '1 day'
Truncating any date or timestamp to the month level will give you the first of the month containing that date. Adding a month gives you the first of the following month. Then, removing a day will give you the date of the last day of the month of the provided date.
For future searches, Redshift does not accept INTERVAL '1 month'. Instead use dateadd(month, 1, date) as documented here.
To get the end of the month use: DATEADD(DAY, -1, (DATE_TRUNC('month', DATEADD(MONTH, 1, date))))
select to_char(date_trunc('month', now() + '01 Months'::interval) - '01 Days'::interval, 'YYYYmmDD'::text)::numeric as end_period_n
Related
Could you please help me as below statement , i want to migrate from Greenplum to HiveSQL. kindly help me.
(date_trunc('week',idate) - INTERVAL '1 week')::DATE date_from
((date_trunc('week',idate) - INTERVAL '1 week')::DATE + '6 days'::INTERVAL)::DATE
date_trunc('week',idate)::DATE
note: idate is i have to parse the argument like 2021-02-20
If you are looking to
find start of week then use select next_day(date_sub(current_date, 7), 'MON')
add 1 week to the date then use select current_date + interval 7 day
convert date to string then use select to_date(current_date )
Now, from your code, it seems, you are looking for start of week and then deducting 1 week from that.
(date_trunc('week',idate) - INTERVAL '1 week')
This can be re-written in hive like below.
next_day(date_sub(current_date, 7), 'MON') - interval 7 day
I assumed Monday is your start of week. Please validate the SQL before using it.
I want to Increment/Decrement the Year , Day , Month by pulling in the Current Date in Snowflake in a single query ?
For e.g.
Suppose Current System Date - 04082021 I want to make it 05092022. I have tried the dateadd function but I suppose it allows only one part i.e. either year or month or day to be incremented at once.
Is it Possible ? If Not, What are the other alternatives ?
If you want to useDATEADD function, here is the code:
SELECT DATEADD(YEAR,+1,DATEADD(MONTH,+1,DATEADD(DAY,+1,CURRENT_DATE())));
returns 2022-09-05 from 2021-08-04
If you want to keep your format, Please use following code
SELECT TO_CHAR(DATEADD(YEAR,+1,DATEADD(MONTH,+1,DATEADD(DAY,+1,CURRENT_DATE()))),'DDMMYYYY');
I would do the arithmetic in thee parts:
select current_date + interval '1 year' + interval '1 month' + interval '1 day'
You can also use:
select current_date + interval '1 year, 1 month, 1 day'
can anybody convert this oracle expression trunc(add_months(sysdate,3),'Q')-1) to postgresql?
Basically this expression gives you the last day of the current quarter (provided that you remove the last closing parenthese, which otherwise is a syntax error).
In postgres, you could phrase this as:
date_trunc('quarter', current_date) + interval '3 months' - interval '1 day'
This generates a timestamp value, that you can cast if you want a date (this distinction does not exist in Oracle, where a date stores the time component as well).
The Postgres equivalent of your Oracle calculation can be seen below.
select date_trunc('quarter', current_date + interval '3 month') - interval '1 day'
I am writing a Postgres procedure and I want to replace only date part of date with some static no.
eg:-
varDate Date default '2018-05-21';
Say I want to make this date as '2018-05-08';
Can anyone tell how to achieve this.
Till now what i have tried is this
varDate := varDate - interval '1 day' * 21 + interval '1 day' * 8;
The above expression gives me proper results. But is there any shortcut to change only the date part of the date.
As far as I understand you want to change the day of the month to 8.
One way to do this is to "truncate" the date to the start of the month, then add 8 days:
vardate := date_trunc('month', vardate)::date + 8;
date_trunc returns a timestamp that's why the cast ::date is needed.
Another option is to "build" a date based on the existing date:
vardate := make_date(extract(year from vardate)::int, extract(month from vardate)::int, 8);
Another option is to add a number of days to the date so you land on the 8th day:
select vardate::date + (8 - extract(day from vardate) * interval '1 day'
Is it possible to convert e.g. string "201701" to dates '2017-01-01' and '2017-01-31' in PostgreSQL?
So for:
"201701" get '2017-01-01' and '2017-01-31'
"201702" get '2017-02-01' and '2017-02-28'
"201703" get '2017-03-01' and '2017-02-31'
etc
You may use the TO_DATE function, and append the day component using string concatenation, something like this:
SELECT
TO_DATE('201702' || '01', 'YYYYMMDD') AS first,
(TO_DATE('201702' || '01', 'YYYYMMDD') + INTERVAL '1 month') -
INTERVAL '1 day' AS last;
The above trick just adds 01 to form the first of the month. For the last day of the same month, it first adds one month to the first, to get the first of the next month, then rolls back one day to get the last of the current month.
Demo
The above answer by Tim Biegeleisen works, but here's an alternative.
The to_date() function converts a string literal to a date value.
sample usage is : to_date(text,format);
SELECT to_date('201701','YYYYMMDD');
(EDITED)
You can also use date_trunc which is a part of PostgreSQL. It does the same as the above one.
select date_trunc('month', current_date) , date_trunc('month', CURRENT_DATE) + interval '1 month - 1 day';
Code Example Live Demo