I need to pull out first date and last date of the month from the given from_date and to_date as input, For Example:-
I have my psql output table as the following:
Year
Term
Start Date
End Date
2022
Odd
01-02-2022
30-04-2022
2022
Even
01-07-2022
30-09-2022
I need the output as the following:-
Year
Term
Start Date
End Date
2022
Odd
01-02-2022
28-02-2022
2022
Odd
01-03-2022
31-03-2022
2022
Odd
01-04-2022
30-04-2022
2022
Even
01-07-2022
30-07-2022
2022
Even
01-08-2022
31-08-2022
2022
Even
01-09-2022
30-09-2022
I need the ouput in Postgresql, Pls help
Thanks
Your issue boils down to given a period with start and end dates, determine the first and last dates for each month in that period. In Postgres given a date you can determine the first (with date_trunc function) and last of the a month with the expressions:
-- for a given date
date_trunc('month', given_date) -- returns first day of month
date_trunc('month', given_date + interval '1 month' - interval '1 day') -- returns last day of month
Use the first expression above, with generate_series with dates, to create the first of each month in the period. The use the second expression to generate the end of each month. (see demo)
with range_dates (year, term, gsom) as
( select year
, term
, generate_series( date_trunc('month', od.start_date)::date
, date_trunc('month', od.end_date )::date
, interval '1 month'
)::date
from output_data od
)
select year
, term
, gsom start_date
, (gsom + interval '1 month' - interval '1 day')::date end_date
from range_dates
order by term desc, start_date;
Related
I am currently trying to figure out the equivalent functions in PostgresSQL for the below MariaDB functions:
SELECT
CONCAT(DATE_ADD(DATE_ADD(LAST_DAY('2021-02-15 00:00:00'),INTERVAL 1 DAY),INTERVAL -1 MONTH), ' ', '00:00:00') AS first_day_of_the_month,
CONCAT(LAST_DAY('2021-02-15 23:59:59'), ' ' ,'23:59:59') AS last_day_of_the_month;
Expected Result:
first_day_of_the_month | last_day_of_the_month
------------------------|-------------------------------
2021-02-01 00:00:00 | 2021-02-28 23:59:59
I want to get the last day and the first day of the month of a certain date including the time 00:00:00 for the first day and 23:549:59 for the last day.
So far I am comming close to the expected result with these queries:
SELECT
(date_trunc('MONTH', '2021-02-15 00:00:00'::TIMESTAMP) + INTERVAL '1 MONTH - 1 day')::TIMESTAMP
+ '1 days'::INTERVAL - '1 months'::INTERVAL AS first_day_of_the_month,
(date_trunc('MONTH', '2021-02-15 23:59:59'::TIMESTAMP) + INTERVAL '1 MONTH - 1 day')::TIMESTAMP AS last_day_of_the_month;
However, as you can see in the DB-Fiddle for the last_day_of_the_month I am getting 00:00:00 instead of 23:59:59.
How do I need to change the query to get the correct timestamp?
demo:db<>fiddle
/* Option 1: Certain Date */
SELECT
date_trunc('month', timestamp '2021-02-15 00:00:00') AS first_timestamp,
date_trunc('month', timestamp '2021-02-15 23:59:59') + interval '1 month - 1 second' AS last_timestamp;
/* Option 2: Current_date */
SELECT
date_trunc('month', current_date)::timestamp AS first_timestamp,
date_trunc('month', current_date)::timestamp + interval '1 month - 1 second' AS last_timestamp;
First timestamp of month
date_trunc('month', ...) normalizes the date to the first possible timestamp of the month. So it returns 00:00:00 of the first day of the month
Last timestamp of month
If you add a month to the previous result, you get the first day/timestamp (00:00:00) of the next month. If you subtract a day from this, you get the last day of the current month. Of course, you can subtract just a second to get 23:59:59 of the last day of the month as you expected.
For the first day of the month:
date_trunc('month', the_date_column)
For the last day:
date_trunc('month', the_date_column) + interval '1 month' - interval '1 day'
Or if you prefer:
date_trunc('month', date) + interval '1 month - 1 day'
If you really one one second from midnight on the last day, replace day with second in the logic. I don't recommend that. In fact, if you want to define a range of date, define an open range and just use the first day of the next month. Then you can learn about tsrange which builds this into the language.
How to query (SELECT) in Postgresql, so that the results of a column with different dates, are between Sunday and Saturday of the current week.
Query fake example:
SELECT * FROM table WHERE datecolumn BETWEEN CURRENT WEEK
In another query, I have the number of the week in the year. How to make a SELECT for these dates, applying in the WHERE clause the specific week number in the specific year.
Query fake example:
SELECT * FROM table WHERE datecolumn BETWEEN WEEK15 FROM year 2020
Perhaps you can use something like this:
SELECT *
FROM table
WHERE
EXTRACT(week FROM datecolumn) = EXTRACT(week FROM NOW())
AND
EXTRACT(isoyear FROM datecolumn) = EXTRACT(isoyear FROM NOW())
The week is ISO-8601 week number. By definition, ISO weeks start on Mondays and the first week of a year contains January 4 of that year. In other words, the first Thursday of a year is in week 1 of that year.
In the ISO week-numbering system, it is possible for early-January dates to be part of the 52nd or 53rd week of the previous year, and for late-December dates to be part of the first week of the next year.
For example, 2005-01-01 is part of the 53rd week of year 2004, and 2006-01-01 is part of the 52nd week of year 2005, while 2012-12-31 is part of the first week of 2013.
It's recommended to use the isoyear field together with week to get consistent results.
If you need custom (non-ISO) week numbering - you will have to craft your own calculation.
I would recommend the following pair of conditions:
where
date_column >= current_date - extract(dow from current_date) * interval '1 day'
and date_column < current_date - (extract(dow from current_date) - 8) * interval '1 day'
Postgres' date_trunc(week, ...) starts weeks on Monday, so we need something a little more complicated, using extract(dow from ...), which returns 0 on Sundays.
The advantage of this approach is that it is SARGeable, since no function is applied to the column being filtered. This means that this would happily take advantage of an index on the date column.
I would use date_trunc(), but like this. For the current week:
where datecolumn >= date_trunc('week', now()) and
datecolumn < date_trunc('week', now()) + interval '1 week'
For the nth week of the year, this is trickier. I think this does what you want:
where datecolumn >= (date_trunc('week', now()) -
(extract(week from now()) - 1) * interval '1 week' +
<n> * interval '1 week'
) and
datecolumn < (date_trunc('week', now()) -
(extract(week from now()) - 1) * interval '1 week' +
(<n> + 1) * interval '1 week'
)
Both of these are structured so the computations are NOT on the columns, so they are compatible with using indexes.
The below query returns a distinct count of 'members' for a given month and brand (see image below).
select to_char(transaction_date, 'YYYY-MM') as month, brand,
count(distinct UNIQUE_MEM_ID) as distinct_count
from source.table
group by to_char(transaction_date, 'YYYY-MM'), brand;
The data is collected with a 15 day lag after the month closes (meaning September 2016 MONTHLY data won't be 100% until October 15). I am only concerned with monthly data.
The query I would like to build: Until the 15th of this month (October), last month's data (September) should reflect August's data. The current partial month (October) should default to the prior month and thus also to the above logic.
After the 15th of this month, last month's data (September) is now 100% and thus September should reflect September (and October will reflect September until November 15th, and so on).
The current partial month will always = the prior month. The complexity of the query is how to calc prior month.
This query will be ran on a rolling basis so needs to be dynamic.
To be clear, I am trying to build a query where distinct_count for the prior month (until end of current month + 15 days) should reflect (current month - 2) value (for each respective brand). After 15 days of the close of the month, prior month = (current month - 1).
Partial current month defaults to prior month's data. The 15 day value should be variable/modifiable.
First, simplify the query to:
select to_char(transaction_date, 'YYYY-MM') as month, brand,
count(distinct members) as distinct_count
from source.table
group by members, to_char(transaction_date, 'YYYY-MM'), brand;
Then, you are going to have a problem. The problem is that one row (say from Aug 20th) needs to go into two groups. A simple group by won't handle this. So, let's use union all. I think the result is something like this:
select date_trunc('month', transaction_date) as month, brand,
count(distinct members) as distinct_count
from source.table
where (date_trunc('month', transaction_date) < date_trunc('month' current_date) - interval '1 month') or
(day(current_date) > 15 and date_trunc('month', transaction_date) = date_trunc('month' current_date) - interval '1 month')
group by date_trunc('month', transaction_date), brand
union all
select date_trunc('month' current_date) - interval '1 month' as month, brand,
count(distinct members) as distinct_count
from source.table
where (day(current_date) < 15 and date_trunc('month', transaction_date) = date_trunc('month' current_date) - interval '1 month')
group by brand;
Since you already have a working query, I concentrate on the subselect. The condition you can use here is CASE, especially "Searched CASE"
case
when extract(day from current_date) < 15 then
extract(month from current_date - interval '2 months')
else
extract(month from current_date - interval '1 month')
end case
This may be used as part of a where clause, for example.
Here is some sudo code to get the begin date and the end date for your interval.
Begin date:
date DATE_TRUNC('month', CURRENT_DATE - integer 15) - interval '1 month'
This will return the current month only after the 15th day, from there you can subtract a full month to get your starting point.
End Date:
To calculate this, grab the begin date, plus a month, minus a day.
If the source table is partitioned by transaction_date, this syntax (not masking transaction_date with expression) enables partitions eliminatation.
select to_char(transaction_date, 'YYYY-MM') as month
,count (distinct members) as distinct_count
,brand as brand
FROM source.table
where transaction_date between date_trunc('month', current_date) - case when extract (day from current_date) >= 15 then 1 else 2 end * interval '1' month
and date_trunc('month', current_date) - case when extract (day from current_date) >= 15 then 0 else 1 end * interval '1' month - interval '1' day
group by to_char(transaction_date, 'YYYY-MM')
,brand
;
Is there a way to use extract from date in format YYYY-MM-DD how many days were in this month?
example:
for 2016-02-05 it will give 29 (Feb 2016 has 29 days)
for 2016-03-12 it will give 31
for 2015-02-05 it will give 28 (Feb 2015 had 28 days)
I'm using PostgreSQL
EDIT:
LAST_DAY function in postgres is not what i'm looking for. it returns DATE while I expect an Integer
One way to achieve this would be to subtract the beginning of the following month from the beginning of the current month:
db=> SELECT DATE_TRUNC('MONTH', '2016-02-05'::DATE + INTERVAL '1 MONTH') -
DATE_TRUNC('MONTH', '2016-02-05'::DATE);
?column?
----------
29 days
(1 row)
Just needed this today and seems that I came up with pretty much the same as Mureinik, just that I needed it numeric. (PostgreSQL couldn't convert from interval to number directly)
previous month:
select CAST(to_char(date_trunc('month', current_date) - (date_trunc('month', current_date) - interval '1 month'),'dd') as integer)
current month:
select CAST(to_char(date_trunc('month', current_date) + interval '1 month' - date_trunc('month', current_date), 'dd') as integer)
You can try next:
SELECT
DATE_PART('days',
DATE_TRUNC('month', TO_DATE('2016-02-05', 'YYYY-MM-DD'))
+ '1 MONTH'::INTERVAL
- DATE_TRUNC('month', TO_DATE('2016-02-05', 'YYYY-MM-DD'))
);
Note: there date is used twice. And used convert function TO_DATE
In a Postgres 9.1 database, I am trying to generate a series of weeks for a given month but with some constraints. I need all weeks to start on Monday and get cut when they start or end in another month.
Example:
For February, 2013 I want to generate a series like this:
start
------------------------
2013-02-01 00:00:00+00
2013-02-04 00:00:00+00
2013-02-11 00:00:00+00
2013-02-18 00:00:00+00
2013-02-25 00:00:00+00
The query that I have now looks like this:
SELECT GREATEST(date_trunc('week', dates.d),
date_trunc('month',dates.d)) as start
FROM generate_series(to_timestamp(1359676800),to_timestamp(1362095999), '1 week') as dates(d)
This query gets me the first 4 weeks but it's missing the week from the 25th. Is it possible to get the last week?
SELECT generate_series(date_trunc('week', date '2013-02-01' + interval '6 days')
, date_trunc('week', date '2013-02-01' + interval '1 month - 1 day')
, interval '1 week')::date AS day
UNION SELECT date '2013-02-01'
ORDER BY 1;
This variant does not need a subselect, GREATEST or GROUP BY and only generates the required rows. Simpler, faster. It's cheaper to UNION one row.
Add 6 days to the first day of the month before date_trunc('week', ...) to compute the first Monday of the month.
Add 1 month and subtract 1 day before date_trunc('week', ...) to get the last Monday of the month.
This can conveniently be stuffed into a single interval expression: '1 month - 1 day'
UNION (not UNION ALL) the first day of the month to add it unless it's already included as Monday.
Note that date + interval results in timestamp, which is the optimum here. Detailed explanation:
Generating time series between two dates in PostgreSQL
Automation
You can provide the start of the date series in a CTE:
WITH t(d) AS (SELECT date '2013-02-01') -- enter 1st of month once
SELECT generate_series(date_trunc('week', d + interval '6 days')
, date_trunc('week', d + interval '1 month - 1 day')
, interval '1 week')::date AS day
FROM t
UNION SELECT d FROM t
ORDER BY 1;
Or wrap it into a simple SQL function for convenience with repeated calls:
CREATE OR REPLACE FUNCTION f_week_starts_this_month(date)
RETURNS SETOF date AS
$func$
SELECT generate_series(date_trunc('week', $1 + interval '6 days')
, date_trunc('week', $1 + interval '1 month - 1 day')
, interval '1 week')::date AS day
UNION
SELECT $1
ORDER BY 1
$func$ LANGUAGE sql IMMUTABLE;
Call:
SELECT * FROM f_week_starts_this_month('2013-02-01');
You would pass the date for the first day of the month, but it works for any date. You the first day and all Mondays for the following month.
select
greatest(date_trunc('week', dates.d), date_trunc('month',dates.d)) as start
from generate_series('2013-02-01'::date, '2013-02-28', '1 day') as dates(d)
group by 1
order by 1