I have this postgresql query
SELECT pick.min_date, extract('week' from pick.min_date) as week FROM account_invoice inv
left join stock_picking pick on inv.origin=pick.name
WHERE inv.number ='INV/2022/17359'
and the results are
min_date | week
2022-08-11 02:01:00 | 32
What I need for the week column is number 2 because the date (11) is in the 2nd weeks of august. Any help will be great. Thank you
Subtract the week number of the target date from the week number of the first day of the month of the target date and add one. Here is your query with this modification.
SELECT pick.min_date,
extract('week' from pick.min_date) -
extract('week' from date_trunc('month', pick.min_date)) + 1 as week
FROM account_invoice inv
left join stock_picking pick on inv.origin=pick.name
WHERE inv.number ='INV/2022/17359';
Demo
SELECT
extract('week' from '2022-08-11'::date) -
extract('week' from date_trunc('month', '2022-08-11'::date)) + 1;
-- yields 2
I use date part for these extractions
select
((date_part('day', dt::date)::integer - 1) / 7) +1 as currentweekofthemonth,
date_part('week', dt::date) AS weekoftheyear,
date_part('month', dt::date) AS mon,
date_part('year', dt::date) AS yr from
(select '2022-08-11 02:01:00' as dt) as drt ;
OUTPUT :
currentweekofthemonth weekoftheyear mon yr
2 32 8 2022
Related
Let's say I have a table orders with columns amount and order_date.
I want to be able to group this data by quarters and aggregate the amount, the catch however is that the quarters do not start on January 1st but on any given arbitrary date, say July 12th. These quarters are also split in 13 week intervals. From what I see using something like date_trunc such as:
SELECT SUM(orders.amount), DATE_TRUNC('quarter', orders.order_date) AS interval FROM orders WHERE orders.order_date BETWEEN [date_start] AND [date_end] GROUP BY interval
is out of the question as this forces quarters to start on Jan 1st and it has 'hardcoded' quarter starting dates (Apr 1st, Jul 1st, etc).
I have tried using something like:
SELECT SUM(orders.amount),
to_timestamp(floor((extract('epoch' from orders.order_date / 7862400 )) * 7862400 ) AT TIME ZONE 'UTC' AS interval
FROM orders
WHERE orders.order_date BETWEEN [date_start] AND [date_end]
GROUP BY interval
(where 7862400 is the time interval that I want)
But with this method I cannot figure out how to set the offset for the initial grouping date, in my example I would like it to start from July 12th of each year (then count 13 weeks and start the next quarter, and so on). Hope I was clear and I would appreciate any help!
You can use generate_series() to create the first day of each quarter, join it and group by it.
SELECT quarters.first_day,
quarters.first_day + '13 weeks'::interval last_day,
sum(orders.amount) amount
FROM orders
LEFT JOIN generate_series('2019-07-12'::timestamp,
'2020-07-10'::timestamp,
'13 weeks'::interval) quarters (first_day)
ON quarters.first_day <= orders.order_date
AND quarters.first_day + '13 weeks'::interval > orders.order_date
WHERE orders.order_date BETWEEN [date_start]
AND [date_end]
GROUP BY quarters.first_day,
quarters.first_day + '13 weeks'::interval;
You just need to make sure, that the boundary days you give the generate_series() cover the whole period you want to query, so that depends on your [date_start] and [date_end].
You can generate your own 'quarterly calendar' and use that in place of the Postgers 'quarter' date extraction.
create or replace function quarterly_calendar(annual_date text default extract('YEAR' from current_date)::text)
returns table( quarter integer
, quarter_start_date date
, quarter_end_date date
)
language sql immutable strict leakproof
as $$
with RECURSIVE quarters as
(select 1 qtr, qdt::date q_start_dt, (qdt + interval '90 day' )::date q_end_dt, (qdt+interval '1 year' - interval '1 day')::date last_dt
from ( select date_trunc('year',current_date) + interval '6 month 11 day' qdt) q
union all
select qtr+1, (q_end_dt + interval '1 day')::date, least ((q_end_dt + interval '91 day')::date,last_dt), last_dt
from quarters
where qtr+1 <=5
)
select qtr, q_start_dt, q_end_dt
from quarters;
$$;
-- test
select * from quarterly_calender();
It does actually create 5 quarters. But that is because a year is not a multiple of 13 weeks (or 91 days or 7862400 seconds). In your given year from 12-July-2019 through 11-July-2020 is 2 days (366 days total) over 4 times that interval. You'll have to decide how to handle that 5th quarter. It occurs every year, having either 1 or 2 days. Hope this helps .
There is one table:
ID DATE
1 2017-09-16 20:12:48
2 2017-09-16 20:38:54
3 2017-09-16 23:58:01
4 2017-09-17 00:24:48
5 2017-09-17 00:26:42
..
The result I need is the last 7-days of data with hourly aggregated count of rows:
COUNT DATE
2 2017-09-16 21:00:00
0 2017-09-16 22:00:00
0 2017-09-16 23:00:00
1 2017-09-17 00:00:00
2 2017-09-17 01:00:00
..
I tried different stuff with EXTRACT, DISTINCT and also used the generate_series function (most stuff from similar stackoverflow questions)
This try was the best one currently:
SELECT
date_trunc('hour', demotime) as date,
COUNT(demotime) as count
FROM demo
GROUP BY date
How to generate hourly series for 7 days and fill-in the count of rows?
SQL DEMO
SELECT dd, count("demotime")
FROM generate_series
( current_date - interval '7 days'
, current_date
, '1 hour'::interval) dd
LEFT JOIN Table1
ON dd = date_trunc('hour', demotime)
GROUP BY dd;
To work from now and now - 7 days:
SELECT dd, count("demotime")
FROM generate_series
( date_trunc('hour', NOW()) - interval '7 days'
, date_trunc('hour', NOW())
, '1 hour'::interval) dd
LEFT JOIN Table1
ON dd = date_trunc('hour', demotime)
GROUP BY dd;
I'm trying to write query which shows calendar in format:
mon. thues. wen. th. fri. sat. sun.
First value in current week is always monday.
For instance, today is friday - 24-07-2015, so i need first value to be 20-07-2015 and the last one 26-07-2015
Until now, I came up with this:
SELECT
extract(DOW FROM r.e - s.a) w,
r.e - s.a as dates
FROM generate_series(0,7,1) AS s(a),
(SELECT CURRENT_DATE e) r
WHERE extract(DOW FROM CURRENT_DATE - s.a) >=1
GROUP BY 1,2;
and then I could sort it with distinct and then substract sunday in the same way. But maybe you can suggest some more elegant solution?
Basically I would like to find the date of closest monday and closest sunday.
Ok, I've managed it:
WITH monday AS (
SELECT
r.e - s.a as dow
FROM generate_series(0,6,1) AS s(a),
(SELECT CURRENT_DATE e) r
WHERE extract(DOW FROM CURRENT_DATE - s.a) = 1
), sunday as (
SELECT
r.e + s.a as dow
FROM generate_series(0,6,1) AS s(a),
(SELECT CURRENT_DATE e) r
WHERE extract(DOW FROM CURRENT_DATE + s.a) = 0
), columns AS(
SELECT n::date
FROM generate_series((select dow from monday), (select dow from sunday), $$1 day$$) n
) SELECT n FROM columns;
I have a table with the date columns (start_date, end_date) and I want to calculate the difference between these dates and grouped by the month.
I am able to get the datediff in days, but I do not know how to group this in month, any suggestions?
Table:
id Start_date End_date days
1234 2014-06-03 2014-07-05 32
12345 2014-02-02 2014-05-10 97
Expected results:
month diff_days
2 26
3 30
4 31
5 10
6 27
7 5
I think your expected output numbers are off a little. You might want to double-check.
I use a calendar table myself, but this query uses a CTE and date arithmetic. Avoiding the hard-coded date '2014-01-01' and the interval for 365 days is straightforward, but it makes the query harder to read, so I just used those values directly.
with your_data as (
select date '2014-06-03' as start_date, date '2014-07-05' as end_date union all
select '2014-02-02', '2014-05-10'
), calendar as (
select date '2014-01-01' + (n || ' days')::interval calendar_date
from generate_series(0, 365) n
)
select extract (month from calendar_date) calendar_month, count(*) from calendar
inner join your_data on calendar.calendar_date between start_date and end_date
group by calendar_month
order by calendar_month;
calendar_month count
--
2 27
3 31
4 30
5 10
6 28
7 5
As a rule of thumb, you should never group by the month alone--doing that risks grouping data from different years. This is a safer version that includes the year, and which also restricts output to a single calendar year.
with your_data as (
select date '2014-06-03' as start_date, date '2014-07-05' as end_date union all
select '2014-02-02', '2014-05-10'
), calendar as (
select date '2014-01-01' + (n || ' days')::interval calendar_date
from generate_series(0, 700) n
)
select extract (year from calendar_date) calendar_year, extract (month from calendar_date) calendar_month, count(*) from calendar
inner join your_data on calendar.calendar_date between start_date and end_date
where calendar_date between '2014-01-01' and '2014-12-31'
group by calendar_year, calendar_month
order by calendar_year, calendar_month;
SQL Fiddle
with min_max as (
select min(start_date) as start_date, max(end_date) as end_date
from t
), g as (
select daterange(d::date, (d + interval '1 month')::date, '[)') as r
from generate_series(
(select date_trunc('month', start_date) from min_max),
(select end_date from min_max),
'1 month'
) g(d)
)
select *
from (
select
to_char(lower(r), 'YYYY Mon') as "Month",
sum(upper(r) - lower(r)) as days
from (
select t.r * g.r as r
from
(
select daterange(start_date, end_date, '[]') as r
from t
) t
inner join
g on t.r && g.r
) s
group by 1
) s
order by to_timestamp("Month", 'YYYY Mon')
;
Month | days
----------+------
2014 Feb | 27
2014 Mar | 31
2014 Apr | 30
2014 May | 10
2014 Jun | 28
2014 Jul | 5
Range data types
Range functions and operators
I got the information I needed from my last post about Postgres: Defining the longest streak (in days) per developer.
However now I want know the longest streak per developer regardless of Saturdays or Sundays. For instance, Bob worked from Thursday 18, Friday 19, Monday 22 and Tuesday 23, hence Bob streak is 4 days.
I understand I can use the DOW window function, which gives me 0 as Sunday , 1 Monday and so on. But
I don’t see how I can apply DOW function in the last solution proposed by Gordon Linoff.
Can some of you help me in this matter? Cheers,
WITH
working_limits AS (
SELECT
MIN(mr_date) AS start_date,
MAX(mr_date) AS end_date
FROM
xxx
),
working_days AS (
SELECT
ROW_NUMBER() OVER () AS day_number,
s.d::date AS date
FROM
GENERATE_SERIES((SELECT start_date FROM working_limits),
(SELECT end_date FROM working_limits),
'1 day') AS s(d)
WHERE
EXTRACT(dow FROM s.d) BETWEEN 1 AND 5),
worked_days AS (
SELECT
ROW_NUMBER() OVER () AS day_number,
developer,
mr_date AS date
FROM
xxx
ORDER BY
developer,
mr_date
)
SELECT
y.developer,
MAX(y.days)
FROM (
SELECT
x.developer,
COUNT(*) AS days
FROM (
SELECT
wngd.date,
wd.developer,
wngd.day_number - wd.day_number AS delta
FROM
working_days wngd INNER JOIN worked_days wd
ON
wngd.date = wd.date) AS x
GROUP BY
x.developer,
x.delta) AS y
GROUP BY
y.developer;