Getting Dates by Selecting a week in oracle - oracle10g

I have a textbox with random numbers from 1 to 52 which are week numbers of a calendar and a drop down which mentions as years.
For example if I select 2 in a textbox with year 2014, then I want the dates to be mentioned as 05-1-2014 - 11-1-2014. Is it possible to do it.
Also I have tried one query which doesnt match my requirement
SELECT date_val, TO_CHAR (date_val, 'ww')
FROM (SELECT TO_DATE ('01-jan-2013', 'DD-MON-YYYY') + LEVEL AS date_val
FROM DUAL
CONNECT BY LEVEL <= 365)
Please help.

Try this. Here 2 is the number of week in the year (FirstSunday+(NumberOfWeek-1)*7 as WeekStart, FirstSunday+ NumberOfWeek*7-1 as WeekEnd) and 2014 is a year:
select
FirstSunday+(2-1)*7 as WeekStart,
FirstSunday+ 2*7-1 as WeekEnd
from
(
Select NEXT_DAY(TO_DATE('01/01/'||'2014','DD/MM/YYYY')-7, 'SUN') as FirstSunday
from dual
)
SQLFiddle demo

Try this too,
SELECT start_date,
start_date + 6 end_day
FROM(
SELECT TRUNC(Trunc(to_date('2014', 'YYYY'),'YYYY')+ 1 * 7,'IW')-1 start_date
FROM duaL
);

Related

extract days of daterange grouped by month postresql

I have a pickupDate and returnDate in my OrderHistory table. I want to extract the sum of rental days of all OrderHistory entries, grouped/ordered by month. A cte seems to be the solution but I don´t get how to implement it in my query since the cte´s i saw were refering to themselves where it says "FROM cte".
I tried something like this:
SELECT
SUM((EXTRACT (DAY FROM("OrderHistory"."returnDate")-("OrderHistory"."pickupDate")))) as traveltime
, to_char("OrderHistory"."pickupDate"::date, 'YYYY-MM') as M
FROM
"OrderHistory"
GROUP BY
M
ORDER BY
M
But the outcome doesn´t split bookings btw two months (e.g. pickupDate=27th march 2022 and returnDate=03rd of april 2022) but will assign the whole 7 days to the month of march, since the returndate is in it. It should show 4 days in march and 3 in april.
Sorry for the probably very stupid question but I am a beginner. (my code is written in postgresql btw)
PostgreSQL naming conventions
Are PostgreSQL column names case-sensitive?
use legal, lower-case names exclusively so double-quoting is not
needed.
Final result in db fiddle
Add daterange column.
alter table order_history add column date_ranges daterange;
update order_history
with a(m_begin, m_end, pickup_date) as
(select date_trunc('month', pickup_date)::date,
(date_trunc('month', pickup_date) + interval '1 month - 1 day')::date,
pickup_date from order_history)
update order_history set date_ranges =
daterange(a.m_begin, a.m_end,'[]') from a
where a.pickup_date = order_history.pickup_date;
then final query:
WITH A AS(
select
pickup_date,
return_date,
return_date - pickup_date as total,
case when return_date <# date_ranges then (return_date - pickup_date)
else ( date_trunc('month', pickup_date) + interval '1 month - 1 day')::date - pickup_date
end partial_mth
from order_history),
b as (SELECT *, a.total - partial_mth parital_not_mth FROM a)
select *,
case when to_char(pickup_date,'YYYY-MM') = to_char(return_date,'YYYY-MM')
then
sum(partial_mth) over(partition by to_char(pickup_date,'YYYY-MM')) +
sum(parital_not_mth) over (partition by to_char(return_date,'YYYY-MM'))
else sum(partial_mth) over(partition by to_char(pickup_date,'YYYY-MM'))
end
from b;
After trying different things I think I found the best answer to my question, that I want to share with the community:
WITH hier as (
SELECT
"OrderHistory"."pickupDate" as start_date
, "OrderHistory"."returnDate" as end_date
, to_char("OrderHistory"."pickupDate"::date, 'YYYY-MM') as M
FROM
"OrderHistory"
GROUP BY
1, 2, 3
ORDER BY
3
), calendar as (
select date '2022-01-01' + (n || ' days')::interval calendar_date
from generate_series(0, 365) n
)
select
to_char(calendar_date::date, 'YYYY-MM')
, count(*) as tage_gebucht
from calendar
inner join hier on calendar.calendar_date between start_date and end_date
where calendar_date between '2022-01-01' and '2022-12-31'
group by 1
order by 1;
I think this is the simplest solution I came up with.

Using 'over' function results in column "table.id" must appear in the GROUP BY clause or be used in an aggregate function

I'm currently writing an application which shows the growth of the total number of events in my table over time, I currently have the following query to do this:
query = session.query(
count(Event.id).label('count'),
extract('year', Event.date).label('year'),
extract('month', Event.date).label('month')
).filter(
Event.date.isnot(None)
).group_by('year', 'month').all()
This results in the following output:
Count
Year
Month
100
2021
1
50
2021
2
75
2021
3
While this is okay on it's own, I want it to display the total number over time, so not just the number of events that month, so the desired outpout should be:
Count
Year
Month
100
2021
1
150
2021
2
225
2021
3
I read on various places I should use a window function using SqlAlchemy's over function, however I can't seem to wrap my head around it and every time I try using it I get the following error:
sqlalchemy.exc.ProgrammingError: (psycopg2.errors.GroupingError) column "event.id" must appear in the GROUP BY clause or be used in an aggregate function
LINE 1: SELECT count(event.id) OVER (PARTITION BY event.date ORDER...
^
[SQL: SELECT count(event.id) OVER (PARTITION BY event.date ORDER BY EXTRACT(year FROM event.date), EXTRACT(month FROM event.date)) AS count, EXTRACT(year FROM event.date) AS year, EXTRACT(month FROM event.date) AS month
FROM event
WHERE event.date IS NOT NULL GROUP BY year, month]
This is the query I used:
session.query(
count(Event.id).over(
order_by=(
extract('year', Event.date),
extract('month', Event.date)
),
partition_by=Event.date
).label('count'),
extract('year', Event.date).label('year'),
extract('month', Event.date).label('month')
).filter(
Event.date.isnot(None)
).group_by('year', 'month').all()
Could someone show me what I'm doing wrong? I've been searching for hours but can't figure out how to get the desired output as adding event.id in the group by would stop my rows from getting grouped by month and year
The final query I ended up using:
query = session.query(
extract('year', Event.date).label('year'),
extract('month', Event.date).label('month'),
func.sum(func.count(Event.id)).over(order_by=(
extract('year', Event.date),
extract('month', Event.date)
)).label('count'),
).filter(
Event.date.isnot(None)
).group_by('year', 'month')
I'm not 100% sure what you want, but I'm assuming you want the number of events up to that month for each month. You're going to first need to calculate the # of events per month and also sum them with the postgresql window function.
You can do that with in a single select statement:
SELECT extract(year FROM events.date) AS year
, extract(month FROM events.date) AS month
, SUM(COUNT(events.id)) OVER(ORDER BY extract(year FROM events.date), extract(month FROM events.date)) AS total_so_far
FROM events
GROUP BY 1,2
but it might be easier to think about if you split it into two:
SELECT year, month, SUM(events_count) OVER(ORDER BY year, month)
FROM (
SELECT extract(year FROM events.date) AS year
, extract(month FROM events.date) AS month
, COUNT(events.id) AS events_count
FROM events
GROUP BY 1,2
)
but not sure how to do that in SqlAlchemy

Postgres: longest streak per developer regardless of Saturdays and Sundays

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;

Last 12 months, group by week

I have a table with a column REGDATE, a registration date (YYYY-MM-DD HH:MM:SS). I would like to show an histogram (ExtJS) in order to understand in which period of the years users are signing up. I would like to do this for the past twelve months with respect to the current date and to group dates by week.
Any hints?
FWIW in PostgreSQL, Karaszi has an answer that works, but there is a faster query:
SELECT date_trunc('week', REGDATE) AS "Week" , count(*) AS "No. of users"
FROM <<TABLE>>
WHERE REGDATE > now() - interval '12 months'
GROUP BY 1
ORDER BY 1;
I based this off the work of Ben Goodacre
in MySQL:
SELECT COUNT(*), DATE_FORMAT(regdate, "%X%V") AS regweek FROM table GROUP BY regweek;
or
SELECT COUNT(*), YEARWEEK(NOW(), 2) as regweek FROM table GROUP BY regweek;
in PostgreSQL:
SELECT COUNT(*), EXTRACT(YEAR FROM regdate)::text || EXTRACT(WEEK FROM regdate)::text AS regweek FROM table GROUP BY regweek;
Maybe this?
select to_char(REGDATE,'WW') "Week number",
count(*) "number of signups",
from YOUR_TABLE
where REGDATE > current_date-365
group by to_char(REGDATE,'WW')
order by to_char(REGDATE,'WW')
Hint: (SQL)
SELECT CONVERT (VARCHAR(7), REGDATE, 120) AS [RegistrationMonth]
FROM ...
GROUP BY CONVERT (VARCHAR(7), REGDATE, 120)
ORDER BY CONVERT (VARCHAR(7), REGDATE, 120)

deriving calendar month from week number

I've had a hunt around for something similar to this but can't find anything.
I have a query that provides the number of transactions that have occurred each day and need to group by year, month, week BUT of course some months span multiple week numbers, eg. Sept. & Oct. 2009.
Take for example week 39 last year (September & October). Thursday is the 1st October therefore 4 days of that week fall in Oct., therefore the volume of transactions for the last 3 days of Sept. should be added to the first week of October's totals? Clear?
For example:
VOLUME----TRANSACTION----YEAR----MONTH----WEEK
1264.1730----53----2009----September----37
2739.7200---109----2009----September----38
522.5500-----21----2009----October----39
1196.6450----51----2009----September----39
2827.9550---113----2009----October----40
2730.4050---110----2009----October----41
3763.7200---154----2009----October----42
3425.6250---137----2009----October----43
3551.8100---143----2009----November--44
2788.0150---113----2009----November--45
The problem is that the calendar is awkward, and there's not much you can do about it. As far as I can see, you have three choices:
Group by year and month. Display the week or weeks in the result but don't group by them.
Group by year and weeks. Display the month or months in the result but don't group by them.
Group by year, month, week, and accept that some of the groups contain less than one week's data. (i.e. what you have now)
From your description it seems like you want option 2:
SELECT year, MIN(month), week, SUM(transaction)
FROM Table1
GROUP BY year, week
Something like this would do:
-- For weeks starting Sunday and ending Saturday, the US default:
SET DATEFIRST 7
-- Alternatively, for weeks starting Saturday and ending Friday:
--SET DATEFIRST 6
SELECT
[Date]
, DATENAME(WEEKDAY,[Date]) AS [DayOfWeek]
, DATEADD(DAY,1-DATEPART(WEEKDAY,[Date]),[Date]) AS WeekStarting
, DATEADD(DAY,7-DATEPART(WEEKDAY,[Date]),[Date]) AS WeekEnding
FROM (
SELECT CONVERT(DATETIME,'20100124') UNION ALL
SELECT CONVERT(DATETIME,'20100125') UNION ALL
SELECT CONVERT(DATETIME,'20100126') UNION ALL
SELECT CONVERT(DATETIME,'20100127') UNION ALL
SELECT CONVERT(DATETIME,'20100128') UNION ALL
SELECT CONVERT(DATETIME,'20100129') UNION ALL
SELECT CONVERT(DATETIME,'20100130') UNION ALL
SELECT CONVERT(DATETIME,'20100131') UNION ALL
SELECT CONVERT(DATETIME,'20100201') UNION ALL
SELECT CONVERT(DATETIME,'20100202') UNION ALL
SELECT CONVERT(DATETIME,'20100203') UNION ALL
SELECT CONVERT(DATETIME,'20100204') UNION ALL
SELECT CONVERT(DATETIME,'20100205') UNION ALL
SELECT CONVERT(DATETIME,'20100206')
) a ([Date])
Then, convert your week start or end date to a month:
SELECT *
, WeekStartingMonthStart = DATEADD(DAY,1-DAY(WeekStarting),WeekStarting)
, WeekStartingMonthEnd = DATEADD(DAY,-1,DATEADD(MONTH,1,DATEADD(DAY,1-DAY(WeekStarting),WeekStarting)))
, WeekEndingMonthStart = DATEADD(DAY,1-DAY(WeekEnding),WeekEnding)
, WeekEndingMonthEnd = DATEADD(DAY,-1,DATEADD(MONTH,1,DATEADD(DAY,1-DAY(WeekEnding),WeekEnding)))
FROM (
SELECT
[Date]
, DATENAME(WEEKDAY,[Date]) AS [DayOfWeek]
, DATEADD(DAY,1-DATEPART(WEEKDAY,[Date]),[Date]) AS WeekStarting
, DATEADD(DAY,7-DATEPART(WEEKDAY,[Date]),[Date]) AS WeekEnding
FROM (
SELECT CONVERT(DATETIME,'20100124') UNION ALL
SELECT CONVERT(DATETIME,'20100125') UNION ALL
SELECT CONVERT(DATETIME,'20100126') UNION ALL
SELECT CONVERT(DATETIME,'20100127') UNION ALL
SELECT CONVERT(DATETIME,'20100128') UNION ALL
SELECT CONVERT(DATETIME,'20100129') UNION ALL
SELECT CONVERT(DATETIME,'20100130') UNION ALL
SELECT CONVERT(DATETIME,'20100131') UNION ALL
SELECT CONVERT(DATETIME,'20100201') UNION ALL
SELECT CONVERT(DATETIME,'20100202') UNION ALL
SELECT CONVERT(DATETIME,'20100203') UNION ALL
SELECT CONVERT(DATETIME,'20100204') UNION ALL
SELECT CONVERT(DATETIME,'20100205') UNION ALL
SELECT CONVERT(DATETIME,'20100206')
) a ([Date])
) a