I'm currently trying to get the first and last day of any year. I have data from 1950 and I want to get the first day of the year in the dataset to the last day of the year in the dataset (note that the last day of the year might not be December 31rst and same with the first day of the year).
Initially I thought I could use a CTE and call DATEPART with the day of the year selection, but this wouldn't partition appropriately. I also tried a CTE self-join, but since the last day or first day of the year might be different, this also yields inaccurate results.
For instance, using the below actually generates some MINs in the MAX and vice versa, though in theory it should only grab the MAX date for the year and the MIN date for the year:
;WITH CT AS(
SELECT Points
, Date
, DATEPART(DY,Date) DA
FROM Table
WHERE DATEPART(DY,Date) BETWEEN 363 AND 366
OR DATEPART(DY,Date) BETWEEN 1 AND 3
)
SELECT MIN(c.Date) MinYear
, MAX(c.Date) MaxYear
FROM CT c
GROUP BY YEAR(c.Date)
You want something like this for the first day of the year:
dateadd(year, datediff(year,0, c.Date), 0)
and this for the last day of the year:
--first day of next year -1
dateadd(day, -1, dateadd(year, datediff(year,0, c.Date) + 1, 0)
try this
for getting first day ,last day of the year && firstofthe next_year
SELECT
DATEADD(yy, DATEDIFF(yy,0,getdate()), 0) AS Start_Of_Year,
dateadd(yy, datediff(yy,-1, getdate()), -1) AS Last_Day_Of_Year,
DATEADD(yy, DATEDIFF(yy,0,getdate()) + 1, 0) AS FirstOf_the_NextYear
so putting this in your query
;WITH CT AS(
SELECT Points
, Date
, DATEPART(DY,Date) DA
FROM Table
WHERE DATEPART(DY,Date) BETWEEN
DATEPART(day,DATEADD(yy, DATEDIFF(yy,0,getdate()), 0)) AND
DATEPART(day,dateadd(yy, datediff(yy,-1, getdate()), -1))
)
SELECT MIN(c.Date) MinYear
, MAX(c.Date) MaxYear
FROM CT c
GROUP BY YEAR(c.Date)
I should refrain from developing in the evenings because I solved it, and it's actually quite simple:
SELECT MIN(Date)
, MAX(Date)
FROM Table
GROUP BY YEAR(Date)
I can put these values into a CTE and then JOIN on the dates and get what I need:
;WITH CT AS(
SELECT MIN(Date) Mi
, MAX(Date) Ma
FROM Table
GROUP BY YEAR(Date)
)
SELECT c.Mi
, m.Points
, c.Ma
, f.Points
FROM CT c
INNER JOIN Table m ON c.Mi = m.Date
INNER JOIN Table f ON c.Ma = f.Date
Related
I am running an analysis on medication prescribing practices. We want to identify whether someone has been on a class of medications for 60 days out of a 90 day quarter. We have a start and end date for each prescription, and the bounds of the quarter (e.g., 4/1/2022 – 6/30/2022). For each prescription I’ve calculated the number of days between the start and end date (only including days that fall within the bounds of the quarter). There are many instances in which multiple drugs within the same class are prescribed someone might try one antidepressant but not like it, so be given another in the same class.
My original strategy was just to total up number of days for each class of medication and see if it’s 60 or over. The days don’t have to be consecutive, but if they overlap, days during an overlap period shouldn’t count twice (which they would in a simple sum).
For instance in the data table below, patient 1 in row 1 should be included as they are over 60 days. Patient 2 should also get in (rows 2 and 3) because the non-overlapping total (57+8) within the same med class gets them to over 60 days. However, patient 3 should NOT get in, even though the total of 32 + 32 is over 60 because the intervals overlap. This means that they were really on the medication class for only 32 days – this is an instance where someone might be on two different antidepressants simultaneously.
It’s not sufficient to just sum the days in the interval, but I also have to include some way to examine whether the intervals are overlapping and only add days if an interval for a given medication class falls outside another interval for that same class.
Row num Patid Med class Start date End date Interval
1 1 A 2022-04-28 2022-09-12 63
2 2 B 2022-05-03 2022-06-29 57
3 2 B 2022-04-21 2022-04-29 8
4 3 A 2022-01-19 2022-05-03 32
5 3 A 2022-01-19 2022-05-03 32
I’m having a hard time figuring out how to do this. Note, I'm limited to just using SQL for this.
Code that produced the above data. I would embed this in another query to generate a total interval but need to deal with the overlap issue.
DECLARE #startdt DATE;
DECLARE #enddt DATE;
SET #startdt='4/1/2022'
SET #enddt='6/30/2022'
--for q4 fy2022-23 (4/1/2022-6/30/2022)`
SELECT DISTINCT
rx.patid, d.medication_category as medcat, start_date, end_date,
-- case statement to capture days within quarter only
CASE WHEN start_date<#startdt and end_date>#enddt then 90
WHEN start_date<#startdt and end_date>=#startdt then datediff(d,#startdt,end_date)
WHEN start_date>=#startdt and end_date>#enddt then datediff(d,start_date,#enddt)
ELSE datediff(d,start_date,end_date)
END as interval
FROM rx
INNER JOIN Drug_names_categories d
ON rx.drugname=d.drugname
WHERE start_date<'7/1/2022' and end_date>'3/30/2022'
AND rx.patid IS NOT NULL
AND d.medication_category IS NOT NULL
AND d.medication_category <>''
You can accomplish what you want by generating a calendar table (using a Common Table Expression) of individual days within the test range, joining those days with the prescriptions with overlapping days, and then counting distinct days for each patient and medication category combination.
Something like:
DECLARE #startdt DATE = '2022-04-01';
DECLARE #enddt DATE = '2022-06-30';
DECLARE #threshold INT = 60;
WITH Days AS (
SELECT #startdt AS Day
UNION ALL
SELECT DATEADD(day, 1, Day)
FROM Days
WHERE Day < #enddt
)
SELECT
rx.patid, d.medication_category as medcat,
COUNT(DISTINCT DD.Day) AS days_medicated,
MIN(DD.Day) AS start_date,
MAX(DD.Day) AS end_date
FROM rx
INNER JOIN Drug_names_categories d
ON rx.drugname = d.drugname
INNER JOIN Days DD
ON DD.Day BETWEEN rx.start_date AND rx.end_date
WHERE rx.start_date <= #enddt AND #startdt <= rx.end_date
GROUP BY rx.patid, d.medication_category
HAVING COUNT(DISTINCT DD.Day) >= #threshold
ORDER BY rx.patid, start_date;
If using SQL Server 2022 or later, the Days generator can be simplified by using the new GENERATE_SERIES() function:
WITH Days AS (
SELECT DATEADD(day, S.value, #startdt) AS Day
FROM GENERATE_SERIES(0, DATEDIFF(day, #Startdt, #enddt)) S
)
See this db<>fiddle for an example with some sample data.
I would do this using a date/calendar table, then it's pretty easy.
If you don't already have a date table, this link is one of many that describe how to create one easily ( https://www.mssqltips.com/sqlservertip/4054/creating-a-date-dimension-or-calendar-table-in-sql-server/ )
Here's the script from this link (in case the link dies)
DECLARE #StartDate date = '20100101';
DECLARE #CutoffDate date = DATEADD(DAY, -1, DATEADD(YEAR, 30, #StartDate));
;WITH seq(n) AS
(
SELECT 0 UNION ALL SELECT n + 1 FROM seq
WHERE n < DATEDIFF(DAY, #StartDate, #CutoffDate)
),
d(d) AS
(
SELECT DATEADD(DAY, n, #StartDate) FROM seq
),
src AS
(
SELECT
TheDate = CONVERT(date, d),
TheDay = DATEPART(DAY, d),
TheDayName = DATENAME(WEEKDAY, d),
TheWeek = DATEPART(WEEK, d),
TheISOWeek = DATEPART(ISO_WEEK, d),
TheDayOfWeek = DATEPART(WEEKDAY, d),
TheMonth = DATEPART(MONTH, d),
TheMonthName = DATENAME(MONTH, d),
TheQuarter = DATEPART(Quarter, d),
TheYear = DATEPART(YEAR, d),
TheFirstOfMonth = DATEFROMPARTS(YEAR(d), MONTH(d), 1),
TheLastOfYear = DATEFROMPARTS(YEAR(d), 12, 31),
TheDayOfYear = DATEPART(DAYOFYEAR, d)
FROM d
)
SELECT *
INTO MyDateTable
FROM src
ORDER BY TheDate
OPTION (MAXRECURSION 0);
No that you have your new date table you can join to it to get the list of dates that are within the start and end date, something like
SELECT DISTINCT COUNT(TheDate)
FROM rx
INNER JOIN MyDateTable dt on dt BETWEEN rx.start_date AND rx.end_date
INNER JOIN Drug_names_categories d ON rx.drugname=d.drugname
WHERE start_date<'7/1/2022' and end_date>'3/30/2022'
AND rx.patid IS NOT NULL
AND d.medication_category IS NOT NULL
AND d.medication_category <>''
Obviously this is simple example but you could extend this easily to include all the details you need, the point is that you now have a list of dates or distinct list of dates which you can work with easily.
You could also simply the date range applied by referencing the TheQuarter and TheYear columns. If this is a common task consider extending the date table to contain a comound YearQurater columns (e.g. 2023Q1/202301 etc)
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.
i need to get the balance for the 1st of each month from a table ordered by date, if the 1st is missing from the dataset for a certain month then for that month i want the next available dates data.
I have tried many things but I tried the following to place a case in the where statement which just gives me the first and the second any ideas, maybe an over statement
select date_
, balance
from mytable
where case when extract(day from date_) = 1 then extract(day from date_) = 1 else (extract (day from date_) = 2 )end
group by date_
order by date_ desc
you can use window function row_number:
select * from
(
select * , row_number() over (partition by date_trunc('month',date_) order by date_ ) rn
) t
where rn = 1
DISTINCT ON should do the trick:
SELECT DISTINCT ON (EXTRACT (day FROM date_))
date_, balance
FROM mytable
ORDER BY EXTRACT (day FROM date_), date_;
That will get the first date for each month.
How do I find the first, second, third and fourth saturday of the month?
Ex.: I want to end up with this format...
Blockquote
YYYY, MM, Week#1
Blockquote
YYYY, MM, Week#2
Blockquote
Thanks,
This solution does not depend on Datefirst setting.
declare #d datetime = getdate();
select
dateadd(dd, n, firstSaturday)
from (
select
firstSaturday = dateadd(day, 7-(##datefirst+datepart(weekday, dateadd(day,-1, convert(char(6),#d,112)+'01')))%7, dateadd(day,-1, convert(char(6),#d,112)+'01'))
) t
cross apply (values (0), (7), (14), (21)) q(n)
Here is one way to do it, using a stacked cte to create an inline tally table, with another cte on top of that to generate the months calendar.
Please note that you can change the GETDATE() in the first row code to any date you want (even if it's in the middle of the month) and the code will produce all the Saturdays in that month.
-- Get the current month's start date
DECLARE #MonthStart datetime = DATEADD(MONTH, (DATEDIFF(MONTH, 0, GETDATE())), 0)
;WITH lv0 AS (SELECT 0 g UNION ALL SELECT 0)
,lv1 AS (SELECT 0 g FROM lv0 a CROSS JOIN lv0 b) -- 4
,lv2 AS (SELECT 0 g FROM lv1 a CROSS JOIN lv1 b) -- 16
,lv3 AS (SELECT 0 g FROM lv2 a CROSS JOIN lv2 b) -- 256
,Tally (n) AS (SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) FROM lv3)
-- gets all the dates in the current month
,CurrentMonth AS (SELECT TOP (32) dateadd(day, n-1, #MonthStart) As TheDate
FROM Tally
WHERE MONTH(dateadd(day, n-1, #MonthStart)) = MONTH(#MonthStart)
ORDER BY n)
-- gets all the Saturday dates of the current month
SELECT TheDate, DATENAME(WEEKDAY, TheDate)
FROM CurrentMonth
WHERE DATEPART(WEEKDAY, TheDate) = 7 -- Depending on server settings!
If you already have a numbers table, you can use it instead of the stacked cte. If you don't know what is a numbers table and why you should have one, read The "Numbers" or "Tally" Table: What it is and how it replaces a loop by Jeff Moden
You can try this.
SET DATEFIRST 1
DECLARE #MonthId INT = 5
DECLARE #FirstDayOfTheMonth DATE = CONCAT(YEAR(GETDATE()), RIGHT(CONCAT('00', #MonthId),2), '01')
DECLARE #SaturdayId INT = 6
SELECT
DATEADD(DAY, #SaturdayId + WK.ID - DATEPART(WEEKDAY, #FirstDayOfTheMonth), #FirstDayOfTheMonth)
FROM ( VALUES(0),(7),(14),(21),(28)) AS WK(ID)
WHERE
MONTH(DATEADD(DAY, #SaturdayId + WK.ID- DATEPART(WEEKDAY, #FirstDayOfTheMonth),#FirstDayOfTheMonth)) = #MonthId
Here is my code but its showing null while today is friday. But I would like to get last working day.
-- Insert statements for procedure here
--Below is the param you would pass
DECLARE #dateToEvaluate date=GETDATE();
--Routine
DECLARE #startDate date=CAST('1/1/'+CAST(YEAR(#dateToEvaluate) AS char(4)) AS date); -- let's get the first of the year
WITH
tally(n) AS (SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL))-1 FROM sys.all_columns),
dates AS (
SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) AS dt_id,
DATEADD(DAY,n,#startDate) AS dt,
DATENAME(WEEKDAY,DATEADD(DAY,n,#startdate)) AS dt_name
FROM tally
WHERE n<366 --arbitrary
AND DATEPART(WEEKDAY,DATEADD(DAY,n,#startDate)) NOT IN (6)
AND DATEADD(DAY,n,#startDate) NOT IN (SELECT CAST(HolidayDate AS date) FROM Holiday)),
curr_id(id) AS (SELECT dt_id FROM dates WHERE dt=#dateToEvaluate)
SELECT d.dt
FROM dates AS d
CROSS JOIN
curr_id c
WHERE d.dt_id+1=c.id
The code below will take any date and "walk backward" to find the previous week day (M-F) which is not in the #holidays table.
declare #currentdate datetime = '2015-03-22'
declare #holidays table (holiday datetime)
insert #holidays values ('2015-03-20')
;with cte as (
select
#currentdate k
union all
select
dateadd(day, -1, k)
from cte
where
k = #currentdate
or ((datepart(dw, k) + ##DATEFIRST - 1 - 1) % 7) + 1 > 5 --determine day of week independent of culture
or k in (select holiday from #holidays)
)
select min(k) from cte
The dates table doesn't have any FRIDAY dates in it. Change the NOT IN (6) to NOT IN (1, 7). This will remove Saturday and Sundays from the dates table.