How to subtract dates from previous date on same patient?

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How to subtract dates from previous date on same patient?

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RTaylor2208 SSChampion Points: 13296 More actions · Answer 1

I can't take credit for the function. I had that in useful scripts stored on my machine. But its definitely from a blog online.

The rest was me though 🙂

MCITP SQL 2005, MCSA SQL 2012

NineIron SSChampion Points: 12529 More actions · Answer 2

NineIron

SSChampion

Points: 12529

June 23, 2015 at 7:43 am

#1807357

Folks. This works like a charm.

Thanx.

Jeff Moden SSC Guru Points: 1000539 More actions · Answer 3

RTaylor2208 (6/23/2015)
I can't take credit for the function. I had that in useful scripts stored on my machine. But its definitely from a blog online.
The rest was me though 🙂

I think everyone ends up writing one of those workday scripts at one time or another, myself included. It's how I got my start writing on SQL Server Central...

http://qa.sqlservercentral.com/articles/Advanced+Querying/calculatingworkdays/1660/

--Jeff Moden

RBAR is pronounced "ree-bar" and is a "Modenism" for Row-By-Agonizing-Row.
First step towards the paradigm shift of writing Set Based code:
________Stop thinking about what you want to do to a ROW... think, instead, of what you want to do to a COLUMN.
"Change is inevitable... change for the better is not".

Helpful Links:
How to post code problems
How to Post Performance Problems
Create a Tally Function (fnTally)
Intro to Tally Tables and Functions

sachin.patke Grasshopper Points: 15 More actions · Answer 4

Hi. Here's a quick script that returns all dates between two dates provided and whether that date is a weekend or not. you could use this in your scenario as you deem fit. Hope this helps.

--

DECLARE @startDate DATETIME='02/02/2015' -- mm/dd/yyyy

DECLARE @endDate DATETIME= GETDATE() -- mm/dd/yyyy

;WITH Calender (CalanderDate, dayOfTheWeek) AS

(

SELECT @startDate AS CalanderDate, datepart(weekday,@startDate) as dayOfTheWeek

UNION ALL

SELECT CalanderDate + 1, datepart(weekday,CalanderDate + 1) as dayOfTheWeek FROM Calender

WHERE CalanderDate + 1 <= @endDate

)

SELECT [Date] = CONVERT(VARCHAR(10),CalanderDate,25),

(case when dayOfTheWeek in (1,7) then 'WeekEnd' else 'NotAWeekend' end) [IsWeekEnd]

FROM Calender

OPTION (MAXRECURSION 0)

Jeff Moden SSC Guru Points: 1000539 More actions · Answer 5

sachin.patke (6/24/2015)
Hi. Here's a quick script that returns all dates between two dates provided and whether that date is a weekend or not. you could use this in your scenario as you deem fit. Hope this helps.
--
DECLARE @startDate DATETIME='02/02/2015' -- mm/dd/yyyy
DECLARE @endDate DATETIME= GETDATE() -- mm/dd/yyyy
;WITH Calender (CalanderDate, dayOfTheWeek) AS
(
SELECT @startDate AS CalanderDate, datepart(weekday,@startDate) as dayOfTheWeek
UNION ALL
SELECT CalanderDate + 1, datepart(weekday,CalanderDate + 1) as dayOfTheWeek FROM Calender
WHERE CalanderDate + 1 <= @endDate
)
SELECT [Date] = CONVERT(VARCHAR(10),CalanderDate,25),
(case when dayOfTheWeek in (1,7) then 'WeekEnd' else 'NotAWeekend' end) [IsWeekEnd]
FROM Calender
OPTION (MAXRECURSION 0)

Ah... be careful now. That's not actually "quick". That uses a recursive CTE that counts and, even for a small number of rows, performance and resource usage is comparatively horrible. Please see the following article for why such things should be avoided even for a small number of rows.

http://qa.sqlservercentral.com/articles/T-SQL/74118/

--Jeff Moden

RBAR is pronounced "ree-bar" and is a "Modenism" for Row-By-Agonizing-Row.
First step towards the paradigm shift of writing Set Based code:
________Stop thinking about what you want to do to a ROW... think, instead, of what you want to do to a COLUMN.
"Change is inevitable... change for the better is not".

Helpful Links:
How to post code problems
How to Post Performance Problems
Create a Tally Function (fnTally)
Intro to Tally Tables and Functions

NineIron SSChampion Points: 12529 More actions · Answer 6

NineIron

SSChampion

Points: 12529

June 25, 2015 at 4:21 am

#1807987

Thanx for the articles.