How to find the difference between dates within the same column using SQL? - sql

Question

I am trying to solve the following challenge:
1) If a patient visits the ER within 48 hours, I want to flag that as 1.
2) If the same patient visits the ER again after 48 hours, I want to flag that as 2.
3) Each subsequent visit must be flagged as 3, 4, 5 etcetera after the first 48 hours.
Here is what my table looks like:
PATIENT_ID ADMIT_DATE LOCATION
---------- ---------- --------
33 1/10/2014 ER
33 1/11/2014 ER
33 1/15/2014 ER
33 1/17/2014 ER
45 2/20/2014 OBS
45 2/21/2014 OBS
45 2/25/2014 OBS
45 2/30/2014 OBS
45 2/32/2014 OBS
And here is what the desired result should look like:
PATIENT_ID ADMIT_DATE LOCATION FLAG
---------- ---------- -------- ----
33 1/10/2014 ER 1
33 1/15/2014 ER 2
33 1/17/2014 ER 3
45 2/20/2014 OBS 1
45 2/25/2014 OBS 2
45 2/30/2014 OBS 3
45 2/32/2014 OBS 4
I have started something like this but could not complete it:
SELECT PATIENT_ID, ADMIT_DATE, LOCATION,
CASE WHEN MIN(ADMIT_DATE)-MAX(ADMIT_DATE)<48 THEN 1 ELSE 0 AS FLAG
FROM MYTABLE
GROUP BY PATIENT_ID, ADMIT_DATE, LOCATION
Can someone please help?

Answer 1

You can achieve this easy using LAG, DATEDIFF and ROWNUMBER functions. The LAG function helps you to get the previous ADMIT_DATE value. Then you can calculate the difference in hours using the DATEDIFF function. Finally, using ROWNUMBER you can simple rank your results.
This is full working example:
SET NOCOUNT ON
GO
DECLARE #DataSource TABLE
(
[ATIENT_ID] TINYINT
,[ADMIT_DATE] DATE
,[LOCATION] VARCHAR(3)
)
INSERT INTO #DataSource ([ATIENT_ID], [ADMIT_DATE], [LOCATION])
VALUES (33, '1-10-2014', 'ER')
,(33, '1-11-2014', 'ER')
,(33, '1-15-2014', 'ER')
,(33, '1-17-2014', 'ER')
,(45, '2-15-2014', 'OBS')
,(45, '2-16-2014', 'OBS')
,(45, '2-20-2014', 'OBS')
,(45, '2-25-2014', 'OBS')
,(45, '2-27-2014', 'OBS')
;WITH DataSource ([ATIENT_ID], [ADMIT_DATE], [LOCATION], [DIFF_IN_HOURS]) AS
(
SELECT [ATIENT_ID]
,[ADMIT_DATE]
,[LOCATION]
,DATEDIFF(
HOUR
,LAG([ADMIT_DATE], 1, NULL) OVER (PARTITION BY [ATIENT_ID], [LOCATION] ORDER BY [ADMIT_DATE] ASC)
,[ADMIT_DATE]
)
FROM #DataSource
)
SELECT [ATIENT_ID]
,[ADMIT_DATE]
,[LOCATION]
,ROW_NUMBER() OVER (PARTITION BY [ATIENT_ID], [LOCATION] ORDER BY [ADMIT_DATE] ASC)
FROM DataSource
WHERE [DIFF_IN_HOURS] >= 48
OR [DIFF_IN_HOURS] IS NULL -- these are first records
SET NOCOUNT OFF
GO
Note, I have fixed your sample data as it was wrong.
This is alternative solution without LAG function:
;WITH TempDataSource ([ATIENT_ID], [ADMIT_DATE], [LOCATION], [Rank]) AS
(
SELECT [ATIENT_ID]
,[ADMIT_DATE]
,[LOCATION]
,ROW_NUMBER() OVER (PARTITION BY [ATIENT_ID], [LOCATION] ORDER BY [ADMIT_DATE] ASC)
FROM #DataSource
),
DataSource ([ATIENT_ID], [ADMIT_DATE], [LOCATION], [DIFF_IN_HOURS]) AS
(
SELECT DS1.[ATIENT_ID]
,DS1.[ADMIT_DATE]
,DS1.[LOCATION]
,DATEDIFF(HOUR, DS2.[ADMIT_DATE], DS1.[ADMIT_DATE])
FROM TempDataSource DS1
LEFT JOIN TempDataSource DS2
ON DS1.[Rank] - 1 = DS2.[Rank]
AND DS1.[ATIENT_ID] = DS2.[ATIENT_ID]
AND DS1.[LOCATION] = DS2.[LOCATION]
)
SELECT [ATIENT_ID]
,[ADMIT_DATE]
,[LOCATION]
,ROW_NUMBER() OVER (PARTITION BY [ATIENT_ID], [LOCATION] ORDER BY [ADMIT_DATE] ASC)
FROM DataSource
WHERE [DIFF_IN_HOURS] >= 48
OR [DIFF_IN_HOURS] IS NULL -- these are first records

Answer 2

SELECT Patient_id,Admit_date, Location,
CASE WHEN DATEDIFF (HH , min(admit_date) , max(admit_date)) < 48 THEN count(flag)+1 ELSE 0 End As Flag
FROM tbl_Patient
GROUP BY PATIENT_ID, ADMIT_DATE, LOCATION

Answer 3

you can use DATEDIFF() available in sql-server like
SELECT DATEDIFF(hour,startDate,endDate) AS 'Duration'
You can visit http://msdn.microsoft.com/en-IN/library/ms189794.aspx