simplify a SQL case statement in a case expression - sql

Question

How would I simplify this case statement in T-SQL? It provides the desired result, but it's very unwieldy and hard to read. I have to use the inner case statement to convert a Julian date (aka 6 digit number) into a regular date format.
Basically i'm doing a datediff( getdate(), case statement). Getdate() just returns the time now (ie. 2/27/2020) and the case statement converts a julian date (ie. 123456) into a normal date (ie, 1/1/2020).
Here's the expect output if the query was ran today on Feb 27.
Select CASE
WHEN Datediff(day, Getdate(), CASE
WHEN a.wadpl = 0
THEN NULL
ELSE Dateadd(d, Substring(Cast(wadpl AS VARCHAR(6)), 4, 3) - 1, CONVERT(DATETIME, CASE
WHEN LEFT(Cast(wadpl AS VARCHAR(6)), 1) = '1'
THEN '20'
ELSE '21'
END + Substring(Cast(wadpl AS VARCHAR(6)), 2, 2) + '-01-01'))
END) < 0
THEN 'Overdue Now'
WHEN Datediff(day, Getdate(), CASE
WHEN a.wadpl = 0
THEN NULL
ELSE Dateadd(d, Substring(Cast(wadpl AS VARCHAR(6)), 4, 3) - 1, CONVERT(DATETIME, CASE
WHEN LEFT(Cast(wadpl AS VARCHAR(6)), 1) = '1'
THEN '20'
ELSE '21'
END + Substring(Cast(wadpl AS VARCHAR(6)), 2, 2) + '-01-01'))
END) <= 30
THEN 'Coming due in 01-30 days'
ELSE 'Not Overdue'
END [Overdue Status]
FROM Table_X

Answer 1

Here is a easy one to understand, assuming a.wadpl is an integer:
SELECT CASE
WHEN DATEDIFF(DAY, GETDATE(), DATEADD(DAY, a.wadpl % 1000, DATEADD(YEAR,a.wadpl / 1000,'1899-12-31'))) <0 THEN 'Overdue now'
WHEN DATEDIFF(DAY, GETDATE(), DATEADD(DAY, a.wadpl % 1000, DATEADD(YEAR,a.wadpl / 1000,'1899-12-31'))) <= 30 THEN 'Coming due in 01-30 days'
ELSE 'Not Overdue'
END [Overdue Status]
FROM Table_X
or you can simplify by using a subquery (or you can use a WITH):
SELECT CASE
WHEN Age <0 THEN 'Overdue now'
WHEN Age <= 30 THEN 'Coming due in 01-30 days'
ELSE 'Not Overdue'
END [Overdue Status]
FROM (
SELECT DATEDIFF(DAY,GETDATE(),
DATEADD(DAY,wadpl%1000,DATEADD(YEAR,wadpl/1000,'1899-12-31'))) Age, *
FROM Table_X) a
This will of course cause you to do this arithmetic for each row, and you can't easily use any indexes. If you were asking about aggregates, then I would suggest doing the opposite, and pre-calculating the dates and use those in your query instead. You might also want to consider putting a persisted computed column on table_x:
ALTER TABLE TABLE_X
ADD wadpl_dt AS
(DATEADD(DAY,wadpl%1000,DATEADD(YEAR,wadpl/1000,'1899-12-31'))) PERSISTED;
Now you can just refer to table_x.wadpl_dt whenever you want the datetime, and your query would become:
SELECT CASE
WHEN Age <0 THEN 'Overdue now'
WHEN Age <= 30 THEN 'Coming due in 01-30 days'
ELSE 'Not Overdue'
END [Overdue Status]
FROM (
SELECT DATEDIFF(DAY,GETDATE(), wadpl_dt) Age, *
FROM Table_X) a
Here is the easy way to convert a date to what you refer to as the julian date:
SELECT (DATEPART(YEAR,GETDATE())-1900) * 1000 + DATEPART(DAYOFYEAR, GETDATE())
And this is how you can use it:
DECLARE #overdue int;
DECLARE #next30 int;
SET #overdue = (SELECT (DATEPART(YEAR,GETDATE())-1900) * 1000 + DATEPART(DAYOFYEAR, GETDATE()));
SET #next30 = (SELECT (DATEPART(YEAR,GETDATE()+30)-1900) * 1000 + DATEPART(DAYOFYEAR, GETDATE()+30));
SELECT CASE
WHEN wadpl < #overdue THEN 'Overdue now'
WHEN wadpl <= #next30 THEN 'Coming due in 01-30 days'
ELSE 'Not Overdue'
END [Overdue Status]
FROM Table_X