SQL Server : convert Julian Date to YYYY-MM-DD

Question

I have searched far and wide, but I can't seem find a way to convert julian to yyyy-mm-dd.

Here is the format of my julian:

The Julian format consists of the year, the first two digits, and the day within the year, the last three digits.

For example, 95076 is March 17, 1995. The 95 indicates the year and the 076 indicates it is the 76th day of the year.

15260

I have tried this but it isn't working:

dateadd(d,(convert(int,LAST_CHANGED_DATE) % 1000)-1, convert(date,(convert(varchar,convert(int,LAST_CHANGED_DATE) /1000 + 1900) + '/1/1'))) as GrgDate

Answer 1

You can select each part of the date using datepart()

SELECT DATEPART(yy, 95076), DATEPART(dy, 95076)

+++EDIT: I misunderstood something. Here's my correction: +++++

SELECT DATEADD(day, CAST(RIGHT('95076',3) AS int) – 1, CONVERT(datetime,LEFT('95076',2) + '0101', 112))

Answer 2

Edit: leaving this answer for Oracle and MySQL users
This will not work in T-SQL.

Use this:

MAKEDATE(1900 + d / 1000, d % 1000)

For example:

SELECT MAKEDATE(1900 + 95076 / 1000, 95076 % 1000)

This returns March, 17 1995 00:00:00.

SQLFiddle

Answer 3

I concatenated 20 to my JD and then ran DATEADD(YEAR, LAST_CHANGE_DATE / 1000 - 1900, LAST_CHANGE_DATE % 1000 - 1)

this got me the result. Thank you!!!

Answer 4

FOR SQL Users DECLARE @jdate VARCHAR(10) SET @jdate = 117338 SELECT dateadd(dd, (@jdate - ((@jdate/1000) * 1000)) - 1, dateadd(yy, @jdate/1000, 0))

Answer 5

This will definitely work in all case.

DECLARE @date int
SET @date = 21319
SELECT DATEADD(dd, RIGHT(@date,LEN(@date)-3)-1, DATEADD(yy,LEFT(@date,1)*100 +RIGHT(LEFT(@date,3),2),'1 Jan 1900'))