SQL - Julien Date (CYYDDD) to date - sql

Question

Unfortunately, this is my first approach with SQL!
I am creating with the following code a query between an oracle DB and Excel (Power Query).
select "$Table"."Order" as "Order",
"$Table"."NR" as "Nr",
"$Table"."JDDATE" as "JDDATE"
from "POOLDB2"."3112" "$Table"
WHERE "Key" >118001
AND "CodeAA" = 1
This code works!
Now I want to format the Julian Date (CYYDDD) - for example 118001 for the 01.01.2019 - to a normal date format.
Does anyone know, how to implement this into the code above?
Maybe something like :
select "$Table"."Order" as "Order",
"$Table"."NR" as "Nr",
DATEADD(DAY, JDDATE % 1000 - 1, DATEADD(year, JDDATE/1000, 0))
"$Table"."JDDATE" as "JDDATE"
from "POOLDB2"."3112" "$Table"
WHERE "Key" >118001
AND "CodeAA" = 1
Best regards

Answer 1

There are many different formats for Julian Date... In your use case, this should do it :
with t as (select 118001 jd from dual)
select to_char( to_date(to_char(1901 + floor(jd / 1000)),'YYYY') + mod(jd,1000) - 1, 'dd.mm.yyyy' ) from t
Yields : 01.01.2019

Answer 2

For Oracle,
select to_char(sysdate,'J') from dual; --To Julian Date
select to_date(2456143,'J') from dual; --To Normal Date
must work.
Edit: Sorry I didn't see oracle tag.
Edit: For the requested behavior by OP
select to_date(to_char(1901 + floor(118001 / 1000)),'YYYY') from dual;

Answer 3

You can use the 118001 value you have, split into separate year and day sections, by adding to the nominal starting date 1900-01-01 (based on your comment that 118001 is actually 2018-01-01, not 2019-01-01):
select date '1900-01-01'
+ floor(118001 / 1000) * interval '1' year
+ (mod(118001, 1000) - 1) * interval '1' day
from dual;
DATE'1900-
----------
2018-01-01
or by startng the fixed date a day earlier you can remove the explicit -1:
select date '1899-12-31'
+ floor(118019 / 1000) * interval '1' year
+ mod(118019, 1000) * interval '1' day
from dual;
DATE'1899-
----------
2018-01-19
This avoids having to build up a longer string to convert to a date, though you could do that (modifying #GMB's approach) as:
select to_date(to_char(1900 + floor(118001 / 1000)) || '-01-01', 'YYYY-MM-DD')
+ (mod(118001, 1000) - 1)
from dual;
You need to specify the month, at least, in the to_date() call as Oracle defaults to the current month if that is not supplied. That behaviour is tucked away in the documentation:
If you specify a date value without a time component, then the default time is midnight. If you specify a date value without a date, then the default date is the first day of the current month.
The first part of that is fairly well known and makes sense ; the second part is a bit less obvious, and doesn't make it clear that it applies to partial dates too - so ifyou don't supply a year then the current year is used; if you don't supply a month then the current month is used; but if you don't supply a day then the 1st is used.
You can see what it's doing with some test conversions:
select to_date('2018-12-25', 'YYYY-MM-DD') as demo_a,
to_date('12:34:56', 'HH24:MI:SS') as demo_b,
to_date('2019', 'YYYY') as demo_c,
to_date('07-04', 'MM-DD') as demo_d,
to_date('2019-01', 'YYYY-MM') as demo_e
from dual;
DEMO_A DEMO_B DEMO_C DEMO_D DEMO_E
------------------- ------------------- ------------------- ------------------- -------------------
2018-12-25 00:00:00 2018-12-01 12:34:56 2019-12-01 00:00:00 2018-07-04 00:00:00 2019-01-01 00:00:00