How to get the data of nearest date when comparing two tables - db2

Question

I have two tables that need to be joined on the nearest date (nearest before date).
Screenshot of My Requirement
For example: In Table1 Date is 6/19/2018(M/DD/YYYY) then I would like to get the data of nearest before date from the Table2(If table has 07/19/2018, 06/20/2018 and 06/16/2018, I would like to get 06/16/2018 record information).
I have multiple records in table1 and want to get the nearest date record in form from the table2. Please see the image for more info about my requirement. Thank you in advance for your help.

Answer 1

Assuming that you must do it for every customer distinctly (customer column is a key in the example). If you have another key (let's say customer, item_name, item_name column is not shown, add it manually in this case), then change the corresponding predicates (to a[2].customer=x.customer and a[2].item_name=x.item_name in the example). If you don't want to do it for each customer, just remove the predicates a[2].customer=x.customer and.
You can run the statement below AS IS to check.
with
xyz (customer, req_del_date) as (values
('ABC', date('2018-06-19'))
, ('ABC', date('2018-09-04'))
, ('ABC', date('2018-04-24'))
, ('ABC', date('2018-03-17'))
)
, abc (customer, actual_del_date) as (values
('ABC', date('2018-11-20'))
, ('ABC', date('2018-06-12'))
, ('ABC', date('2018-05-09'))
, ('ABC', date('2018-04-27'))
, ('ABC', date('2018-04-14'))
, ('ABC', date('2017-12-31'))
, ('ABC', date('2017-12-30'))
)
select x.customer, x.req_del_date, a.actual_del_date, a.diff_days
from xyz x, table (
select a.customer, a.actual_del_date
, days(x.req_del_date) - days(actual_del_date) diff_days -- just for test
-- other columns from abc if needed
from abc a
where a.customer=x.customer and x.req_del_date>=a.actual_del_date
and (days(x.req_del_date) - days(a.actual_del_date)) =
(
select min(days(x.req_del_date) - days(a2.actual_del_date))
from abc a2
where a2.customer=x.customer and x.req_del_date>=a2.actual_del_date
)
) a;