Get rows with difference of dates being one - sql

Question

I have the following table and rows defined in SQLFiddle
I need to select rows from products table where difference between two rows start_date and
nvl(return_date,end_date) is 1. i.e. start_date of current row and nvl(return_date,end_date) of previous row should be one
For example
PRODUCT_NO TSH098 and PRODUCT_REG_NO FLDG, the END_DATE is August, 15 2012 and
PRODUCT_NO TSH128 and PRODUCT_REG_NO FLDG start_date is August, 16 2012, so the difference is only of a day.
How can I get the desired output using sql.
Any help is highly appreciable.
Thanks

Answer 1

You can use lag analytical function to get access to a row at a given physical offset prior to the current position. According to your sorting order it might look like this(not so elegant though).
select *
from products p
join (select *
from(select p.Product_no
, p.Product_Reg_No
, case
when (lag(start_date, 1, start_date) over(order by product_reg_no)-
nvl(return_date, end_date)) = 1
then lag(start_date, 1, start_date)
over(order by product_reg_no)
end start_date
, End_Date
, Return_Date
from products p
order by 2,1 desc
)
where start_date is not null
) s
on (p.start_date = s.start_date or p.end_date = s.end_date)
order by 2, 1 desc
SQL FIddle DEMO

Answer 2

In SQL, date + X adds X days to the date. So you can:
select *
from products
where start_date + 1 = nvl(end_date, return_date)
If the dates could contain a time part, use trunc to remove the time part:
select *
from products
where trunc(start_date) + 1 = trunc(nvl(end_date, return_date))
Live example at SQL Fiddle.

Answer 3

I am under the impression you only want the matching dates differing by 1 day if the product reg no matches. So I simply joint it and I think this is what you want
select p1.product_reg_no,
p1.product_no product_no_1,
p2.product_no product_no_2,
p1.start_date start_date_1,
nvl(p2.return_date,p2.end_date) return_or_end_date_2
from products p1
join products p2 on (p1.product_reg_no = p2.product_reg_no)
where p1.start_date-1 = nvl(p2.return_date,p2.end_date)
SQL Fiddle
If I was wrong with the grouping then just leave the join condition away which with the given example products table brings the same result
select p1.product_reg_no,
p1.product_no product_no_1,
p2.product_no product_no_2,
p1.start_date start_date_1,
nvl(p2.return_date,p2.end_date) return_or_end_date_2
from products p1, products p2
where p1.start_date-1 = nvl(p2.return_date,p2.end_date)
SQL Fiddle 2
Now you say the difference is 1 day. I automatically assumed that start_date is 1 day higher than the nvl(return_date,end_date). Also I assumed that the date is always midnight. But to have all that also excluded you can work with trunc and go in both directions:
select p1.product_reg_no,
p1.product_no product_no_1,
p2.product_no product_no_2,
p1.start_date start_date_1,
nvl(p2.return_date,p2.end_date) return_or_end_date_2
from products p1, products p2
where trunc(p1.start_date)-1 = trunc(nvl(p2.return_date,p2.end_date))
or trunc(p1.start_date)+1 = trunc(nvl(p2.return_date,p2.end_date))
SQL Fiddle 3
And this all works because dates (not timestamp) can be calculated by adding and subtracting.
EDIT: Following your comment you want return_date or end_date to be compared and equal dates are also wanted:
select p1.product_reg_no,
p1.product_no product_no_1,
p2.product_no product_no_2,
p1.start_date start_date_1,
p2.return_date return_date_2,
p2.end_date end_date_2
from products p1, products p2
where trunc(p1.start_date) = trunc(p2.return_date)
or trunc(p1.start_date)-1 = trunc(p2.return_date)
or trunc(p1.start_date)+1 = trunc(p2.return_date)
or trunc(p1.start_date) = trunc(p2.end_date)
or trunc(p1.start_date)-1 = trunc(p2.end_date)
or trunc(p1.start_date)+1 = trunc(p2.end_date)
SQL Fiddle 4

Answer 4

The way to compare the current row with the previous row is to user the LAG() function. Something like this:
select * from
(
select p.*
, lag (end_date) over
(order by start_date )
as prev_end_date
, lag (return_date) over
(order by start_date )
as prev_return_date
from products p
)
where (trunc(start_date) - 1) = trunc(nvl(prev_return_date, prev_end_date))
order by 2,1 desc
However, this will not return the results you desire, because you have not defined a mechanism for defining a sort order. And without a sort order the concept of "previous row" is meaningless.
However, what you can do is this:
select p1.*
, p2.*
from products p1 cross join products p2
where (trunc(p2.start_date) - 1) = trunc(nvl(p1.return_date, p1.end_date))
order by 2, 1 desc
This SQL queries your table twice, filtering on the basis of dates. The each row in the result set contains a record from each table. If a given start_date matches more than one end_date or vice versa you will get records for multiple hits.

Answer 5

You mean like this?
SELECT T2.*
FROM PRODUCTS T1
JOIN PRODUCTS T2 ON (
nvl(T1.end_date, T1.return_date) + 1 = T2.start_date
);
In your SQL Fiddle example, it returns:
PRODUCT_NO PRODUCT_REG_NO START_DATE END_DATE RETURN_DATE
TSH128 FLDG August, 16 2012 00:00:00-0400 September, 15 2012 00:00:00-0400 (null)
TSH125 SCRW August, 08 2012 00:00:00-0400 September, 07 2012 00:00:00-0400 (null)
TSH137 SCRW September, 08 2012 00:00:00-0400 October, 07 2012 00:00:00-0400 (null)
TSH128 is returned for the reasons you already explained.
TSH125 is returned because TSH116 end_date is August, 07 2012.
TSH137 is returned because TSH125 end_date is September, 07 2012.
If you want to compare only rows within the same product_reg_no, it's easy to add that to the JOIN condition. If you want both "directions" of the 1-day difference, it's easy to add that too.