SQL: Get next date value per row - sql

Question

I've a table containing a date column.
ID | Date
----|-----------
1 | 2000-01-01
2 | 2000-02-01
3 | 2000-02-01
4 | 2000-03-01
I need a select that returns for each row, the ID, the Date and the smallest date (of all dates in the table) that is larger than the current date.
ID | Date | Next date
----+------------+------------
1 | 2000-01-01 | 2000-02-01
2 | 2000-02-01 | 2000-03-01
3 | 2000-02-01 | 2000-03-01
4 | 2000-03-01 | (NULL)
My first approach was
SELECT id, date, LEAD (date, 1) OVER (ORDER BY date NULLS LAST) AS next_date
FROM t
But this only works, if the values in column DATE are unique.
Any ideas?

Answer 1

You could use an analytic function with a windowing clause. lead() doesn't support a windowing clause, so you need use one that does like min() or first_value():
FIRST_VALUE ("Date")
OVER (ORDER BY "Date" RANGE BETWEEN 1 FOLLOWING AND UNBOUNDED FOLLOWING)
The default windowing clause is RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW, which would give all your rows the same value of 2000-01-01 and using a ROWS window would run into the same problem you're having with lead() with duplicate dates (ID 2 would still get 2000-02-01; and ID 4 would get 2000-03-01 instead of null if you you used ROWS BETWEEN CURRENT ROW... rather than 1 FOLLOWING).
Demo using this range:
with t (ID, "Date") as (
select 1, date '2000-01-01' from dual
union all select 2, date '2000-02-01' from dual
union all select 3, date '2000-02-01' from dual
union all select 4, date '2000-03-01' from dual
)
select id, "Date", FIRST_VALUE ("Date") OVER (ORDER BY "Date"
RANGE BETWEEN 1 FOLLOWING AND UNBOUNDED FOLLOWING) AS next_date
FROM t;
ID Date NEXT_DATE
---------- ---------- ----------
1 2000-01-01 2000-02-01
2 2000-02-01 2000-03-01
3 2000-02-01 2000-03-01
4 2000-03-01
Only rows where the date value is higher than the current row are considered. And this still only has to hit the table once.
(I've put "Date" in double-quotes because date is a reserved word; from your sample data it looks like a quoted identifier, but it isn't quoted in your query, so it's probably just got a more sensible name really...)

Answer 2

select * , (select min(t2.date) from table t2 where t2.date > t1.date)
from Table t1
Above code is in sql server

Answer 3

To answer my own question. ;-) (Just to show another option to people stumbling across this post)
Another solution would be using a subselect:
SELECT t.id,
t.date,
(SELECT MIN (t.date)
FROM t t2
WHERE t2.date > t.date)
AS next_date
FROM t;

Answer 4

One approach would be to create a CTE containing the distinct dates and their immediate lead values. Then, join this CTE to your original table on the date to get the final result.
WITH cte AS (
SELECT t.date,
LEAD(t.date, 1) OVER (ORDER BY t.date NULLS LAST) AS next_date
FROM (SELECT DISTINCT date FROM yourTable) t
)
SELECT
t1.ID,
t1.date,
t2.next_date
FROM yourTable t1
INNER JOIN cte t2
ON t1.date = t2.date

Answer 5

Here is another approach, without "distinct":
select
ted.id,
ted.date_col,
(select
min(ted2.date_col)
from
test_date_v ted2
where
ted2.id != ted.id and
ted2.date_col > ted.date_col) next_date_col
from
test_date_v ted;