Oracle SQL - left join record with closest datetime - sql

Question

I have 2 tables.
table1:
item
end time
1
2022-11-23 08:12:00
1
2022-11-23 09:12:00
2
2022-11-22 13:12:00
3
2022-11-22 14:12:00
table2:
item
value
last_dt
1
11
2022-11-23 09:12:00
1
12
2022-11-23 08:30:00
1
13
2022-11-24 08:30:00
2
21
2022-11-22 13:12:00
3
31
2022-11-22 14:12:00
3
32
2022-11-22 14:30:00
i would like to left join table1 to table2 by comparing the table1's end_time with table2's last_dt.
below is the expected result.
item
end time
value
1
2022-11-23 08:12:00
12
1
2022-11-23 09:12:00
11
2
2022-11-22 13:12:00
21
3
2022-11-22 14:12:00
31

Answer 1

You may use lateral join with fetch first row only to select the closest value per row. But it effectively will perform a nested loop (which would be fast in case of index on item, last_dt and small table1).
select *
from table1
left join lateral (
select value
from table2
where table2.item = table1.item
order by abs(table2.last_dt - table1.end_time) asc
fetch first row only
) val
on 1 = 1
Alternatively you may use first aggregate function and order by time difference. It would work in old Oracle versions (at least from 10g) also.
select
table1.item,
table1.end_time,
max(table2.value) keep(dense_rank first order by abs(table2.last_dt - table1.end_time)) as value
from table1
left join table2
on table2.item = table1.item
group by
table1.item,
table1.end_time
For your sample data both will return this result:
ITEM
END_TIME
VALUE
1
2022-11-23 08:12:00
12
2
2022-11-22 13:12:00
21
3
2022-11-22 14:12:00
31
db<>fiddle

Answer 2

Preparing
-- ms-sql-syntax
create table table1(item# int, end_time datetime);
insert table1 select 1, '2022-11-23T08:12:00';
insert table1 select 2, '2022-11-22T13:12:00';
insert table1 select 3, '2022-11-22T14:12:00';
create table table2 (item# int, value int, end_time datetime);
insert table2 select 1, 11, '2022-11-23T09:12:00';
insert table2 select 1, 12, '2022-11-23T08:30:00';
insert table2 select 1, 13, '2022-11-24T08:30:00';
insert table2 select 2, 21, '2022-11-22T13:12:00';
insert table2 select 3, 31, '2022-11-22T14:12:00';
insert table2 select 3, 32, '2022-11-22T14:30:00';
expected result
item# end_time value
1 2022-11-23 08:12:00 12
2 2022-11-22 13:12:00 21
3 2022-11-22 14:12:00 31
Second
You do not need "left join". You need "outer apply" or "cross apply"
https://oracle-base.com/articles/12c/lateral-inline-views-cross-apply-and-outer-apply-joins-12cr1#cross-apply-join
It should be something like this:
-- it should be oracle syntax. not sure
SELECT
t1.item#, t1.end_time, t2.value
FROM table1 AS t1
CROSS APPLY (
SELECT value
FROM table2 AS t2ca
WHERE rownum = 1
ORDER BY ABS(#DATEDIFF('SS', t2ca.end_time, t1.end_time))
) AS t2
-- ms-sql-syntax. exactly
SELECT
t1.item#, t1.end_time, t2.value
FROM table1 AS t1
CROSS APPLY (
SELECT top 1 value
FROM table2 AS t2ca
ORDER BY ABS(DATEDIFF(second, t2ca.end_time, t1.end_time))
) AS t2
TOP 1 WITH TIES -- ms-sql-syntax
-- TOP 1 WITH TIES -- ms-sql-syntax
SELECT TOP 1 WITH TIES
t1.item#
, t1.end_time
, t2.value
FROM table1 AS t1
CROSS JOIN table2 AS t2
ORDER BY
ROW_NUMBER() OVER (
PARTITION BY t1.item#, t1.end_time
ORDER BY ABS(DATEDIFF(second, t2.end_time, t1.end_time))
);
SUBQUERY and window-ROW_NUMBER() -- ms-sql-syntax
-- SUBQUERY and window-ROW_NUMBER() -- ms-sql-syntax
SELECT
item#
, end_time
, value
FROM (
SELECT
t1.item#
, t1.end_time
, t2.value
, ROW_NUMBER() OVER (
PARTITION BY t1.item#, t1.end_time
ORDER BY ABS(DATEDIFF (second, t2.end_time, t1.end_time))
) AS __rn__
FROM table1 AS t1
CROSS JOIN table2 AS t2
) AS ordering
WHERE __rn__ = 1