Join table by id and nearest date for every date - sql

Question

I have 2 tables:
TABLE 1
id date_measured value 1
1 01/01/2017 5
1 02/20/2017 6
1 04/01/2017 5
2 03/02/2017 5
2 04/02/2017 3
TABLE 2
id date_measured value 2
1 01/06/2017 5
1 03/01/2017 6
2 02/01/2017 5
2 03/09/2017 7
2 04/05/2017 4
I want to join it such that each id matches and the closest date matches so:
id date_measured1 value 1 date_measured2 value 2
1 01/01/2017 5 01/06/2017 5
1 02/20/2017 6 03/01/2017 6
2 02/01/2017 5 02/01/2017 5
2 03/02/2017 5 03/09/2017 7
2 04/02/2017 3 04/05/2017 4
etc. IE for each id for each date measured take the closest measured date in the other table and make it a row. Something closeish to
SELECT *
FROM table1 a
INNER JOIN table2 b
ON a.id = b.id
AND <date from a is closest date from b>
But I have no idea how to do the second part. Any suggestions?

Answer 1

In standard SQL, you can get the date using a correlated subquery:
select t1.*,
(select t2.date_measured
from table2 t2
where t2.id = t1.id
order by abs(t2.date_measured - t1.date_measured) asc
fetch first 1 row only
) as t2_date_measured
from table1 t1;
You can then join back to table2 to get additional information from that row.
The above is generic SQL (not necessarily standard SQL). Date/time functions tend to be peculiar to each database; so - may not work for the difference. Not all databases support fetch first 1 row only, but almost all support some mechanism for doing the same thing.

Answer 2

If you have window functions use ROW_NUMBER():
SQL DEMO I use postgresql so date function may vary on your rdbms
WITH cte as (
SELECT *,
t1.id as t1_id,
t1.date_measured as t1_date,
t1.value1,
t2.id as t2_id,
t2.date_measured as t2_date,
t2.value2,
date_part('day', age(t1.date_measured, t2.date_measured)) as days,
ROW_NUMBER() OVER (PARTITION BY t1.id, t1.date_measured
ORDER BY abs(date_part('day', age(t1.date_measured, t2.date_measured)))
) as rn
FROM table1 t1
JOIN table2 t2
ON t1.id = t2.id
)
SELECT *
FROM cte
WHERE rn = 1
ORDER BY t1_id, t1_date