如何使用前 1 在子查询中获取两列？

Question

Here's the thing. I have these two tables:

table A:

id    col1    date_x
A     xxxx    2020-02-02
B     yyyy    2020-02-02
C     zzzz    2020-02-02

table B

id    col2    date_y
A     yyyy    2020-01-02
A     yyyy    2020-02-02
A     yyyy    2020-03-02

I wanted to bring col2 when date_y is the highest possible but it has to be lower than date_x.

This is what I've done:

select *,
       (
         select top 1 col2
         from table_B
         where table_B.date_y < a.date_x 
         and table_B.id = a.id
       ) as col2                   
from table_A a

Now, I wanted to bring date_y as well, in order to do some validation.

What is the best way of doing this? I thought about creating another (select top 1...) but this seems very inefficient. Another join would also be inefficient.

Answer 1

You can join the tables on your conditions and use MAX() and FIRST_VALUE() window functions to get the date_y and col2 values:

select distinct a.*,
       first_value(b.col2) over (partition by a.id order by b.date_y desc, b.col2) col2,
       max(b.date_y) over (partition by a.id) date_y
from tableA a left join tableB b
on b.id = a.id and b.date_y < a.date_x

You may change the LEFT join to an INNER join if you want only matched rows from the 2 tables.
See the demo .

Answer 2

Your approach using a correlated subquery is OK - and Redshift supports top (although I prefer limit , that is more widely supported in other databases).

However you are missing an order by clause in the subquery - without it, you get an unpredictable row out of those that satisfy the where clause, which is not what you want.

I would recommend:

select 
    a.*,
    (
        select col2
        from table_B b
        where b.date_y < a.date_x and b.id = a.id
        order by b.date_y desc
        limit 1
    ) as col2                   
 from table_A a

For performance, consider an index on table_B(id, date_y, col2) .