如果表A中的表A值不可用，则从表B中选择先前的可用值

Question

我想显示表A中与表B中的日期匹配的所有日期。 如果表A中没有表A日期，则应选择以前可用的日期。

例如，

我有表A

Date1
"2017-04-10"
"2017-04-11"
"2017-04-12"
"2017-04-13"
"2017-04-14"
"2017-04-15"
"2017-04-16"
"2017-04-17"
"2017-04-18"
"2017-04-19"
"2017-04-20"

和表B为

Date2
"2017-04-10"
"2017-04-11"
"2017-04-12"
"2017-04-13"
"2017-04-18"
"2017-04-19"
"2017-04-20"

我看的结果是，

Date1        | NewDate
"2017-04-10" | "2017-04-10"
"2017-04-11" | "2017-04-11"
"2017-04-12" | "2017-04-12"
"2017-04-13" | "2017-04-13"
"2017-04-14" | "2017-04-13"
"2017-04-15" | "2017-04-13"
"2017-04-16" | "2017-04-13"
"2017-04-17" | "2017-04-13"
"2017-04-18" | "2017-04-18"
"2017-04-19" | "2017-04-19"
"2017-04-20" | "2017-04-20"

有人可以帮帮我吗？

Answer 1

一种方法使用横向连接或相关子查询：

select a.*,
       (select b.date
        from b
        where b.date <= a.date
        order by b.date desc
        limit 1
       ) b
from a;

如果您有大量数据，那么以下内容可能会更有效：

select a.date, b_date
from (select a.date,
             max(b_date) over (order by date, b_date) as b_date
      from ((select a.date as date, null as b_date
             from a
            ) union all
            (select b.date as date, b.date as b_date
             from b
            ) b
           ) ab
      ) ab
where b_date is not null;