连接 2 个表，其中右侧日期是小于或等于左侧日期的最大日期

Question

我有 2 table 我想加入 Oracle SQL:

Table A:
DATE        VALUE
02-May-19   19.25
03-May-19   19.35
04-May-19   19.37
06-May-19   19.45
11-May-19   19.30
01-Jun-19   18.79
01-Jul-19   19.33

Table B:
DATE        VALUE
02-May-19   11.08
01-Jun-19   11.09
01-Jul-19   11.11

我可以左外连接它们以引入日期匹配的两个值：

select a."date" as "DATE", a.value as a_value, b.value as b_value
from a 
left outer join b on b."date" = a."date"
;

DATE        A_VALUE B_VALUE
02-May-19   19.25   11.08
03-May-19   19.35   
04-May-19   19.37   
06-May-19   19.45   
11-May-19   19.30   
01-Jun-19   18.79   11.09
01-Jul-19   19.33   11.11

我想做的是 table 中没有日期的地方 B 取最好的 b.value where b."date" <= a ."date"。像这样：

DATE        A_VALUE B_VALUE
02-May-19   19.25   11.08
03-May-19   19.35   11.08
04-May-19   19.37   11.08
06-May-19   19.45   11.08
11-May-19   19.30   11.08
01-Jun-19   18.79   11.09
01-Jul-19   19.33   11.11

Answer 1

在连接条件中使用 month() 和 year() 函数

select a.date, a.value, b.value 
from a 
left outer join b 
on EXTRACT(month FROM b.date)=EXTRACT(month FROM a.date) 
and EXTRACT(year FROM b.date)=EXTRACT(yearFROM a.date)

Answer 2

也许最简单的方法是相关子查询：

select a.*,
       (select max(b.value) keep (dense_rank first order by b.date desc)
        from b
        where b.date <= a.date
       ) b_value
from a;

Answer 3

在您的查询中，您可以添加带有忽略空值选项的滞后函数：

select a."date"
     , a."value"
     , coalesce(b."value", lag(b."value") ignore nulls over (order by a."date")) as best_match_value
from a 
left outer join b on b."date" = a."date"
order by a."date";

连接 2 个表，其中右侧日期是小于或等于左侧日期的最大日期

Join 2 tables where right date is greatest date less than or equal to left date

sql

oracle

join

date

outer-join