Question

假设我有一些数据并进行查询，我得到了类似的东西 -

-----------------------------
trun(date) | location | sum |
-----------------------------
14-June-11 | B        | 5   |
-----------------------------
13-June-11 | B        | 5   |
-----------------------------
14-June-11 | C        | 5   |
-----------------------------
13-June-11 | C        | 5   |
-----------------------------

SELECT TRUNC(DATE_TIME),MIN(LOCATION) AS LOCATION, SUM(CREDIT) AS SUM FROM 
(SELECT * FROM TABLE A
WHERE
 A.DATE_TIME >= TO_DATE('13/JUN/2011','dd/mon/yyyy')
 AND A.DATE_TIME <= TO_DATE('15/JUN/2011','dd/mon/yyyy'))
GROUP BY TRUNC(DATE_TIME), LOCATION

还有另一张表B，其中有一个位置列表

----
A  |
----
B  |
----
C  |
----

我想要这样的东西 -

-----------------------------
trun(date) | location | sum |
-----------------------------
14-June-11 | A        | 0   |
-----------------------------
14-June-11 | B        | 5   |
-----------------------------
14-June-11 | C        | 5   |
-----------------------------
13-June-11 | A        | 0   |
-----------------------------
13-June-11 | B        | 5   |
-----------------------------
13-June-11 | C        | 5   |
-----------------------------

我尝试使用表B中的右连接，但是我无法为6月14日和11月13日创建2个单独的记录。任何建议或帮助将不胜感激。

Answer 1

无需单独阅读tablea即可获取不同的date_time值。相反，这是经常被忽视的分区外连接的工作。

以下是使用该功能的答案。（William Robertson的答案中的表格和插页将有助于设置它。）

select a.date_time
     , b.location
     , coalesce(a.asum,0) as asum
from  ( SELECT trunc(a.date_time) date_time, 
               a.location, 
               sum(a.credit) as asum 
        FROM tablea a 
        WHERE  a.date_time between date '2011-06-13' and date '2011-06-15' or a.date_time is null
        GROUP BY trunc(a.date_time), a.location ) a 
        -- This is the key part here...
        PARTITION BY (date_time)
right join tableb b on a.location = b.location
order by 1 desc, 2;

PARTITION BY关键字的作用是使外连接为date_time的每个不同值单独操作，根据需要为每个值创建空外连接行。

+-----------+----------+------+
| DATE_TIME | LOCATION | ASUM |
+-----------+----------+------+
| 14-JUN-11 | A        |    0 |
| 14-JUN-11 | B        |    5 |
| 14-JUN-11 | C        |    5 |
| 13-JUN-11 | A        |    0 |
| 13-JUN-11 | B        |    5 |
| 13-JUN-11 | C        |    5 |
+-----------+----------+------+

Answer 2

这个怎么样：

library(tidyverse) # for `dplyr` and `tidyr`
df %>% 
  group_by(claimid) %>% 
  mutate(dates = list(as.Date(startdate:enddate, origin = "1970-01-01"))) %>%
  select(1, 4) %>% 
  unnest %>%
  ungroup

# # A tibble: 82 x 2
#   claimid      dates
#    <fctr>     <date>
# 1    123A 2018-01-01
# 2    123A 2018-01-02
# 3    123A 2018-01-03
# 4    123A 2018-01-04
# 5    123A 2018-01-05
# 6    123A 2018-01-06
# 7    125B 2017-05-20
# 8    125B 2017-05-21
# 9    125B 2017-05-22
# 10   125B 2017-05-23
# # ... with 72 more rows

测试数据：

with dates as
     ( select distinct trunc(date_time) as date_time from tablea )
select trunc(d.date_time)
     , b.location
     , coalesce(sum(credit),0) as sum
from   dates d
       cross join tableb b
       left join tablea a
            on  a.location = b.location
            and trunc(a.date_time) = d.date_time
where  a.date_time between date '2011-06-13' and date '2011-06-15' or a.date_time is null
group by d.date_time, b.location
order by 1 desc, 2;

Oracle中的外部联接表 - 按每个日期单独加入

2 个答案: