聚合成默认值的更好方法

Question

对于此示例，我有三个表（个人、业务和 ind_to_business）。个人拥有关于人的信息。商业有关于商业的信息。 ind_to_business 包含有关哪些人与哪些业务相关联的信息。以下是他们的 DDL：

CREATE TABLE individual
(
 ID INTEGER PRIMARY KEY,
 NAME VARCHAR2(100) NOT NULL,
 ENTERPRISE_ID VARCHAR2(25) NOT NULL UNIQUE
);
CREATE TABLE business
(
 ID INTEGER PRIMARY KEY,
 NAME VARCHAR2(100) NOT NULL,
 ENTERPRISE_ID VARCHAR2(25) NOT NULL UNIQUE
);
CREATE TABLE ind_to_business
(
  ID INTEGER PRIMARY KEY,
  IND_ID REFERENCES individual(id),
  BUS_ID REFERENCES business(id),
  START_DT DATE NOT NULL,
  END_DT DATE
);

我正在寻找为每个人显示一行的最佳方式。如果它们与一项业务相关联，我想显示该业务的ENTERPRISE_ID。如果它们与多个业务相关联，我想显示默认值“多个”。它们将始终与企业相关联，因此不需要LEFT JOIN。它们也可以多次与企业相关联（离开和回来）。同一业务的多条记录将被聚合。

所以对于以下示例数据：

个人：

+----+------------+---------------+
| ID |    NAME    | ENTERPRISE_ID |
+----+------------+---------------+
|  1 | John Smith | 53a23B7       |
|  2 | Jane Doe   | 63f2a35       |
+----+------------+---------------+

业务：

+----+----------+---------------+
| ID |   NAME   | ENTERPRISE_ID |
+----+----------+---------------+
|  3 | ABC Corp | 2a34d9b       |
|  4 | XYZ Inc  | 34bf21e       |
+----+----------+---------------+

ind_to_business

+----+--------+--------+-------------+-------------+
| ID | IND_ID | BUS_ID |  START_DT   |   END_DT    |
+----+--------+--------+-------------+-------------+
|  5 |      1 |      3 | 01-JAN-2000 | 31-DEC-2002 |
|  6 |      1 |      3 | 01-JAN-2015 |             |
|  7 |      2 |      3 | 01-JAN-2000 |             |
|  8 |      2 |      4 | 01-MAR-2006 | 05-JUN-2010 |
|  9 |      2 |      4 | 15-DEC-2019 |             |
+----+--------+--------+-------------+-------------+

我希望得到以下输出：

+---------+------------+------------+
| IND_ID  |    NAME    | LINKED_BUS |
+---------+------------+------------+
| 53a23B7 | John Smith | 2a34d9b    |
| 63f2a35 | Jane Doe   | Multiple   |
+---------+------------+------------+

这是我当前的查询：

SELECT DISTINCT
       sub.ind_id,
       sub.name,
       DECODE(sub.bus_count, 1, sub.bus_id, 'Multiple') AS LINKED_BUS
FROM (SELECT i.enterprise_id AS IND_ID, 
             i.name,
             b.enterprise_id AS BUS_ID,
             COUNT(DISTINCT b.enterprise_id) OVER (PARTITION BY i.id) AS BUS_COUNT
      FROM individual i
      INNER JOIN ind_to_business i2b ON i.id = i2b.ind_id
      INNER JOIN business b ON i2b.bus_id = b.id) sub;

我的查询有效，但这是在大型数据集上运行并且需要很长时间才能运行。我想知道是否有人对如何改进这一点有任何想法，以免浪费太多处理（即需要对最终结果执行 DISTINCT 或在内联视图中执行 COUNT(DISTINCT) 仅使用该值在上面的DECODE 中）。

我还为这个问题创建了一个 DBFiddle。 (Link)

提前感谢您的任何意见。

Answer 1

无需使用DISTINCT 两次。您可以使用subquery factoring 并将内联视图放在WITH 子句中，并将数据集DISTINCT 放在子查询本身中。

WITH data AS
(
  SELECT distinct 
       i.enterprise_id AS IND_ID, 
       i.name,
       b.enterprise_id AS BUS_ID
  FROM individual i
  JOIN ind_to_business i2b ON i.id = i2b.ind_id
  JOIN business b ON i2b.bus_id = b.id
)
SELECT ind_id,
       name,
       case 
         when count(*) = 1 then MIN(bus_id)
         else 'Multiple' 
       end AS LINKED_BUS
FROM data
GROUP BY ind_id, name;

IND_ID     NAME       LINKED_BUS               
---------- ---------- -------------------------
53a23B7    John Smith 2a34d9b                  
63f2a35    Jane Doe   Multiple

Answer 2

您可以尝试使用相关子查询。这消除了对外部distinct 的需要：

SELECT 
    i.enterprise_id ind_id,
    i.name,
    (
        SELECT DECODE(COUNT(DISTINCT b.enterprise_id), 1, MIN(bus_id), 'Multiple')
        FROM ind_to_business i2b
        INNER JOIN business b ON i2b.bus_id = b.id
        WHERE i2b.ind_id = i.id
    ) linked_bus
FROM individual i

Answer 3

您可以加入每个人的聚合 ind_to_business。一种方法：

select i.id, i.name, coalesce(b.enterprise_id, 'Multiple')
from individual i
join
(
  select
    ind_id,
    case when min(bus_id) = max(bus_id) then min(bus_id) else null end as bus_id
  from ind_to_business
  group by ind_id
) ib on ib.ind_id = i.id
left join business b on b.id = ib.bus_id
order by i.id;

Answer 4

首先您应该进行子查询以获取所有需要的维度，然后使用CASE 语句进行所有最终聚合。

select
    ind_id,
    name,
    case
        when count(*) > 1 then 'Multiple'
        else ind_id
    end as linked_bus
from
(
    select 
        distinct i.enterprise_id as ind_id, 
        i.name,
        b.enterprise_id as bus_id
    from individual i

    join ind_to_business i2b 
    on i.id = i2b.ind_id

    join business b 
    on i2b.bus_id = b.id
) vals

group by
    ind_id,
    name
order by
    ind_id