SQL JOIN 乘以结果集 - 笛卡尔积

Question

>我有一个基表，我想再连接 2 个表来获取我需要的所有数据。在运行查询时，我得到了一个笛卡尔乘积，因为其中 2 个表(主表和我正在连接的另一个表(正在通过non-unique键连接。

下面是一个示例：

发票 (i(

id    order_id       name             comment
-----------------------------------------------------------------
1     500            Males            Mice
2     500            Females          Mice
3     500            Courier Fee      Within City

订单 (o(

id    order_number
------------------------------
500   AN5246516264

订单项目 (oi(

id    order_id       strain_id    species_id      comments
-----------------------------------------------------------------
1100  500            858          9876            Mice
1101  500            858          9876            Mice
1102  500            NULL         NULL            Within City

使用上面定义的表，这是我的查询：

查询

SELECT
    i.name, i.comment,
    o.order_number,
    oi.strain_id, oi.species_id
FROM invoice i
LEFT JOIN order o
    ON i.order_id = o.id
LEFT JOIN order_items oi
    ON o.id = oi.order_id

运行查询后，我得到一个笛卡尔乘积，如下所示(不一定按顺序(：

name         comment        order_number     strain_id    species_id
-----------------------------------------------------------------------
Males        Mice           AN5246516264     858          9876 ---> I want this row
Males        Mice           AN5246516264     858          9876
Males        Within City    AN5246516264     NULL         NULL
Females      Mice           AN5246516264     858          9876
Females      Mice           AN5246516264     858          9876 ---> I want this row
Females      Within City    AN5246516264     NULL         NULL
Courier Fee  Mice           AN5246516264     858          9876
Courier Fee  Mice           AN5246516264     858          9876
Courier Fee  Within City    AN5246516264     NULL         NULL ---> I want this row

我了解这里发生的事情，我知道为什么它会产生笛卡尔乘积，但我不知道如何解决我的问题。

我只想将order_number、strain_id和species_id附加到Invoice (i)表中。

是的，表结构以一种奇怪的方式构建。我确实相信数据库设计师在制作表格时喝醉了，但我现在无法改变这一点。

有什么建议吗？

编辑

我已经查看了所有 3 个表，但找不到任何其他列来连接它们。我只剩下order_id.

Answer 1

您可以先使用函数 row_number() 对表中的行进行编号i和oi，然后将其用作连接条件的一部分：

with i as  (select row_number() over (partition by order_id order by id) rn, i.*  
              from invoice i),
     oi as (select row_number() over (partition by order_id order by id) rn, oi.* 
              from order_items oi)
select i.name, i.comments, o.order_number, oi.strain_id, oi.species_id
  from i left join orders o on i.order_id = o.id
  left join oi on oi.order_id = o.id and oi.rn = i.rn

测试数据和输出：

create table invoice (id number(4), order_id number(4), 
  name varchar2(15), comments varchar2(20));
insert into invoice values (1, 500, 'Males', 'Mice');
insert into invoice values (2, 500, 'Females' ,'Mice');
insert into invoice values (3, 500, 'Courier Fee', 'Within City');
--
create table orders (id number(4), order_number varchar2(15));
insert into orders values (500, 'AN5246516264');
--
create table order_items(id number(5), order_id number(4), 
  strain_id number(5), species_id number(5), comments varchar2(20));
insert into order_items values(1100, 500, 858,  9876, 'Mice');
insert into order_items values(1101, 500, 858,  9876, 'Mice');
insert into order_items values(1102, 500, NULL, NULL, 'Within City');

输出：

NAME            COMMENTS             ORDER_NUMBER    STRAIN_ID SPECIES_ID
--------------- -------------------- --------------- --------- ----------
Males           Mice                 AN5246516264          858       9876
Females         Mice                 AN5246516264          858       9876
Courier Fee     Within City          AN5246516264