查找缺失的对 (Oracle SQL)

Posted 2023-05-09

【中文标题】查找缺失的对 (Oracle SQL)

【英文标题】：Find missing pairs (Oracle SQL)

【发布时间】：2021-04-13 22:12:25

【问题描述】：

假设我有一个名为customer_accounts 的表，如下所示：

表：CUSTOMER_ACCOUNTS

+------------------+--------------+
| CUSTOMER_ID (PK) |  ACCOUNT_ID  | 
+------------------+--------------+
|        1         |          5   |
|        1         |          6   | 
|        1         |          8   | 
+------------------+--------------+

我还有另一个名为 distinct_account_pairs 的表，如下所示：

表：DISTINCT_ACCOUNT_PAIRS（之前）

+------------------+---------------+---------------+
| CUSTOMER_ID (PK) |  ACCOUNT_ID1  | ACCOUNT_ID2   |
+------------------+---------------+---------------+
|        1         |          6    |       5       |
+------------------+---------------+---------------+

如何使用每个 CUSTOMER_ID 的缺失对填充 distinct_account_pairs 表？在此示例中，distinct_account_pairs 中的 PK CUSTOMER_ID 1 缺少 5 和 8、6 和 8，因此之后应该看起来像这样。请注意 distinct_account_pairs 可能有 account_id1

表：DISTINCT_ACCOUNT_PAIRS（之后）

+------------------+---------------+---------------+
| CUSTOMER_ID (PK) |  ACCOUNT_ID1  | ACCOUNT_ID2   |
+------------------+---------------+---------------+
|        1         |          6    |       5       |
|        1         |          5    |       8       |
|        1         |          6    |       8       |
+------------------+---------------+---------------+

【问题讨论】：

我假设 PK 的意思是“主键”？您确实意识到，您在CUSTOMER_ACCOUNTS 中标记（PK）的列实际上不是该表中的“主键”吗？不可能 - 它有重复的值。

关于您的问题：因此，第二个表中已经存在的对必须保持原样（即使ACCOUNT_ID1 > ACCOUNT_ID2，但 missing 对都可以添加在所有情况下都使用ACCOUNT_ID1 < ACCOUNT_ID2？另外，您需要更改第二个表（可能通过INSERT 语句），而不仅仅是编写SELECT 查询以获得所需的结果？

【参考方案1】：

你可以用self join来做，不存在：

 create table CUSTOMER_ACCOUNTS ( CUSTOMER_ID int,  ACCOUNT_ID  int);

 insert into CUSTOMER_ACCOUNTS values(        1         ,          5   );
 insert into CUSTOMER_ACCOUNTS values(        1         ,          6   ); 
 insert into CUSTOMER_ACCOUNTS values(        1         ,          8   );

 create table DISTINCT_ACCOUNT_PAIRS ( CUSTOMER_ID int,  ACCOUNT_ID1  int, ACCOUNT_ID2   int);
 insert into DISTINCT_ACCOUNT_PAIRS values(        1         ,          6    ,       5       );

插入查询：

 insert into DISTINCT_ACCOUNT_PAIRS 
 select a.CUSTOMER_ID,a.ACCOUNT_ID,b.ACCOUNT_ID from CUSTOMER_ACCOUNTS a inner join CUSTOMER_ACCOUNTS b
 on a.CUSTOMER_ID = b.CUSTOMER_ID and a.ACCOUNT_ID < b.ACCOUNT_ID
 and not exists 
               (select 1 from DISTINCT_ACCOUNT_PAIRS d where d.CUSTOMER_ID=a.CUSTOMER_ID
               and ((d.ACCOUNT_ID1=a.ACCOUNT_ID and d.ACCOUNT_ID2=b.ACCOUNT_ID)
               or (d.ACCOUNT_ID1=b.ACCOUNT_ID and d.ACCOUNT_ID2=a.ACCOUNT_ID))
               );

查询：从 DISTINCT_ACCOUNT_PAIRS 中选择

 SELECT * FROM DISTINCT_ACCOUNT_PAIRS;

输出：

CUSTOMER_ID	ACCOUNT_ID1	ACCOUNT_ID2
1	6	5
1	6	8
1	5	8

dbhere

【参考方案2】：

您将生成所有对，然后比较它们：

select ca1.customer_id, ca1.account_id, ca2.account_id
from customer_accounts ca1 join
     customer_accounts ca2
     on ca1.customer_id = ca2.customer_id and
        ca1.account_id < ca2.account_id left join
     distinct_account_pairs dap
     on dap.customer_id = ca1.customer_id and
        dap.account_id1 in (ca1.account_id, ca2.account_id) and
        dap.account_id2 in (ca1.account_id, ca2.account_id) 
where dap.customer_id is null;