Oracle SQL联接删除和复制行

我在Oracle DB中有三个表

01. TBL_ITS_INS_CLAIM_TRANSACTIONS

╔═════════════════════════════╗
║ Id  PolicyMasterID  Ref_No  ║
╠═════════════════════════════╣
║ 01    106              P100 ║
║ 02    106              P200 ║
╚═════════════════════════════╝





02. TBL_ITS_INS_CLAIM_SETTLEMENT

╔══════════════════════════════╗
║ Id   CLAIM_TRANSACTIONS_ID   ║
╠══════════════════════════════╣
║ 01    01                     ║
║ 02    01                     ║
║ 03    01                     ║
╚══════════════════════════════╝


03. TBL_ITS_INS_CLAIM_SUBROGATION

╔════════════════════════════╗
║ Id   CLAIM_TRANSACTIONS_ID ║
╠════════════════════════════╣
║ 01    01                   ║
║ 02    01                   ║
║ 03    01                   ║
╚════════════════════════════

我尝试过的是，我在下面写了获取结果的查询……但它重复了行。 我编辑了以下代码：

SELECT TBL_ITS_INS_CLAIM_TRANSACTIONS.ID as "Claim Trans ID",      
  TBL_ITS_INS_CLAIM_SETTLEMENT.ID as "SettlementId",
  TBL_ITS_INS_CLAIM_SUBROGATION.ID as "SUBROGATION ID"
FROM TBL_ITS_INS_CLAIM_TRANSACTIONS
LEFT JOIN TBL_ITS_INS_CLAIM_SUBROGATION
ON TBL_ITS_INS_CLAIM_TRANSACTIONS.ID = TBL_ITS_INS_CLAIM_SUBROGATION.CLAIM_TRANSACTIONS_ID
LEFT JOIN TBL_ITS_INS_CLAIM_SETTLEMENT
ON TBL_ITS_INS_CLAIM_TRANSACTIONS.ID = TBL_ITS_INS_CLAIM_SETTLEMENT.CLAIM_TRANSACTIONS_ID

以上查询的结果

╔══════════════════════════════════════════════════╗
║ Claim_Trans_ID    SettlementId   SUBROGATION _ID ║
╠══════════════════════════════════════════════════╣
║ 106                   01              01         ║
║ 106                   01              02         ║
║ 106                   02              01         ║
║ 106                   02              02         ║
║ 106                   03              01         ║
║ 106                   03              02         ║
╚══════════════════════════════════════════════════╝

我想得到下面的结果

╔══════════════════════════════════════════════════╗
║ Claim_Trans_ID    SettlementId   SUBROGATION _ID ║
╠══════════════════════════════════════════════════╣
║ 106                     01            01         ║
║ 106                     02            02         ║
║ 106                     03                       ║
╚══════════════════════════════════════════════════╝

---

我怎样才能得到上面这样的结果呢？

只是一个评论，但太长了

我很感谢您在格式化数据方面所做的努力，但这对于人们复制您的案例和测试某些代码来说并不是很有用

发布示例数据的最佳方式（IMHO）是发布人们可以轻松运行以创建案例的内容，而无需编辑

在您的问题中，我会将样本数据发布为：

create table TBL_ITS_INS_CLAIM_TRANSACTIONS(Id, PolicyMasterID, Ref_No) as (
  select '01', 106, 'P100' from dual union all
  select '02', 106, 'P100' from dual
);
create table TBL_ITS_INS_CLAIM_SETTLEMENT(ID, CLAIM_TRANSACTIONS_ID) as (
  select '01', '01' from dual union all
  select '02', '01' from dual union all
  select '03', '01' from dual
);
create table TBL_ITS_INS_CLAIM_SUBROGATION (Id, CLAIM_TRANSACTIONS_ID) as (
select '01', '01' from dual union all
select '02', '01' from dual union all
select '03', '01' from dual
);

创建表后，很容易注意到表结构与查询不匹配：

SQL> SELECT TBL_ITS_INS_CLAIM_TRANSACTIONS.ID as "Claim Trans ID",
  2    TBL_ITS_INS_CLAIM_SETTLEMENT.ID as "SettlementId",
  3    TBL_ITS_INS_CLAIM_SUBROGATION.ID as "SUBROGATION ID"
  4  FROM TBL_ITS_INS_CLAIM_TRANSACTIONS
  5  LEFT JOIN TBL_ITS_INS_CLAIM_SUBROGATION
  6  ON TBL_ITS_INS_CLAIM_TRANSACTIONS.ID = TBL_ITS_INS_CLAIM_SUBROGATION.CLAIM_TRAN_SUM_ID
  7  LEFT JOIN TBL_ITS_INS_CLAIM_SETTLEMENT
  8  ON TBL_ITS_INS_CLAIM_TRANSACTIONS.ID = TBL_ITS_INS_CLAIM_SETTLEMENT.CLAIM_TRAN_SUM_ID;
ON TBL_ITS_INS_CLAIM_TRANSACTIONS.ID = TBL_ITS_INS_CLAIM_SUBROGATION.CLAIM_TRAN_SUM_ID
                                       *
ERROR at line 6:
ORA-00904: "TBL_ITS_INS_CLAIM_SUBROGATION"."CLAIM_TRAN_SUM_ID": invalid
identifier

---

您的结果与您的查询不匹配。当您没有在select子句中选择策略主Id时，为什么会显示它。此外，唯一匹配的事务id是01。为什么会显示其他结果？您能更好地解释一下所需结果背后的逻辑吗？今天的提示：使用表别名！（使查询更易于编写和读取。）什么不起作用？我只是解释说，您发布的表结构与您的查询不匹配，因此您应该编辑您的问题以解决此问题。我不明白你为什么接受这个答案。你需要的结果背后的逻辑我还不清楚。请试着更好地解释为什么你期望得到这样的结果。此外，更广泛的样本数据集也可能有用