Question

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                   ║
╚════════════════════════════

我试过的是，我在下面写了查询获取结果...但它的重复行。我编辑了代码：

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

导致上述查询

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

我希望得到如下结果

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

我怎样才能得到如上所述的结果

Answer 1

只是评论，但太长了。

我感谢您在格式化数据方面的努力，但这对于人们重现您的案例并测试一些代码并不是那么有用。

发布样本数据的最佳方式（恕我直言）发布了一些人们可以轻松运行以创建案例的内容，而无需进行编辑。

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，很容易注意到您的表结构与您的查询不匹配：

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

Oracle SQL Join删除和重复行

1 个答案: