Question

我有两个独立的oracle表，每个表都有一个层次结构。它们通过ACCOUNT_TYPE键相关联。表1的定义就是这样。

CREATE TABLE ACCOUNTS(
    ACCOUNT_CODE VARCHAR2(6),
    ACCOUNT_PRED VARCHAR2(6),
    ACCOUNT_TYPE VARCHAR2(6),
    ACCOUNT_TITLE VARCHAR2(100)
    );

表2定义就是这样

 CREATE TABLE ACCOUNT_TYPES(
    ACCOUNT_TYPE  VARCHAR2(6),
    ACCOUNT_NUMBER_PRED VARCHAR2(6),
    ACCOUNT_TITLE VARCHAR(100)
    );

表1包含以下数据

Insert into HR.ACCOUNTS (ACCOUNT_CODE,ACCOUNT_PRED,ACCOUNT_TYPE,ACCOUNT_TITLE) values ('0001',null,'11','xxxx');
Insert into HR.ACCOUNTS (ACCOUNT_CODE,ACCOUNT_PRED,ACCOUNT_TYPE,ACCOUNT_TITLE) values ('0042','0070','13','xxxx');
Insert into HR.ACCOUNTS (ACCOUNT_CODE,ACCOUNT_PRED,ACCOUNT_TYPE,ACCOUNT_TITLE) values ('0054','0110','13','xxxx');
Insert into HR.ACCOUNTS (ACCOUNT_CODE,ACCOUNT_PRED,ACCOUNT_TYPE,ACCOUNT_TITLE) values ('0056','0070','13','xxxx');
Insert into HR.ACCOUNTS (ACCOUNT_CODE,ACCOUNT_PRED,ACCOUNT_TYPE,ACCOUNT_TITLE) values ('0070',null,'13','xxxx');
Insert into HR.ACCOUNTS (ACCOUNT_CODE,ACCOUNT_PRED,ACCOUNT_TYPE,ACCOUNT_TITLE) values ('0110',null,'13','xxxx');
Insert into HR.ACCOUNTS (ACCOUNT_CODE,ACCOUNT_PRED,ACCOUNT_TYPE,ACCOUNT_TITLE) values ('0172','0171','13','xxxx');
Insert into HR.ACCOUNTS (ACCOUNT_CODE,ACCOUNT_PRED,ACCOUNT_TYPE,ACCOUNT_TITLE) values ('0060','0001','11','XXXX');

表二包含以下数据

REM INSERTING into ACCOUNT_TYPES
SET DEFINE OFF;
Insert into ACCOUNT_TYPES (ACCOUNT_TYPE,ACCOUNT_NUMBER_PRED,ACCOUNT_TITLE) values ('10',null,'xxxx');
Insert into ACCOUNT_TYPES (ACCOUNT_TYPE,ACCOUNT_NUMBER_PRED,ACCOUNT_TITLE) values ('11','10','xxxx');
Insert into ACCOUNT_TYPES (ACCOUNT_TYPE,ACCOUNT_NUMBER_PRED,ACCOUNT_TITLE) values ('12','10','xxxx');
Insert into ACCOUNT_TYPES (ACCOUNT_TYPE,ACCOUNT_NUMBER_PRED,ACCOUNT_TITLE) values ('13','10','xxxx');
Insert into ACCOUNT_TYPES (ACCOUNT_TYPE,ACCOUNT_NUMBER_PRED,ACCOUNT_TITLE) values ('14','10','xxxx');
Insert into ACCOUNT_TYPES (ACCOUNT_TYPE,ACCOUNT_NUMBER_PRED,ACCOUNT_TITLE) values ('15','10','xxxx');
Insert into ACCOUNT_TYPES (ACCOUNT_TYPE,ACCOUNT_NUMBER_PRED,ACCOUNT_TITLE) values ('16','10','xxxx');
Insert into ACCOUNT_TYPES (ACCOUNT_TYPE,ACCOUNT_NUMBER_PRED,ACCOUNT_TITLE) values ('17','10','xxxx');
Insert into ACCOUNT_TYPES (ACCOUNT_TYPE,ACCOUNT_NUMBER_PRED,ACCOUNT_TITLE) values ('18','10','xxxx');
Insert into ACCOUNT_TYPES (ACCOUNT_TYPE,ACCOUNT_NUMBER_PRED,ACCOUNT_TITLE) values ('19','10','xxxx');
Insert into ACCOUNT_TYPES (ACCOUNT_TYPE,ACCOUNT_NUMBER_PRED,ACCOUNT_TITLE) values ('1A','10','xxxx');

我可以像这样运行层次结构查询

SELECT lpad(' ', (level -1) * 3) || ACCOUNT_CODE  AS ACCOUNT_CODE,
             ACCOUNT_TITLE TITLE,
             ACCOUNT_PRED PRED,
             ACCOUNT_TYPE ATYPE
     FROM    ACCOUNTS
  CONNECT BY PRIOR ACCOUNT_CODE = ACCOUNT_PRED
  START WITH       ACCOUNT_PRED  IS NULL

和另一个像这样

SELECT lpad(' ', (level -1) * 3) ||ACCOUNT_TYPE ,
             ACCOUNT_TITLE,
             ACCOUNT_NUMBER_PRED
     FROM    ACCOUNT_TYPES
  CONNECT BY PRIOR ACCOUNT_TYPE = ACCOUNT_NUMBER_PRED
  START WITH  ACCOUNT_NUMBER_PRED  IS NULL;

查询一个基本上会返回这些值

我试图在层次结构中首先获取帐户类型所以相反，我试图产生这种结果。

任何人都可以帮我制作一个查询，该查询基本上将帐户类型包含在第一级，然后在其下方显示将在这些帐户下报告的帐户集。

非常感谢任何帮助。

Answer 1

如果帐户和帐户类型的键是析取的，则此查询会使层次结构中的最高帐户类型级别丰富您的数据集。（如果帐户类型的键可以相同 - 添加一些前缀到不同的。）

select ACCOUNT_CODE, nvl(ACCOUNT_PRED,ACCOUNT_TYPE) ACCOUNT_PRED,  ACCOUNT_TITLE from   ACCOUNTS
union all
select ACCOUNT_TYPE ACCOUNT_CODE, null ACCOUNT_PRED, ACCOUNT_TITLE from   ACCOUNT_TYPES where  ACCOUNT_TYPE in (
  select ACCOUNT_TYPE from   ACCOUNTS where ACCOUNT_PRED is NULL)

请注意，上半部分使用 NVL 从顶级（null）切换到相应的帐户类型。第二部分添加了缺少的帐户类型级别。

将此作为源使用只需应用分层查询。

with acc as (     
select ACCOUNT_CODE, nvl(ACCOUNT_PRED,ACCOUNT_TYPE) ACCOUNT_PRED,  ACCOUNT_TITLE from   ACCOUNTS
union all
select ACCOUNT_TYPE ACCOUNT_CODE, null ACCOUNT_PRED, ACCOUNT_TITLE from   ACCOUNT_TYPES where  ACCOUNT_TYPE in (
  select ACCOUNT_TYPE from   ACCOUNTS where ACCOUNT_PRED is NULL)
)
SELECT lpad(' ', (level -1) * 3) || ACCOUNT_CODE  AS ACCOUNT_CODE, 
             ACCOUNT_TITLE TITLE,
             ACCOUNT_PRED PRED 
     FROM    ACC 
  CONNECT BY PRIOR ACCOUNT_CODE = ACCOUNT_PRED
  START WITH     ACCOUNT_PRED  IS NULL;

哪个产生了预期的结果：

ACCOUNT_CODE    TITLE  PRED 
--------------- ------ ------
11              xxxx          
   0001         xxxx   11     
      0060      XXXX   0001   
13              xxxx          
   0070         xxxx   13     
      0042      xxxx   0070   
      0056      xxxx   0070   
   0110         xxxx   13     
      0054      xxxx   0110

Answer 2

进行查询，为您提供层次结构所需的数据：

select ACCOUNT_CODE, nvl(ac.ACCOUNT_PRED, ac.ACCOUNT_TYPE)
from ACCOUNT_TYPES at
join ACCOUNTS ac on (at.ACCOUNT_TYPE = ac.ACCOUNT_TYPE)

这会给你一个帐号＆＃34;表＆＃34; （数据集）您的pred可能是一个帐户类型。现在，您可以在两个表之间建立联合，并对结果进行层次结构：

select *
from 
 (SELECT ACCOUNT,PRED
 FROM    ACCOUNT_TYPES

 union all

 select ACCOUNT, nvl(ac.ACCOUNT_PRED, ac.ACCOUNT_TYPE) as PRED
 from ACCOUNT_TYPES at
 join ACCOUNTS ac on (at.ACCOUNT_TYPE = ac.ACCOUNT_TYPE))
CONNECT BY PRIOR ACCOUNT = PRED
START WITH  PRED  IS NULL

说实话，我没有运行它，所以我不确定它是否有效，但这个想法应该没问题。希望它有所帮助。

具有两个表

2 个答案: