带有两个表的 Sql 层次结构查询答案

【问题标题】：Sql Hierarchy query with two tables带有两个表的 Sql 层次结构查询
【发布时间】：2016-10-02 22:09:44
【问题描述】：

我有两个单独的 oracle 表，每个表都有一个层次结构。它们由 ACCOUNT_TYPE 键关联。表1的定义是这样的。

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

表2定义是这样的

 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

还有一个这样的

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

我正在尝试在层次结构中首先获取帐户类型所以相反，我试图产生这个结果。

任何人都可以帮我生成一个查询，该查询基本上包括作为第一级的帐户类型，然后在它们之下是将在这些帐户下报告的一组帐户。

任何帮助将不胜感激。

【问题讨论】：

删除 MySQL 标签，因为您使用的是 Oracle 支持的 CONNECT WITH。

标签： sql oracle

【解决方案1】：

如果帐户和帐户类型的键是分离的，则此查询将使用层次结构上的顶级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 - 添加一些前缀来区分）。

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简单地从顶层（null）切换到相应的帐户类型。第二部分添加缺少的帐户类型级别。

使用它作为来源，只需应用您的分层查询。

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;

按预期产生结果：

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

【讨论】：

感谢 Marmite，真是太棒了！非常感谢您的回复。

【解决方案2】：

进行查询，为您预测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)

这将为您提供一个帐户“表”（数据集），您的 pred 可能是帐户类型。现在您可以在两个表之间建立一个联合，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，我还没有运行它，所以我不确定它是否有效，但这个想法应该没问题。希望有帮助。

【讨论】：