2

我希望我能够解释困扰我的问题。我有以下分层数据集（这只是 34K 记录的子集）

PARENT_ID   CHILD_ID          EXAM
TUDA12802   TUDA12982         N 
TUDA12982   TUDA12984         J    
TUDA12984   TUDA999           J
TUDA12982   TUDA12983         N
TUDA12983   TUDA15322         J
TUDA12983   TUDA15323         J

这是树的表示

TUDA12982 N
- TUDA12984 J
--  TUDA999 J
- TUDA12983 N
--  TUDA15322 J
--  TUDA15323 J

我需要的是一个包含exam=N 和底层exam = 'J' 记录的所有记录的列表，这些记录可以嵌套。

select *
from test1 
connect by prior child_id = parent_id
start with child_id = 'TUDA12982'
order siblings by child_id;

给我

PARENT_ID      CHILD_ID          EXAM
TUDA12802   TUDA12982         N 
TUDA12982   TUDA12984         J    
TUDA12984   TUDA999           J
TUDA12982   TUDA12983         N
TUDA12983   TUDA15323         J
TUDA12983   TUDA15322         J

但我需要的是

TUDA12802   TUDA12982         N 
TUDA12982   TUDA12984         J 
TUDA12984   TUDA999           J

当我遇到 EXAM = 'N' 记录时，需要停止遍历。

我需要类似“停止”条款的东西。

select *
from test1 
connect by prior child_id = parent_id
start with child_id = 'TUDA12982'
stop with exam = 'N'
order siblings by child_id;

如何才能做到这一点？

4

2 回答 2

4

罗伯特，

您可以通过在 connect by 子句中添加“exam = 'J'”来做到这一点：

SQL> create table test1(parent_id,child_id,exam)
  2  as
  3  select 'TUDA12802', 'TUDA12982', 'N' from dual union all
  4  select 'TUDA12982', 'TUDA12984', 'J' from dual union all
  5  select 'TUDA12984', 'TUDA999', 'J' from dual union all
  6  select 'TUDA12982', 'TUDA12983', 'N' from dual union all
  7  select 'TUDA12983', 'TUDA15322', 'J' from dual union all
  8  select 'TUDA12983', 'TUDA15323', 'J' from dual
  9  /

Tabel is aangemaakt.

SQL>  select parent_id
  2        , child_id
  3        , exam
  4        , level
  5        , lpad(' ',2*level) || sys_connect_by_path(parent_id||'-'||child_id,'/') scbp
  6     from test1
  7    start with exam = 'N'
  8  connect by prior child_id = parent_id
  9      and exam = 'J'
 10  /

PARENT_ID CHILD_ID  E  LEVEL SCBP
--------- --------- - ------ ----------------------------------------------------------------------
TUDA12802 TUDA12982 N      1   /TUDA12802-TUDA12982
TUDA12982 TUDA12984 J      2     /TUDA12802-TUDA12982/TUDA12982-TUDA12984
TUDA12984 TUDA999   J      3       /TUDA12802-TUDA12982/TUDA12982-TUDA12984/TUDA12984-TUDA999
TUDA12982 TUDA12983 N      1   /TUDA12982-TUDA12983
TUDA12983 TUDA15322 J      2     /TUDA12982-TUDA12983/TUDA12983-TUDA15322
TUDA12983 TUDA15323 J      2     /TUDA12982-TUDA12983/TUDA12983-TUDA15323

6 rijen zijn geselecteerd.

问候，罗布。

于 2009-09-18T12:25:33.450 回答
0

听起来像一个简单的查询，它获取所请求的项目，它的“J”孩子就是你想要的，所以这行不通：

select *
from test1 
where child_id = 'TUDA12982'
or exam = 'J'
connect by prior child_id = parent_id
start with child_id = 'TUDA12982'
order siblings by child_id;

我没有 Oracle，所以我无法测试它是否有效，但根据我对语法的理解以及我刚刚在 Google 上搜索的内容，它看起来会有效。

于 2009-09-18T11:42:59.197 回答