Question

假设我有桌面服务：

     Name      | ID | PARENT_ID | LEVEL | 
------------------------------------------- 
 s1            | 1  | null      | 0     | 
 s2            | 2  | 1         | 1     | 
 s3            | 3  | 1         | 2     |      
 s4            | 4  | 2         | 2     | 
 s5            | 5  | 3         | 3     | 
 s6            | 6  | 4         | 3     |

我希望获得s6（id = 6）的1级父级，它应该返回s2，有没有办法进行递归查询，直到达到某个级别？

Answer 1

你可以上去树而不是下去 - 从leaf（id = 6）到root（在这个反向情况下本身就是一个叶子，connect_by_isleaf = 1），并取一个“父” “使用prior运算符的那片叶子。

upd ：误解了您对LEVEL的要求（在Oracle分层查询中，它是指定行的层次深度的动态伪列）。如果您希望将结果集限制为具有自定义预填充LEVEL列的特定值的行，则只需将其添加到where条件。

SQL Fiddle

Oracle 11g R2架构设置：

CREATE TABLE t ("NAME" varchar2(2), "ID" int, "PARENT_ID" int, "LVL" int) ; INSERT ALL INTO t ("NAME", "ID", "PARENT_ID", "LVL") VALUES ('s1', 1, NULL, 0) INTO t ("NAME", "ID", "PARENT_ID", "LVL") VALUES ('s2', 2, 1, 1) INTO t ("NAME", "ID", "PARENT_ID", "LVL") VALUES ('s3', 3, 1, 2) INTO t ("NAME", "ID", "PARENT_ID", "LVL") VALUES ('s4', 4, 2, 2) INTO t ("NAME", "ID", "PARENT_ID", "LVL") VALUES ('s5', 5, 3, 3) INTO t ("NAME", "ID", "PARENT_ID", "LVL") VALUES ('s6', 6, 4, 3) SELECT * FROM dual ;

查询1 ：

select id as id, name as name from t where lvl = 1 connect by id = prior parent_id start with id = 6

<强> Results ：

| ID | NAME | |----|------| | 2 | s2 |

Answer 2

这可以通过分层查询实现：

create table tq84_h (
  id number,
  parent_id number,
  level_    number
);

insert into tq84_h values (1, null, 0);
insert into tq84_h values (2, 1   , 1);
insert into tq84_h values (3, 1   , 2);
insert into tq84_h values (4, 2   , 2);
insert into tq84_h values (5, 3   , 3);
insert into tq84_h values (6, 4   , 3);

select 
  parent_id
from
  tq84_h
where
  level_ = 2
start with
  id = 6
connect by
  prior parent_id = id and
  level_>1
;

Oracle SQL递归查询以查找第一级的父级

2 个答案: