Question

让我们假设我有这样的样本数据：

| Name     | ID | PARENT_ID |
-----------------------------
| a1       | 1  | null      |
| b2       | 2  | null      |
| c3       | 3  | null      |
| a1.d4    | 4  | 1         |
| a1.e5    | 5  | 1         |
| a1.d4.f6 | 6  | 4         |
| a1.d4.g7 | 7  | 4         |
| a1.e5.h8 | 8  | 5         |
| a2.i9    | 9  | 2         |
| a2.i9.j10| 10 | 9         |

我想从accountId = 1开始选择所有记录，因此预期结果为：

| Name     | ID | PARENT_NAME | PARENT_ID | 
-------------------------------------------
| a1       | 1  | null        | null      |
| a1.d4    | 4  | a1          | 1         |
| a1.e5    | 5  | a1          | 1         |
| a1.d4.f6 | 6  | a1.d4       | 4         |
| a1.d4.g7 | 7  | a1.d4       | 4         |
| a1.e5.h8 | 8  | a1.e5       | 5         |

我目前能够进行递归选择，但是后来我无法从父引用中访问数据，因此我无法返回parent_name。我正在使用的代码（适用于简单的例子）：

SELECT id, parent_id, name
FROM tbl 
  START WITH id = 1 
  CONNECT BY PRIOR id = parent_id

我应该使用哪种SQL来进行上述检索？

未来寻求者的其他关键词：用于选择同一表中父键表示的分层数据的SQL

Answer 1

使用：

    SELECT t1.id, 
           t1.parent_id, 
           t1.name,
           t2.name AS parent_name,
           t2.id AS parent_id
      FROM tbl t1
 LEFT JOIN tbl t2 ON t2.id = t1.parent_id
START WITH t1.id = 1 
CONNECT BY PRIOR t1.id = t1.parent_id

Answer 2

如何使用PRIOR，

所以

SELECT id, parent_id, PRIOR name
   FROM tbl 
START WITH id = 1 
CONNECT BY PRIOR id = parent_id`

或者如果你想获得根名称

SELECT id, parent_id, CONNECT_BY_ROOT name
   FROM tbl 
START WITH id = 1 
CONNECT BY PRIOR id = parent_id

Answer 3

使用新的嵌套查询语法

with q(name, id, parent_id, parent_name) as (
    select 
      t1.name, t1.id, 
      null as parent_id, null as parent_name 
    from t1
    where t1.id = 1
  union all
    select 
      t1.name, t1.id, 
      q.id as parent_id, q.name as parent_name 
    from t1, q
    where t1.parent_id = q.id
)
select * from q

Answer 4

你想这样做吗？

SELECT id, parent_id, name, 
 (select Name from tbl where id = t.parent_id) parent_name
FROM tbl t start with id = 1 CONNECT BY PRIOR id = parent_id

修改基于OMG的另一种选择（但我认为这将同样表现）：

select t1.id, t1.parent_id, t1.name, t2.name AS parent_name, t2.id AS parent_id from (select id, parent_id, name from tbl start with id = 1 connect by prior id = parent_id) t1 left join tbl t2 on t2.id = t1.parent_id

Answer 5

这有点麻烦，但我相信这应该有用（没有额外的连接）。这假定您可以选择一个永远不会出现在相关字段中的字符，作为分隔符。

你可以在没有嵌套select的情况下做到这一点，但我发现这有点干净，有四个对SYS_CONNECT_BY_PATH的引用。

select id, 
       parent_id, 
       case 
         when lvl <> 1 
         then substr(name_path,
                     instr(name_path,'|',1,lvl-1)+1,
                     instr(name_path,'|',1,lvl)
                      -instr(name_path,'|',1,lvl-1)-1) 
         end as name 
from (
  SELECT id, parent_id, sys_connect_by_path(name,'|') as name_path, level as lvl
  FROM tbl 
  START WITH id = 1 
  CONNECT BY PRIOR id = parent_id)

自引用表（Oracle）上的SQL递归查询

5 个答案: