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编辑：我通过引入位置实体提供了附加信息，以明确我为什么尝试使用子查询

在 oracle 11g 数据库中，我有层次结构的元素表，最终将包含几百万行。每行都有指向其父行的索引外键，并且不允许循环。元素也有名称和类型。除此之外，还有另一个实体 - location，它类似于元素（分层，具有指向父级+名称的外键）。顶部元素（你的根）可以在位置（它们通过LocationId连接）。所以有2个实体：

地点：

  • 编号 [NUMBER(9,0), PK]
  • ParentId [NUMBER(9,0), FK]
  • 名称 [VARCHAR2(200)]

元素：

  • 编号 [NUMBER(9,0), PK]
  • LocationId [NUMBER(9,0), FK]
  • ParentId [NUMBER(9,0), FK]
  • 类型 ID [NUMBER(9,0), FK]
  • 名称 [VARCHAR2(200)]

现在假设表包含以下数据，例如：

地点：

Id   | ParentId | Name
----------------------------------
100  |   null   | TopLocation
101  |   100    | Level1Location
102  |   101    | Level2Location

元素：

Id | LocationId | ParentId | TypeId | Name
----------------------------------------------------
1  |    102     |   null   |    10  | TopParent
2  |   null     |     1    |    11  | Level1Child
3  |   null     |     2    |    11  | Level2Child

我要做的是为元素编写查询，除了基本的 4 个元素列之外，它还返回父 ID、名称和类型 ID 的完整路径 + 顶部元素 位置ID 和名称的完整路径。因此，如果我获取ID 为 3 的元素（此条件也可能因此处未指定的多列而变得复杂）查询将不得不返回：

Id | ParentId | TypeId | Name        | IdsPath | TypeIdsPath | NamesPath                           | LocIdsPath   | LocNamesPath
---------------------------------------------------------------------------------------------------------------------------------------------------------------
3  |    2     |    11  | Level2Child | /3/2/1  |   /11/11/10 |  /Level2Child/Level1Child/TopParent | /102/101/100 | /Level2Location/Level1Location/TopLocation

首先，我编写了oracle 分层查询，它返回位置和元素的所需路径

地点

select
    SYS_CONNECT_BY_PATH(Id, '/') IdsPath,
    SYS_CONNECT_BY_PATH(Name, '/') NamesPath
from 
    loc
where
     connect_by_isleaf = 1
CONNECT BY PRIOR ParentId = e.Id
start with Id = 102

元素

select
    SYS_CONNECT_BY_PATH(Id, '/') IdsPath,
    SYS_CONNECT_BY_PATH(TypeId, '/') TypeIdsPath,
    SYS_CONNECT_BY_PATH(Name, '/') NamesPath
from 
    ele
where
     connect_by_isleaf = 1
CONNECT BY PRIOR ParentId = e.Id
start with Id = 3

当我想将这些查询用作加入基本选择的子查询时，问题就开始了 - 不能用连接条件替换start with condition，因为分层查询比全表扫描：

select
    e.*,
    elePath.IdsPath,
    elePath.TypeIdsPath,
    elePath.NamesPath,
    locPath.IdsPath as LocIdsPath,
    locPath.NamesPath as LocNamesPath
from
    ele e
    left join (
        --full table scan!
        select
            CONNECT_BY_ROOT(Id) Id,
            Id as TopEleId,
            SYS_CONNECT_BY_PATH(Id, '/') IdsPath,
            SYS_CONNECT_BY_PATH(TypeId, '/') TypeIdsPath,
            SYS_CONNECT_BY_PATH(Name, '/') NamesPath
        from ele
        where
             connect_by_isleaf = 1
        CONNECT BY PRIOR ParentId = e.Id
    ) elePath on elePath.Id = e.Id
    left join (
        --full table scan!
        select
            CONNECT_BY_ROOT(Id) Id,
            SYS_CONNECT_BY_PATH(Id, '/') IdsPath,
            SYS_CONNECT_BY_PATH(Name, '/') NamesPath
        from loc
        where
             connect_by_isleaf = 1
        CONNECT BY PRIOR ParentId = e.Id
    ) locPath on locPath.Id = elePath.TopEleId
where
    e.Id = 3

我也不能做标量子查询，因为查询必须返回多个路径，而不仅仅是一个。有什么建议么？我是在朝着正确的方向前进，还是应该在元素表中添加一些字段并缓存我需要的所有路径？（它们不会经常更新）

谢谢！

4

1 回答 1

3

您反向遍历层次结构，只需使用connect_by_root()运算符获取根行的列值。

clear screen;
column IdPath format a11;
column TypeIdPathformat a11
column NamePath format a35;

with t1(id1, parent_id, type_id, Name1) as(
  select 1, null, 10, 'TopParent' from dual union all
  select 2, 1   , 11, 'Level1Child' from dual union all
  select 3, 2   , 11, 'Level2Child' from dual
)
select connect_by_root(id1)                as id1
     , connect_by_root(parent_id)          as ParentId
     , connect_by_root(type_id)            as Typeid
     , connect_by_root(name1)              as name1
     , sys_connect_by_path(id1, '/')       as IdPath
     , sys_connect_by_path(type_id, '/')   as TypeIdPath
     , sys_connect_by_path(name1, '/')     as NamePath 
 from t1
where connect_by_isleaf = 1
start with id1 = 3
connect by id1 = prior parent_id

结果：

 id1 ParentId TypeId  Name1        IdPath  TypeIdPath NamePath                         
 ---------------------------------------------------------------------------       
 3        2      11   Level2Child /3/2/1  /11/11/10  /Level2Child/Level1Child/TopParent

编辑#1

获得所需输出的一种方法是使用标量子查询：

with Locations(Id1, ParentId, Name1) as(
  select 100,  null, 'TopLocation' from dual union all
  select 101,  100 , 'Level1Location' from dual union all
  select 102,  101 , 'Level2Location' from dual
),
elements(id1, LocationId, parent_id, type_id, Name1) as(
  select 1, 102,  null, 10, 'TopParent' from dual union all
  select 2, null, 1   , 11, 'Level1Child' from dual union all
  select 3, null, 2   , 11, 'Level2Child' from dual
)
select e.*
     , (select sys_connect_by_path(l.id1, '/')
          from locations l
          where connect_by_isleaf = 1
          start with l.id1 = e.locationid
          connect by l.id1 = prior parentid)        as LocIdPath
     , (select sys_connect_by_path(l.name1, '/')
          from locations l
          where connect_by_isleaf = 1
          start with l.id1 = e.locationid
          connect by l.id1 = prior parentid)        as LocNamePath
  from ( select connect_by_root(id1)                as id1
              , connect_by_root(parent_id)          as ParentId
              , connect_by_root(type_id)            as Typeid
              , connect_by_root(name1)              as name1
              , sys_connect_by_path(id1, '/')       as IdPath
              , sys_connect_by_path(type_id, '/')   as TypeIdPath
              , sys_connect_by_path(name1, '/')     as NamePath  
              , locationid
          from elements
         where connect_by_isleaf = 1
         start with id1 = 3
       connect by id1 = prior parent_id ) e

结果：

ID1   PARENTID     TYPEID NAME1       IDPATH      TYPEIDPATH  NAMEPATH                            LOCATIONID LOCIDPATH     LOCNAMEPATH                               
---------- ---------- ----------- ----------- ----------- ----------------------------------- ---------- ------------- -------------------------------------------
  3          2         11 Level2Child /3/2/1      /11/11/10   /Level2Child/Level1Child/TopParent         102 /102/101/100  /Level2Location/Level1Location/TopLocation
于 2014-03-02T12:17:55.277 回答