Question

我有一个文件夹树结构，可以扩展到三个层次。所有级别都存储Excel工作表。

为此，我有两个不同的数据库表，第一个存储Excel工作表名称和ID以及文件夹ID，而第二个表存储所有具有各自名称，ID和父文件夹ID的文件夹。

我想显示每个Excel工作表名称及其文件夹，父文件夹和父父文件夹的名称，即使为空。

我尝试用左外部自我连接来做到这一点，但失败了。

表1

DocID      DocName  FolderId 
12345      Abc.xlx  98765
12346      rst.xlx  98764
123457     jkl.xlx  98763

表2

FolderID   FolderName  ParentFolderId 
98765      lmn         98764
98764      pqr         98763
98763      dcg         null

所需结果：

DocName  ChildFolder   Parentfolder   Grandparentfolder
abc      lmn           pqr            dcg 
rst      pqr           dcg            null
jkl      dcg           null           null

Answer 1

如果只需要三个文件夹，则可以像下面这样保留三遍联接数据：

-- data
with 
  sheets(DocID, DocName, FolderId) as (
    select 12345, 'Abc.xlx',  98765 from dual union all
    select 12346, 'rst.xlx',  98764 from dual union all
    select 123457, 'jkl.xlx', 98763 from dual ),
  folders(FolderID, FolderName, ParentFolderId) as (
    select 98765, 'lmn', 98764 from dual union all
    select 98764, 'pqr', 98763 from dual union all
    select 98763, 'dcg', null  from dual)
-- end of data

select docname, f1.foldername child, f2.foldername parent, f3.foldername grand 
  from sheets s
  left join folders f1 on s.folderid = f1.folderid
  left join folders f2 on f2.folderid = f1.parentfolderid
  left join folders f3 on f3.folderid = f2.parentfolderid

结果：

DOCNAME CHILD PARENT GRAND
------- ----- ------ -----
Abc.xlx lmn   pqr    dcg
jkl.xlx dcg          
rst.xlx pqr   dcg

您还可以使用递归查询，尤其是在嵌套文件夹更多的情况下。

Answer 2

@PonderStibbons展示了如何使用左外部联接进行此操作。您还可以使用分层查询并调整结果：

with cte (FolderName, BaseFolderId, lvl) as (
  select FolderName, connect_by_root(FolderId), level
  from table2
  connect by FolderId = prior ParentFolderId
)
select *
from (
  select t1.DocName,
    c.FolderName,
    c.lvl
  from table1 t1
  join cte c on c.BaseFolderId = t1.FolderId
)
pivot (
  max(FolderName) for lvl in (1 as ChildFolder, 2 as ParentFolder, 3 as GrandparentFolder)
);

DOCNAME CHILDFOLDER PARENTFOLDER GRANDPARENTFOLDER
------- ----------- ------------ -----------------
jkl.xlx dcg                                       
Abc.xlx lmn         pqr          dcg              
rst.xlx pqr         dcg

或使用递归子查询分解而不是层次查询：

with rcte (FolderName, ParentFolderId, BaseFolderId, lvl) as (
  select FolderName, ParentFolderId, FolderId, 1 from table2
  union all
  select t2.FolderName, t2.ParentFolderId, r.BaseFolderId, r.lvl + 1
  from rcte r
  join table2 t2 on t2.FolderId = r.ParentFolderId
)
select *
from (
  select t1.DocName,
    r.FolderName,
    r.lvl
  from table1 t1
  join rcte r on r.BaseFolderId = t1.FolderId
)
pivot (
  max(FolderName) for lvl in (1 as ChildFolder, 2 as ParentFolder, 3 as GrandparentFolder)
);

DOCNAME CHILDFOLDER PARENTFOLDER GRANDPARENTFOLDER
------- ----------- ------------ -----------------
jkl.xlx dcg                                       
Abc.xlx lmn         pqr          dcg              
rst.xlx pqr         dcg

db<>fiddle

随着您添加更多级别的文件夹，这种方法可能会更好；但是它们都必须在数据透视子句中定义，因此它仍然不完全灵活。

如果您还需要文件夹ID，则将其包含在层次结构/ CTE中，并同时旋转两者：

with rcte (FolderName, FolderId, ParentFolderId, BaseFolderId, lvl) as (
  select FolderName, FolderId, ParentFolderId, FolderId, 1 from table2
  union all
  select t2.FolderName, t2.FolderId, t2.ParentFolderId, r.BaseFolderId, r.lvl + 1
  from rcte r
  join table2 t2 on t2.FolderId = r.ParentFolderId
)
select *
from (
  select t1.DocName,
    r.FolderId,
    r.FolderName,
    r.lvl
  from table1 t1
  join rcte r on r.BaseFolderId = t1.FolderId
)
pivot (
  max(FolderName), max(FolderId) as id
  for lvl in (1 as ChildFolder, 2 as ParentFolder, 3 as GrandparentFolder)
);

DOCNAME CHILDFOLDER CHILDFOLDER_ID PARENTFOLDER PARENTFOLDER_ID GRANDPARENTFOLDER GRANDPARENTFOLDER_ID
------- ----------- -------------- ------------ --------------- ----------------- --------------------
jkl.xlx dcg                  98763                                                                    
Abc.xlx lmn                  98765 pqr                    98764 dcg                              98763
rst.xlx pqr                  98764 dcg                    98763

Updated db<>fiddle

连续获取层次结构中的每个父级

2 个答案: