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用SQL CTE打印树

  •  8
  • leppie  · 技术社区  · 14 年前

    方案如下:

    CREATE TABLE [Structure](
        [StructureId] [uniqueidentifier] NOT NULL,
        [SequenceNumber] [int] NOT NULL, -- order for siblings, unique per parent
        [ParentStructureId] [uniqueidentifier] NULL,
     CONSTRAINT [Structure_PK] PRIMARY KEY CLUSTERED 
    (
        [StructureId] ASC
    )
    ) ON [PRIMARY]
    
    ALTER TABLE [Structure]  WITH CHECK ADD  CONSTRAINT [Structure_FK1] 
    FOREIGN KEY([ParentStructureId])
    REFERENCES [Structure] ([StructureId])
    

    目前,我可以用下面的CTE获得所有的逻辑数据,但我想直接以深度优先的方式打印它。

    WITH SCTE (StructureId, Level, Seq, ParentId)
    AS
    (
      SELECT StructureId,  0, SequenceNumber, [ParentStructureId]
        FROM Structure
        WHERE [ParentStructureId] IS NULL 
              AND StructureId = 'F6C5F016-1270-47C1-972F-349C32DFC92A'
    
      UNION ALL
    
      SELECT Structure.StructureId, Level + 1, SequenceNumber, ParentStructureId
      FROM Structure
      INNER JOIN SCTE ON SCTE.StructureId = Structure.ParentStructureId
    )
    
    SELECT * FROM SCTE
    ORDER BY Level, ParentId, Seq
    

    输出如下(此处截断):

    StructureId                     Level   Seq ParentId
    F6C5F016-1270-47C1-972F-349C32DFC92A    0   0   NULL
    D2E34429-401A-4A49-9E18-E81CCA0FB417    1   0   F6C5F016-1270-47C1-972F-349C32DFC92A
    0CC5E16C-9194-40CA-9F72-1CED2972D7CA    1   1   F6C5F016-1270-47C1-972F-349C32DFC92A
    1ECD1D30-EB85-42B0-969F-75794343E3B4    1   2   F6C5F016-1270-47C1-972F-349C32DFC92A
    EEC3A981-B790-4600-8CD1-F15972CD9230    2   0   0CC5E16C-9194-40CA-9F72-1CED2972D7CA
    4406F639-2F58-4918-A9EF-A4B0F379BEA0    2   1   0CC5E16C-9194-40CA-9F72-1CED2972D7CA
    FCAF7870-C606-4AA6-85EE-57B90B1B0CC3    2   2   0CC5E16C-9194-40CA-9F72-1CED2972D7CA
    855DF5FB-1593-4E5B-8EF9-3770B45F89D6    2   3   0CC5E16C-9194-40CA-9F72-1CED2972D7CA
    3D16DF32-C04F-49B4-B0D9-5BDC9104F810    2   4   0CC5E16C-9194-40CA-9F72-1CED2972D7CA
    A1084D00-0198-47D9-87E0-BB8234233F14    2   5   0CC5E16C-9194-40CA-9F72-1CED2972D7CA
    CE443C0D-376F-46EC-9914-32C6B7200DB1    2   6   0CC5E16C-9194-40CA-9F72-1CED2972D7CA
    0DEA587D-4FCF-414C-AD71-FB00829F8082    2   7   0CC5E16C-9194-40CA-9F72-1CED2972D7CA
    CC9FC8D3-254A-486B-8DC4-07E57627476C    2   0   1ECD1D30-EB85-42B0-969F-75794343E3B4
    215565CC-501F-4850-B8AE-5466DA5E6854    2   1   1ECD1D30-EB85-42B0-969F-75794343E3B4
    D4E6C8E5-5ADD-4AD1-B59B-1A672F66888A    2   2   1ECD1D30-EB85-42B0-969F-75794343E3B4
    796C65BF-4714-4DBF-A97A-2150DBE3098C    2   3   1ECD1D30-EB85-42B0-969F-75794343E3B4
    B39DEB9C-BE42-43B4-9C38-968399D7D1E2    2   4   1ECD1D30-EB85-42B0-969F-75794343E3B4
    6C2F70C6-1DA0-4E1A-BBC1-D7FCAFE6AFEE    2   0   D2E34429-401A-4A49-9E18-E81CCA0FB417
    75D7B43B-C971-46B4-BC42-58C3605ADD79    2   1   D2E34429-401A-4A49-9E18-E81CCA0FB417
    0B5AAAA0-A69F-431E-86BA-148444D7B1E6    2   2   D2E34429-401A-4A49-9E18-E81CCA0FB417
    CB3CF66B-D83A-45E2-953A-6F0CEE094F5B    2   3   D2E34429-401A-4A49-9E18-E81CCA0FB417
    1D5F69C3-F036-4667-BD75-A0DC1506DB6D    2   4   D2E34429-401A-4A49-9E18-E81CCA0FB417
    71B894F7-B9FC-44DE-AEDB-E6FA026A6082    2   5   D2E34429-401A-4A49-9E18-E81CCA0FB417
    F1DFA1E1-013B-449C-9D9D-14C64E75D418    2   6   D2E34429-401A-4A49-9E18-E81CCA0FB417
    

    如你所见,结果是“宽度优先”,这使得打印一棵树有点像现在一样不可能。

    是否有任何方法(可能有一种简单的方法,但我的SQL技能非常差)以“树打印友好”格式获得结果列表?

    我知道我可以将结果转储到一个程序中并对输出进行编码,但是作为一个练习,我更喜欢在SQL本身中这样做。

    谢谢

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  •  11
  •   Andomar    14 年前

    declare @t table (id int, parent int)
    insert @t (id, parent) values (1, null), (2,1), (3,2), (4,3), (5,null), (6,5)
    
    ; with cte as (
        select  id, parent
        ,       cast(RIGHT(REPLICATE('0',12) + 
                     CONVERT(varchar(12),id),12) as varchar(max)) Path
        from    @t
        where   parent is null
        union all
        select  child.id, child.parent
        ,       parent.Path + RIGHT(REPLICATE('0',12) + 
                                    CONVERT(varchar(12),child.id),12) as Path
        from    @t child
        join    cte parent
        on      parent.id = child.parent
    )
    select  *
    from    cte
    order by
            Path
    

    char(x)