架构如下:
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,我可以通过以下CTE获取所有逻辑数据,但我想以深度优先的方式直接打印它.
WITH SCTE (StructureId,Level,Seq,ParentId) AS ( SELECT StructureId,SequenceNumber,[ParentStructureId] FROM Structure WHERE [ParentStructureId] IS NULL AND StructureId = 'F6C5F016-1270-47C1-972F-349C32DFC92A' UNION ALL SELECT Structure.StructureId,Level + 1,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
正如您所看到的,结果是“广度优先”,这使得打印树像现在一样不可能.
是否有任何方法(可能有一种微不足道的方式,但我的sql技能非常差)以“树形打印友好”格式获得结果列表?
我知道我可以将结果转储到程序中并对输出进行编码,但作为练习,我更喜欢在sql本身中执行此操作.
谢谢
解决方法
评论后编辑.您可以将路径添加到节点,并对其进行排序:
declare @t table (id int,parent int) insert @t (id,parent) values (1,null),(2,1),(3,2),(4,3),(5,(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
这将首先打印根,然后按顺序打印.如果您的id可能大于12位,请增加char(x)强制转换中的数字.