日期:2014-05-18 浏览次数:20529 次
--> 测试数据:[tb] IF OBJECT_ID('[tb]') IS NOT NULL DROP TABLE [tb] GO CREATE TABLE [tb] ( [ID] INT , [父ID] INT , [IP] VARCHAR(9) ) INSERT [tb] SELECT 1, 0, '127.0.0.1' UNION ALL SELECT 2, 0, '127.0.0.2' UNION ALL SELECT 3, 0, '127.0.0.5' UNION ALL SELECT 4, 1, NULL UNION ALL SELECT 5, 1, NULL UNION ALL SELECT 6, 2, NULL UNION ALL SELECT 7, 3, NULL UNION ALL SELECT 8, 7, NULL UNION ALL SELECT 9, 4, NULL --------------开始查询-------------------------- ; WITH t AS ( SELECT * FROM [tb] WHERE [父ID] = 0 UNION ALL SELECT b.id, b.[父ID], t.[IP] FROM t , [tb] b WHERE t.[ID] = b.[父ID] ) SELECT * FROM t ----------------结果---------------------------- /* ID 父ID IP ----------- ----------- --------- 1 0 127.0.0.1 2 0 127.0.0.2 3 0 127.0.0.5 7 3 127.0.0.5 8 7 127.0.0.5 6 2 127.0.0.2 4 1 127.0.0.1 5 1 127.0.0.1 9 4 127.0.0.1 (9 行受影响) */
------解决方案--------------------
--> 测试数据:[test] if object_id('[test]') is not null drop table [test] create table [test]([ID] int,[FID] int,[IP] varchar(9)) insert [test] select 1,0,'127.0.0.1' union all select 2,0,'127.0.0.2' union all select 3,0,'127.0.0.5' union all select 4,1,null union all select 5,1,null union all select 6,2,null union all select 7,3,null union all select 8,7,null union all select 9,4,null with t as( select * from test where [FID]=0 union all select a.[ID],a.[FID],b.[IP] from test a inner join t b on a.FID=b.ID ) select * from t order by 1 /* ID FID IP -------------------------------------- 1 0 127.0.0.1 2 0 127.0.0.2 3 0 127.0.0.5 4 1 127.0.0.1 5 1 127.0.0.1 6 2 127.0.0.2 7 3 127.0.0.5 8 7 127.0.0.5 9 4 127.0.0.1 */