您好我有一个站点地图表和一个等级表。
USE [EB_NEW]
GO
/****** Object: Table [dbo].[COMMON.SITEMAP] Script Date: 07/24/2013 06:13:51 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
SET ANSI_PADDING ON
GO
CREATE TABLE [dbo].[COMMON.SITEMAP](
[ID] [int] IDENTITY(1,1) NOT NULL,
[P_ID] [int] NULL,
[C_ID] [int] NULL,
[TYPE] [varchar](50) NULL,
[TITLE] [varchar](50) NULL,
[TASK_URL_ID] [bigint] NULL
) ON [PRIMARY]
GO
SET ANSI_PADDING OFF
GO
SET IDENTITY_INSERT [dbo].[COMMON.SITEMAP] ON
INSERT [dbo].[COMMON.SITEMAP] ([ID], [P_ID], [C_ID], [TYPE], [TITLE], [TASK_URL_ID]) VALUES (2, NULL, 2, N'EB', N'Employee Benefit', NULL)
INSERT [dbo].[COMMON.SITEMAP] ([ID], [P_ID], [C_ID], [TYPE], [TITLE], [TASK_URL_ID]) VALUES (3, 2, 3, N'Company', N'Company Listing', 175)
INSERT [dbo].[COMMON.SITEMAP] ([ID], [P_ID], [C_ID], [TYPE], [TITLE], [TASK_URL_ID]) VALUES (4, 3, 4, N'Company', N'Company Profile Setup', 176)
INSERT [dbo].[COMMON.SITEMAP] ([ID], [P_ID], [C_ID], [TYPE], [TITLE], [TASK_URL_ID]) VALUES (5, 4, 5, N'Company', N'Company Policy Setup', 191)
INSERT [dbo].[COMMON.SITEMAP] ([ID], [P_ID], [C_ID], [TYPE], [TITLE], [TASK_URL_ID]) VALUES (6, 4, 6, N'Company', N'Employee Profile Setup', 178)
INSERT [dbo].[COMMON.SITEMAP] ([ID], [P_ID], [C_ID], [TYPE], [TITLE], [TASK_URL_ID]) VALUES (7, 5, 7, N'Company', N'Company Policy Setup 2', NULL)
INSERT [dbo].[COMMON.SITEMAP] ([ID], [P_ID], [C_ID], [TYPE], [TITLE], [TASK_URL_ID]) VALUES (8, 6, 8, N'Company', N'Employee Profile Setup 2', NULL)
SET IDENTITY_INSERT [dbo].[COMMON.SITEMAP] OFF
结果
ID P_ID C_ID TYPE TITLE TASK_URL_ID
2 NULL 2 EB Employee Benefit NULL
3 2 3 Company Company Listing 175
4 3 4 Company Company Profile Setup 176
5 4 5 Company Company Policy Setup 191
6 4 6 Company Employee Profile Setup 178
7 5 7 Company Company Policy Setup 2 NULL
8 6 8 Company Employee Profile Setup 2 NULL
我使用公用表表达式来确定层次结构数据
WITH ctLevel
AS
(
SELECT
C_ID AS Child
,P_ID AS Parent
,1 AS [Level]
,TASK_URL_ID As taskUrl_ID
,TITLE As title
,TYPE As Type
,ID As ID
FROM
[COMMON.SITEMAP] as common_sitemap
WHERE
P_ID is null and common_sitemap.TYPE ='EB'
UNION ALL
SELECT
C_ID AS Child
,P_ID AS Parent
,[Level] + 1 AS [Level]
,TASK_URL_ID As cTaskUrl_ID
,common_sitemap.TITLE As ctitle
,common_sitemap.TYPE As cType
,common_sitemap.ID As cID
FROM
[COMMON.SITEMAP]as common_sitemap
inner JOIN
ctLevel
ON
(
P_ID = Child
)
WHERE
common_sitemap.P_ID is not null
)
SELECT DISTINCT [Parent] ,[Child], ctLevel.title,ctLevel.Type,
ctLevel.taskUrl_ID,
ctLevel.Level,
common_task_url.TASK_URL
FROM ctLevel
left join
[COMMON.TASK_URL]as common_task_url
on
ctLevel.taskUrl_ID = common_task_url.TASK_URL_ID
order by Level
结果
Parent Child title Type taskUrl_ID Level TASK_URL
----------- ----------- -------------------------------------------------- -------------------------------------------------- -------------------- ----------- -------------------
NULL 2 Employee Benefit EB NULL 1 NULL
2 3 Company Listing Company 175 2 ~/Module/EB/Company/CompanyList.aspx
3 4 Company Profile Setup Company 176 3 ~/Module/EB/Company/CompanyDetail.aspx
4 5 Company Policy Setup Company 191 4 ~/Module/EB/Company/CompanyPolicy.aspx
4 6 Employee Profile Setup Company 178 4 ~/Module/EB/Employee/EmployeeDetail.aspx
5 7 Company Policy Setup 2 Company NULL 5 NULL
6 8 Employee Profile Setup 2 Company
所以我的问题是，是否有可能从最后一个子记录中获取整个相关的层次结构记录。例如 儿童记录 7
结果会得到
Parent Child title Type taskUrl_ID Level TASK_URL
----------- ----------- -------------------------------------------------- -------------------------------------------------- -------------------- ----------- -----------------------
NULL 2 Employee Benefit EB NULL 1 NULL
2 3 Company Listing Company 175 2 ~/Module/EB/Company/CompanyList.aspx
3 4 Company Profile Setup Company 176 3 ~/Module/EB/Company/CompanyDetail.aspx
4 5 Company Policy Setup Company 191 4 ~/Module/EB/Company/CompanyPolicy.aspx
5 7 Company Policy Setup 2 Company NULL 5 NULL
但如果儿童记录 8
Parent Child title Type taskUrl_ID Level TASK_URL
----------- ----------- -------------------------------------------------- -------------------------------------------------- -------------------- ----------- -----------------------
NULL 2 Employee Benefit EB NULL 1 NULL
2 3 Company Listing Company 175 2 ~/Module/EB/Company/CompanyList.aspx
3 4 Company Profile Setup Company 176 3 ~/Module/EB/Company/CompanyDetail.aspx
4 6 Employee Profile Setup Company 178 4 ~/Module/EB/Employee/EmployeeDetail.aspx
6 8 Employee Profile Setup 2 Company NULL 5 NULL
任何灵魂人物？感谢
答案 0 :(得分：0)
有可能。它与您遍历树的方式非常相似。我做了一个小例子：http://sqlfiddle.com/#!3/842ce/13
declare @childID int = 7;
with tree as (
select parent, id from T where parent is null
union all
select t.parent, t.id from T
join tree on T.parent = tree.id
), bottomUp as (
select parent, id from tree where id = @childID
union all
select t.parent, t.id from tree t
join bottomUp on bottomUp.parent = t.id
)
select * from bottomUp
order by id asc
在这里，我正在穿越我走过的树。但是，如果您不想使用，则不必使用第一个树。在bottomUp中将树更改为T，它仍然有效。如果你已经有修剪树的逻辑，你可以保留它。如果您没有任何逻辑，那么如果您只有第二个没有第一个CTE，则速度会更快。