我正在开发一个包含树型导航菜单的DotNetNuke模块。
到目前为止，我有它的工作，从某种意义上说，子节点连接到正确的父节点，但节点兄弟节点仍然无序。有一个名为TabOrder的字段，用于确定兄弟姐妹的顺序，但由于递归，我无法正确排序。
我正在尝试在SQL Server存储过程中执行此操作，这可能是一个错误，但我觉得我非常接近，必须有一个解决方案。有谁知道我做错了什么？
我很感激你有任何想法。提前谢谢。
解决方案：
我终于找到了解决问题的方法。关键是从Root选项卡到Leaf选项卡递归创建Tab Lineage（TabLevel + TabOrder）。一旦创建，我就能正确地订购返回的记录。
然而，当我回来发帖时，我看到了MarkXA的答案，这可能是最好的解决方案。我不知道方法GetNavigationNodes甚至存在。
我认为使用GetNavigationNodes是一种更具前瞻性的解决方案是正确的，但目前我将使用基于SQL的解决方案。 - 我能说什么？我学到了很多困难。
这是：
ALTER procedure [dbo].[Nav_GetTabs]
@CurrentTabID int = 0
AS
--============================================================
--create and populate @TabLineage table variable with Tab Lineage
--
--"Lineage" consists of the concatenation of TabLevel & TabOrder, concatenated recursively from the root to leaf.
--The lineage is VERY important, making it possible to properly order the Tab links in the navigation module.
--This will be used as a lookup table to match Tabs with their lineage.
--============================================================
DECLARE @TabLineage table
(
TabID int,
Lineage varchar(100)
);
WITH TabLineage AS
(
--start with root Tabs
SELECT T.TabID, T.ParentID, CAST(REPLICATE('0', 5 - LEN(CAST(T2.[Level] as varchar(10)) + CAST(T2.TabOrder as varchar(10)))) + CAST(T2.[Level] as varchar(10)) + CAST(T2.TabOrder as varchar(10)) as varchar(100)) AS Lineage
FROM Tabs T
INNER JOIN Tabs T2 ON T.TabID = T2.TabID
INNER JOIN TabPermission TP ON T.TabID = TP.TabID
WHERE T.ParentID IS NULL
AND T.IsDeleted = 0
AND T.IsVisible = 1
AND TP.RoleID = -1
UNION ALL
--continue recursively, from parent to child Tabs
SELECT T.TabID, T.ParentID, CAST(TL.Lineage + REPLICATE('0', 5 - LEN(CAST(T2.[Level] as varchar(10)) + CAST(T2.TabOrder as varchar(10)))) + CAST(T2.[Level] as varchar(10)) + CAST(T2.TabOrder as varchar(10)) as varchar(100)) AS Lineage
FROM Tabs T
INNER JOIN Tabs T2 ON T.TabID = T2.TabID
INNER JOIN TabPermission TP ON T.TabID = TP.TabID
INNER JOIN TabLineage TL ON T.ParentID = TL.TabID
WHERE T.IsDeleted = 0
AND T.IsVisible = 1
AND TP.RoleID = -1
)
--insert results of recursive query into temporary table
INSERT @TabLineage
SELECT TL.TabID, TL.Lineage FROM TabLineage TL ORDER BY TL.Lineage
OPTION (maxrecursion 10); --to increase number of traversed generations, increase "maxrecursion"
--============================================================
--create and populate @Ancestor table variable with @CurrentTab ancestors
--
--"Ancestors" are Tabs following the path from @CurrentTab to the root Tab it's descended from (inclusively).
--These are Tab links we want to see in the navigation.
--============================================================
DECLARE @Ancestor table
(
TabID int
);
WITH Ancestor AS
(
--start with @CurrentTab
SELECT T.TabID, T.ParentID FROM Tabs T WHERE T.TabID = @CurrentTabID
UNION ALL
--continue recursively, from child to parent Tab
SELECT T.TabID, T.ParentID
FROM Ancestor A INNER JOIN Tabs T ON T.TabID = A.ParentID
)
--insert results of recursive query into temporary table
INSERT @Ancestor
SELECT A.TabID FROM Ancestor A
OPTION (maxrecursion 10); --to increase number of traversed generations, increase "maxrecursion"
--============================================================
--retrieve Tabs to display in navigation
--This section UNIONs three query results together, giving us what we want:
-- 1. All Tabs at Level 0.
-- 2. All Tabs in @CurrentTab's lineage.
-- 3. All Tabs which are children of Tabs in @CurrentTab's lineage.
--============================================================
WITH TabNav (TabID, TabLevel, TabName, Lineage) AS
(
--retrieve all Tabs at Level 0 -- (Root Tabs)
(SELECT T.TabID, T.[Level] AS TabLevel, T.TabName, TL.Lineage
FROM Tabs T
INNER JOIN TabPermission TP ON (T.TabID = TP.TabID AND TP.RoleID = -1)
INNER JOIN @TabLineage TL ON T.TabID = TL.TabID
WHERE T.IsDeleted = 0
AND T.IsVisible = 1
AND T.[Level] = 0
UNION
--retrieve Tabs in @CurrentTab's lineage
SELECT T.TabID, T.[Level] AS TabLevel, T.TabName, TL.Lineage
FROM Tabs T
INNER JOIN TabPermission TP ON (T.TabID = TP.TabID AND TP.RoleID = -1)
INNER JOIN @Ancestor A ON T.TabID = A.TabID
INNER JOIN @TabLineage TL ON T.TabID = TL.TabID
WHERE T.IsDeleted = 0
AND T.IsVisible = 1
UNION
--retrieve Tabs which are children of Tabs in @CurrentTab's lineage
SELECT T.TabID, T.[Level] AS TabLevel, T.TabName, TL.Lineage
FROM Tabs T
INNER JOIN TabPermission TP ON (T.TabID = TP.TabID AND TP.RoleID = -1)
INNER JOIN @Ancestor A ON T.ParentID = A.TabID
INNER JOIN @TabLineage TL ON T.TabID = TL.TabID
WHERE T.IsDeleted = 0
AND T.IsVisible = 1)
)
--finally, return the Tabs to be included in the navigation module
SELECT TabID, TabLevel, TabName FROM TabNav ORDER BY Lineage;
--============================================================
答案 0 :(得分：2)
答案是“不要使用SQL”。已经有一个方法DotNetNuke.UI.Navigation.GetNavigationNodes为您执行此操作，如果您使用它，那么如果数据库架构发生更改，您的模块将不会中断。即使你需要做一些GetNavigationNodes无法处理的事情，你最好还是通过API检索页面以防止未来发生。直接进入数据库只是在寻找麻烦：）
答案 1 :(得分：1)
这是一个锅炉板（不是基于给定的OP代码）递归树CTE的示例，它显示了如何对树进行排序：
DECLARE @Contacts table (id varchar(6), first_name varchar(10), reports_to_id varchar(6))
INSERT @Contacts VALUES ('1','Jerome', NULL ) -- tree is as follows:
INSERT @Contacts VALUES ('2','Joe' ,'1') -- 1-Jerome
INSERT @Contacts VALUES ('3','Paul' ,'2') -- / \
INSERT @Contacts VALUES ('4','Jack' ,'3') -- 2-Joe 9-Bill
INSERT @Contacts VALUES ('5','Daniel','3') -- / \ \
INSERT @Contacts VALUES ('6','David' ,'2') -- 3-Paul 6-David 10-Sam
INSERT @Contacts VALUES ('7','Ian' ,'6') -- / \ / \
INSERT @Contacts VALUES ('8','Helen' ,'6') -- 4-Jack 5-Daniel 7-Ian 8-Helen
INSERT @Contacts VALUES ('9','Bill ' ,'1') --
INSERT @Contacts VALUES ('10','Sam' ,'9') --
DECLARE @Root_id varchar(6)
--get all nodes 2 and below
SET @Root_id=2
PRINT '@Root_id='+COALESCE(''''+@Root_id+'''','null')
;WITH StaffTree AS
(
SELECT
c.id, c.first_name, c.reports_to_id, c.reports_to_id as Manager_id, cc.first_name AS Manager_first_name, 1 AS LevelOf
FROM @Contacts c
LEFT OUTER JOIN @Contacts cc ON c.reports_to_id=cc.id
WHERE c.id=@Root_id OR (@Root_id IS NULL AND c.reports_to_id IS NULL)
UNION ALL
SELECT
s.id, s.first_name, s.reports_to_id, t.id, t.first_name, t.LevelOf+1
FROM StaffTree t
INNER JOIN @Contacts s ON t.id=s.reports_to_id
WHERE s.reports_to_id=@Root_id OR @Root_id IS NULL OR t.LevelOf>1
)
SELECT * FROM StaffTree ORDER BY LevelOf, first_name
输出：
@Root_id='2'
id first_name reports_to_id Manager_id Manager_first_name LevelOf
------ ---------- ------------- ---------- ------------------ -----------
2 Joe 1 1 Jerome 1
6 David 2 2 Joe 2
3 Paul 2 2 Joe 2
5 Daniel 3 3 Paul 3
8 Helen 6 6 David 3
7 Ian 6 6 David 3
4 Jack 3 3 Paul 3
(7 row(s) affected)
关键是LevelOf列。在CTE中选择主要父级时，请查看它是如何只是文字1。然后LevelOf列在递归CTE的UNION ALL部分递增。每次到CTE的递归调用（不是行）都将达到UNION ALL一次和增量。它不是那么多。