SQL树结构表

Question

我是SQL新手，想对以下问题提供任何帮助。我有以下代码为节点和树创建表，但不知道如何在我的代码中实现以下规则。

create table Node(
    NodeId int not null, 
    ParentId int null, 
    NodeName varchar(255) not null, 
    constraint PK_Node primary key(NodeId), 
    constraint UK_NodeName unique(NodeName) 
) 
go 

create table Tree(
    NodeId int not null, 
    ParentId int not null, 
    Level int not null, 
    constraint PK_Tree primary key(NodeId, ParentId), 
    constraint UK_Level unique(NodeId, Level) 
) 
go} 

alter table Node 
 
\t add constraint FK_NodeNode foreign key(ParentId) references Node(NodeId) --on delete cascade; 
 

 
alter table Tree 
 
\t add constraint FK_NodeTreeNode foreign key(NodeId) references Node(NodeId) on delete cascade;

设想一个树结构，具有以下性质：

• A）单一根节点
• b）每个节点可以具有多个子节点
• c）节点可能不是d e）所有非根节点都有一个属性，最多的子节点 - 新节点默认为固定默认值
• e）最大子节点属性可随时更改 - 如果新值较小比当前的儿童节点数量多，孩子们分布在沿途所有节点的财产链中。如果该属性设置为零，则该节点成为根节点的直接子节点。
• f）当添加一个新节点时，它将被添加到尽可能接近根节点的节点，按照儿童百分比配额排序递增。如果所有节点都满足其配额，则直接放置在根节点下。

创建表和索引来存储这些信息和程序，以尽可能有效地执行以下操作：

• 我。将节点添加到树
• ii。更改节点的最大子节点属性

Answer 1

并非所有约束都可以在SQL Server中设置。 a，d，e，f的逻辑应该在应用程序中迎合。

树只是Nodes的组合。所以我们只需要定义Node的存储。

create table Node(
    NodeId int not null identity(1,1), -- Identity, auto generate Id for new Node 
    ParentId int null, -- If null, it is the root of the tree 
    MaxChild int not null default(-1), 
    NodeName varchar(255) not null, 
    constraint PK_Node primary key(NodeId), 
    constraint UK_NodeName unique(NodeName), 
    constraint FK_Node foreign key(ParentId) references Node(NodeId) 
    -- How the `Node` to be related and form the tree can be done 
    -- by adding the `Foreign Key` of `Node` itself 
    -- A node cannot be simply deleted if it is a parent. 
) 

添加节点：

-- Add root 
insert Node (NodeName) VALUES ('root') 

-- Add child 
insert Node (ParentId, NodeName) VALUES 
((SELECT NodeId FROM Node WHERE NodeName = 'Root'), 'ChildA') 

insert Node (ParentId, NodeName) VALUES 
((SELECT NodeId FROM Node WHERE NodeName = 'Root'), 'ChildB') 

insert Node (ParentId, NodeName) VALUES 
((SELECT NodeId FROM Node WHERE NodeName = 'ChildA'), 'ChildA-A') 

insert Node (ParentId, NodeName) VALUES 
((SELECT NodeId FROM Node WHERE NodeName = 'ChildA-A'), 'ChildA-A-A') 

为了得到Nodes和儿童的水平算

;WITH 
CountLevel AS 
(
    SELECT *, 1 AS Level FROM Node WHERE ParentId IS NULL 

    UNION ALL 

    SELECT child.*, parent.Level + 1 
    FROM Node child INNER JOIN CountLevel parent ON 
     child.ParentId = parent.NodeId 
), 
CountChild AS 
(
    SELECT NodeId, ParentId, 0 AS ChildCount 
    FROM Node leaf 
    WHERE NOT EXISTS(SELECT * FROM Node WHERE ParentId = leaf.NodeId) 

    UNION ALL 

    SELECT child.ParentId, (SELECT ParentId FROM Node WHERE NodeId = child.ParentId), child.ChildCount + 1 
    FROM CountChild child 
    WHERE child.ParentId IS NOT NULL 
) 
SELECT *, (SELECT SUM(ChildCount) FROM CountChild WHERE NodeId = CountLevel.NodeId) Child FROM CountLevel 

结果

NodeId  ParentId MaxChild NodeName    Level  Child 
----------- ----------- ----------- -------------------- ----------- ----------- 
1   NULL  -1   Root     1   4 
2   1   -1   ChildA    2   2 
3   1   -1   ChildB    2   0 
4   2   -1   ChildA-A    3   1 
5   4   -1   ChildA-A-A   4   0