0

次のスクリプトを使用してテーブルを作成してください。を使用していますが、ファイル Entity Framework 3.5を作成するとテーブルに列が表示されません...なぜですか?? このクエリを実行できないため、エラーが発生しますedmxParentIDNodesDim categories = From c In context.Nodes Where (c.ParentID = "1")ParentID is not a member of HagglerModel.Nodes

     if exists(select name from sysobjects where name = 'NodeInsert' and type = 'tr')
       drop trigger NodeInsert
    go

    if exists(select name from sysobjects where name = 'NodeUpdate' and type = 'tr')
       drop trigger NodeUpdate
    go

    if exists(select name from sysobjects where type = 'u' and name = 'Tree')
        Drop table Tree
    go

    if exists(select name from sysobjects where type = 'u' and name = 'Node')
        Drop table Node
    go

    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
        add constraint FK_NodeNode foreign key(ParentId) references Node(NodeId) --on delete cascade
    go

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

    --alter table Tree
    --  add constraint FK_NodeTreeParent foreign key(ParentId) references Node(NodeId) on delete cascade
    --go

    create trigger NodeInsert on Node for insert as
    begin
        set nocount on

        insert into Tree(NodeId, ParentId, Level)
        select NodeId, NodeId, 0
        from inserted

        insert into Tree(NodeId, ParentId, Level)
        select n.NodeId, t.ParentId, t.Level + 1
        from inserted n, Tree t
        where n.ParentId = t.NodeId
    end
    go

    create trigger NodeUpdate on Node for update as
    if update(ParentId)
    begin
        set nocount on

        declare @child table(NodeId int, Level int)

        insert into @child(NodeId, Level)
        select t.NodeId, t.Level
        from inserted n, Tree t
        where n.NodeId = t.ParentId and t.Level > 0

        delete Tree
        where
            Tree.NodeId in(select NodeId from @child)
            and Tree.ParentId in(
                select t.ParentId
                from inserted n, Tree t
                where n.NodeId = t.NodeId and t.Level > 0
            )

        delete Tree
        where Tree.NodeId in(select NodeId from inserted) and Tree.Level > 0

        insert into Tree(NodeId, ParentId, Level)
        select n.NodeId, t.ParentId, t.Level + 1
        from inserted n, Tree t
        where n.ParentId = t.NodeId

        insert into Tree(NodeId, ParentId, Level)
        select c.NodeId, t.ParentId, t.Level + c.Level
        from inserted n, Tree t, @child c
        where n.NodeId = t.NodeId and t.Level > 0
    end
    go

    insert into Node(NodeId, ParentId, NodeName) values(1, null, 'A')
    insert into Node(NodeId, ParentId, NodeName) values(2, 1, 'B')
    insert into Node(NodeId, ParentId, NodeName) values(3, 1, 'C')
    insert into Node(NodeId, ParentId, NodeName) values(4, 2, 'D')
    insert into Node(NodeId, ParentId, NodeName) values(5, 4, 'E')
    insert into Node(NodeId, ParentId, NodeName) values(6, 4, 'F')
    insert into Node(NodeId, ParentId, NodeName) values(7, 6, 'G')
    select * from Node
    select * from Tree

    --gets all descendants of the Node 2
    select c.*
    from Node n, Tree t, Node c
    where n.NodeName='B'
        and n.NodeId = t.ParentId
        and t.NodeId = c.NodeId

    --gets path to the root from node 7
    select p.*
    from Node n, Tree t, Node p
    where n.NodeName='G'
        and n.NodeId = t.NodeId
        and t.ParentId = p.NodeId

    --changes parent of node 4 from 2 to 1
    update Node set ParentId = 1 where NodeId = 4
    select * from Node
    select * from Tree
4

1 に答える 1

0

あなたの場合、ParentId は外部キーである可能性があると思います。外部キーに直接アクセスすることはできません。

ここにそれを達成する方法に関する投稿があります.. エンティティフレームワーク:外部キープロパティの設定

于 2012-09-10T19:44:03.733 に答える