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实体框架查询检查
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Stack Overflow用户
提问于 2012-09-11 03:33:35
回答 1查看 215关注 0票数 0

请使用以下脚本来创建表。我使用的是Entity Framework 3.5,当我创建edmx文件时,在Nodes table...why中看不到ParentID列。由于这个原因,我不能执行这个查询,我得到了一个错误ParentID is not a member of HagglerModel.Nodes

代码语言:javascript
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     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
EN

回答 1

Stack Overflow用户

发布于 2012-09-11 03:44:04

在你的例子中,我猜ParentId,可能是一个外键。不能直接访问外键。

这里有一篇关于如何实现它的帖子..Entity Framework: Setting a Foreign Key Property

票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/12358269

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