使用关联表中的其他字段首先创建代码,多对多创建代码

hgd*_*ean 282 c# many-to-many entity-framework ef-code-first

我有这种情况:

public class Member
{
    public int MemberID { get; set; }

    public string FirstName { get; set; }
    public string LastName { get; set; }

    public virtual ICollection<Comment> Comments { get; set; }
}

public class Comment
{
    public int CommentID { get; set; }
    public string Message { get; set; }

    public virtual ICollection<Member> Members { get; set; }
}

public class MemberComment
{
    public int MemberID { get; set; }
    public int CommentID { get; set; }
    public int Something { get; set; }
    public string SomethingElse { get; set; }
}
Run Code Online (Sandbox Code Playgroud)

如何使用流畅的API配置我的关联?或者有更好的方法来创建关联表吗?

Sla*_*uma 509

无法与自定义连接表创建多对多关系.在多对多关系中,EF在内部管理连接表并隐藏.它是一个没有模型中的Entity类的表.要使用具有其他属性的此类连接表,您必须实际创建两个一对多关系.它可能看起来像这样:

public class Member
{
    public int MemberID { get; set; }

    public string FirstName { get; set; }
    public string LastName { get; set; }

    public virtual ICollection<MemberComment> MemberComments { get; set; }
}

public class Comment
{
    public int CommentID { get; set; }
    public string Message { get; set; }

    public virtual ICollection<MemberComment> MemberComments { get; set; }
}

public class MemberComment
{
    [Key, Column(Order = 0)]
    public int MemberID { get; set; }
    [Key, Column(Order = 1)]
    public int CommentID { get; set; }

    public virtual Member Member { get; set; }
    public virtual Comment Comment { get; set; }

    public int Something { get; set; }
    public string SomethingElse { get; set; }
}
Run Code Online (Sandbox Code Playgroud)

如果您现在想要查找具有LastName="Smith" 的成员的所有注释,例如您可以编写如下查询:

var commentsOfMembers = context.Members
    .Where(m => m.LastName == "Smith")
    .SelectMany(m => m.MemberComments.Select(mc => mc.Comment))
    .ToList();
Run Code Online (Sandbox Code Playgroud)

...要么...

var commentsOfMembers = context.MemberComments
    .Where(mc => mc.Member.LastName == "Smith")
    .Select(mc => mc.Comment)
    .ToList();
Run Code Online (Sandbox Code Playgroud)

或者要创建一个名为"Smith"的成员列表(我们假设有多个)以及他们的评论,您可以使用投影:

var membersWithComments = context.Members
    .Where(m => m.LastName == "Smith")
    .Select(m => new
    {
        Member = m,
        Comments = m.MemberComments.Select(mc => mc.Comment)
    })
    .ToList();
Run Code Online (Sandbox Code Playgroud)

如果要查找成员的所有注释MemberId= 1:

var commentsOfMember = context.MemberComments
    .Where(mc => mc.MemberId == 1)
    .Select(mc => mc.Comment)
    .ToList();
Run Code Online (Sandbox Code Playgroud)

现在,您还可以按联接表中的属性进行筛选(这在多对多关系中是不可能的),例如:筛选成员1中具有99属性的所有注释Something:

var filteredCommentsOfMember = context.MemberComments
    .Where(mc => mc.MemberId == 1 && mc.Something == 99)
    .Select(mc => mc.Comment)
    .ToList();
Run Code Online (Sandbox Code Playgroud)

由于延迟加载,事情可能会变得更容易.如果您已加载Member,则应该能够在没有明确查询的情况下获取注释:

var commentsOfMember = member.MemberComments.Select(mc => mc.Comment);
Run Code Online (Sandbox Code Playgroud)

我想延迟加载会在幕后自动获取注释.

编辑

只是为了好玩一些例子,更多如何添加实体和关系以及如何在此模型中删除它们:

1)创建该成员的一个成员和两个评论:

var member1 = new Member { FirstName = "Pete" };
var comment1 = new Comment { Message = "Good morning!" };
var comment2 = new Comment { Message = "Good evening!" };
var memberComment1 = new MemberComment { Member = member1, Comment = comment1,
                                         Something = 101 };
var memberComment2 = new MemberComment { Member = member1, Comment = comment2,
                                         Something = 102 };

context.MemberComments.Add(memberComment1); // will also add member1 and comment1
context.MemberComments.Add(memberComment2); // will also add comment2

context.SaveChanges();
Run Code Online (Sandbox Code Playgroud)

2)添加member1的第三条评论:

var member1 = context.Members.Where(m => m.FirstName == "Pete")
    .SingleOrDefault();
if (member1 != null)
{
    var comment3 = new Comment { Message = "Good night!" };
    var memberComment3 = new MemberComment { Member = member1,
                                             Comment = comment3,
                                             Something = 103 };

    context.MemberComments.Add(memberComment3); // will also add comment3
    context.SaveChanges();
}
Run Code Online (Sandbox Code Playgroud)

3)创建新成员并将其与现有comment2相关联:

var comment2 = context.Comments.Where(c => c.Message == "Good evening!")
    .SingleOrDefault();
if (comment2 != null)
{
    var member2 = new Member { FirstName = "Paul" };
    var memberComment4 = new MemberComment { Member = member2,
                                             Comment = comment2,
                                             Something = 201 };

    context.MemberComments.Add(memberComment4);
    context.SaveChanges();
}
Run Code Online (Sandbox Code Playgroud)

4)创建现有member2和comment3之间的关系:

var member2 = context.Members.Where(m => m.FirstName == "Paul")
    .SingleOrDefault();
var comment3 = context.Comments.Where(c => c.Message == "Good night!")
    .SingleOrDefault();
if (member2 != null && comment3 != null)
{
    var memberComment5 = new MemberComment { Member = member2,
                                             Comment = comment3,
                                             Something = 202 };

    context.MemberComments.Add(memberComment5);
    context.SaveChanges();
}
Run Code Online (Sandbox Code Playgroud)

5)再次删除这种关系:

var memberComment5 = context.MemberComments
    .Where(mc => mc.Member.FirstName == "Paul"
        && mc.Comment.Message == "Good night!")
    .SingleOrDefault();
if (memberComment5 != null)
{
    context.MemberComments.Remove(memberComment5);
    context.SaveChanges();
}
Run Code Online (Sandbox Code Playgroud)

6)删除member1及其与评论的所有关系:

var member1 = context.Members.Where(m => m.FirstName == "Pete")
    .SingleOrDefault();
if (member1 != null)
{
    context.Members.Remove(member1);
    context.SaveChanges();
}
Run Code Online (Sandbox Code Playgroud)

该删除的关系MemberComments也因为之间的一个一对多的关系MemberMemberComments与之间CommentMemberComments与级联的安装按约定删除.而这种情况是因为MemberIdCommentIdMemberComment被检测为外键的性能MemberComment导航性能,因为FK属性的类型为非可空的int,需要的关系,并最终导致级联,删除,设置.我认为这个模型很有意义.

  • @hgdean:我发了几个例子,对不起,但是这是一个有趣的模型,关于多对多的问题,连接表中的附加数据时不时出现.现在我下次有东西可以链接到... :) (7认同)
  • 注意:如果您在没有Fluent API的情况下使用此方法,请确保在数据库中检查您只有一个带有`MemberId`和`CommentId`列的复合键,而不是另外的第三列`Member_CommentId`(或类似的东西) - 其中表示您的键的对象之间没有完全匹配的名称 (4认同)
  • @Esteban:没有被覆盖的`OnModelCreating`.该示例仅依赖于映射约定和数据注释. (3认同)
  • @Simon_Weaver(或任何可能知道答案的人)我有类似的情况,但我想拥有该表的"MemberCommentID"主键,这是否可能?我目前正在例外,请看看我的问题,我真的需要帮助...... http://stackoverflow.com/questions/26783934/foreign-key-cycles-or-cascade-paths (3认同)

Est*_*ban 97

Slauma的优秀答案.

我将使用流畅的API映射发布代码来执行此操作.

public class User {
    public int UserID { get; set; }
    public string Username { get; set; }
    public string Password { get; set; }

    public ICollection<UserEmail> UserEmails { get; set; }
}

public class Email {
    public int EmailID { get; set; }
    public string Address { get; set; }

    public ICollection<UserEmail> UserEmails { get; set; }
}

public class UserEmail {
    public int UserID { get; set; }
    public int EmailID { get; set; }
    public bool IsPrimary { get; set; }
}
Run Code Online (Sandbox Code Playgroud)

DbContext派生类上,您可以这样做:

public class MyContext : DbContext {
    protected override void OnModelCreating(DbModelBuilder builder) {
        // Primary keys
        builder.Entity<User>().HasKey(q => q.UserID);
        builder.Entity<Email>().HasKey(q => q.EmailID);
        builder.Entity<UserEmail>().HasKey(q => 
            new { 
                q.UserID, q.EmailID
            });

        // Relationships
        builder.Entity<UserEmail>()
            .HasRequired(t => t.Email)
            .WithMany(t => t.UserEmails)
            .HasForeignKey(t => t.EmailID)

        builder.Entity<UserEmail>()
            .HasRequired(t => t.User)
            .WithMany(t => t.UserEmails)
            .HasForeignKey(t => t.UserID)
    }
}
Run Code Online (Sandbox Code Playgroud)

它与接受的答案具有相同的效果,采用不同的方法,这种方法并不是更好也不是更糟.

编辑: 我已将CreatedDate从bool更改为DateTime.

编辑2:由于时间不够,我已经从我正在开发的应用程序中找到了一个示例,以确保它有效.

  • 谢谢.我按照这个解决方案进行多对多映射. (2认同)

Leo*_*doX 11

@Esteban,你提供的代码是对的,谢谢,但不完整,我已经测试过了."UserEmail"类中缺少属性:

    public UserTest UserTest { get; set; }
    public EmailTest EmailTest { get; set; }
Run Code Online (Sandbox Code Playgroud)

如果有人有兴趣,我发布我测试过的代码.问候

using System.Data.Entity;
using System;
using System.Collections.Generic;
using System.ComponentModel.DataAnnotations;
using System.ComponentModel.DataAnnotations.Schema;
using System.Linq;
using System.Web;

#region example2
public class UserTest
{
    public int UserTestID { get; set; }
    public string UserTestname { get; set; }
    public string Password { get; set; }

    public ICollection<UserTestEmailTest> UserTestEmailTests { get; set; }

    public static void DoSomeTest(ApplicationDbContext context)
    {

        for (int i = 0; i < 5; i++)
        {
            var user = context.UserTest.Add(new UserTest() { UserTestname = "Test" + i });
            var address = context.EmailTest.Add(new EmailTest() { Address = "address@" + i });
        }
        context.SaveChanges();

        foreach (var user in context.UserTest.Include(t => t.UserTestEmailTests))
        {
            foreach (var address in context.EmailTest)
            {
                user.UserTestEmailTests.Add(new UserTestEmailTest() { UserTest = user, EmailTest = address, n1 = user.UserTestID, n2 = address.EmailTestID });
            }
        }
        context.SaveChanges();
    }
}

public class EmailTest
{
    public int EmailTestID { get; set; }
    public string Address { get; set; }

    public ICollection<UserTestEmailTest> UserTestEmailTests { get; set; }
}

public class UserTestEmailTest
{
    public int UserTestID { get; set; }
    public UserTest UserTest { get; set; }
    public int EmailTestID { get; set; }
    public EmailTest EmailTest { get; set; }
    public int n1 { get; set; }
    public int n2 { get; set; }


    //Call this code from ApplicationDbContext.ConfigureMapping
    //and add this lines as well:
    //public System.Data.Entity.DbSet<yournamespace.UserTest> UserTest { get; set; }
    //public System.Data.Entity.DbSet<yournamespace.EmailTest> EmailTest { get; set; }
    internal static void RelateFluent(System.Data.Entity.DbModelBuilder builder)
    {
        // Primary keys
        builder.Entity<UserTest>().HasKey(q => q.UserTestID);
        builder.Entity<EmailTest>().HasKey(q => q.EmailTestID);

        builder.Entity<UserTestEmailTest>().HasKey(q =>
            new
            {
                q.UserTestID,
                q.EmailTestID
            });

        // Relationships
        builder.Entity<UserTestEmailTest>()
            .HasRequired(t => t.EmailTest)
            .WithMany(t => t.UserTestEmailTests)
            .HasForeignKey(t => t.EmailTestID);

        builder.Entity<UserTestEmailTest>()
            .HasRequired(t => t.UserTest)
            .WithMany(t => t.UserTestEmailTests)
            .HasForeignKey(t => t.UserTestID);
    }
}
#endregion
Run Code Online (Sandbox Code Playgroud)


Mau*_*les 5

我想提出一个解决方案,可以实现多对多配置的两种风格。

“捕获”是我们需要创建一个以连接表为目标的视图,因为 EF 验证模式的表最多可以被映射一次EntitySet

这个答案增加了之前答案中已经说过的内容,并且不会覆盖任何这些方法,它建立在它们的基础上。

该模型:

public class Member
{
    public int MemberID { get; set; }

    public string FirstName { get; set; }
    public string LastName { get; set; }

    public virtual ICollection<Comment> Comments { get; set; }
    public virtual ICollection<MemberCommentView> MemberComments { get; set; }
}

public class Comment
{
    public int CommentID { get; set; }
    public string Message { get; set; }

    public virtual ICollection<Member> Members { get; set; }
    public virtual ICollection<MemberCommentView> MemberComments { get; set; }
}

public class MemberCommentView
{
    public int MemberID { get; set; }
    public int CommentID { get; set; }
    public int Something { get; set; }
    public string SomethingElse { get; set; }

    public virtual Member Member { get; set; }
    public virtual Comment Comment { get; set; }
}
Run Code Online (Sandbox Code Playgroud)

配置:

using System.ComponentModel.DataAnnotations.Schema;
using System.Data.Entity.ModelConfiguration;

public class MemberConfiguration : EntityTypeConfiguration<Member>
{
    public MemberConfiguration()
    {
        HasKey(x => x.MemberID);

        Property(x => x.MemberID).HasColumnType("int").IsRequired();
        Property(x => x.FirstName).HasColumnType("varchar(512)");
        Property(x => x.LastName).HasColumnType("varchar(512)")

        // configure many-to-many through internal EF EntitySet
        HasMany(s => s.Comments)
            .WithMany(c => c.Members)
            .Map(cs =>
            {
                cs.ToTable("MemberComment");
                cs.MapLeftKey("MemberID");
                cs.MapRightKey("CommentID");
            });
    }
}

public class CommentConfiguration : EntityTypeConfiguration<Comment>
{
    public CommentConfiguration()
    {
        HasKey(x => x.CommentID);

        Property(x => x.CommentID).HasColumnType("int").IsRequired();
        Property(x => x.Message).HasColumnType("varchar(max)");
    }
}

public class MemberCommentViewConfiguration : EntityTypeConfiguration<MemberCommentView>
{
    public MemberCommentViewConfiguration()
    {
        ToTable("MemberCommentView");
        HasKey(x => new { x.MemberID, x.CommentID });

        Property(x => x.MemberID).HasColumnType("int").IsRequired();
        Property(x => x.CommentID).HasColumnType("int").IsRequired();
        Property(x => x.Something).HasColumnType("int");
        Property(x => x.SomethingElse).HasColumnType("varchar(max)");

        // configure one-to-many targeting the Join Table view
        // making all of its properties available
        HasRequired(a => a.Member).WithMany(b => b.MemberComments);
        HasRequired(a => a.Comment).WithMany(b => b.MemberComments);
    }
}
Run Code Online (Sandbox Code Playgroud)

上下文:

using System.Data.Entity;

public class MyContext : DbContext
{
    public DbSet<Member> Members { get; set; }
    public DbSet<Comment> Comments { get; set; }
    public DbSet<MemberCommentView> MemberComments { get; set; }

    protected override void OnModelCreating(DbModelBuilder modelBuilder)
    {
        base.OnModelCreating(modelBuilder);

        modelBuilder.Configurations.Add(new MemberConfiguration());
        modelBuilder.Configurations.Add(new CommentConfiguration());
        modelBuilder.Configurations.Add(new MemberCommentViewConfiguration());

        OnModelCreatingPartial(modelBuilder);
     }
}
Run Code Online (Sandbox Code Playgroud)

来自 Saluma (@Saluma) 的回答

例如,如果您现在想查找 LastName = "Smith" 成员的所有评论,您可以编写如下查询:

这仍然有效...

var commentsOfMembers = context.Members
    .Where(m => m.LastName == "Smith")
    .SelectMany(m => m.MemberComments.Select(mc => mc.Comment))
    .ToList();
Run Code Online (Sandbox Code Playgroud)

……但现在也可以……

var commentsOfMembers = context.Members
    .Where(m => m.LastName == "Smith")
    .SelectMany(m => m.Comments)
    .ToList();
Run Code Online (Sandbox Code Playgroud)

或者要创建一个名为“Smith”的成员列表(我们假设有多个)以及他们的评论,您可以使用投影:

这仍然有效...

var membersWithComments = context.Members
    .Where(m => m.LastName == "Smith")
    .Select(m => new
    {
        Member = m,
        Comments = m.MemberComments.Select(mc => mc.Comment)
    })
    .ToList();
Run Code Online (Sandbox Code Playgroud)

……但现在也可以……

var membersWithComments = context.Members
    .Where(m => m.LastName == "Smith")
    .Select(m => new
    {
        Member = m,
        m.Comments
    })
        .ToList();
Run Code Online (Sandbox Code Playgroud)

如果您想删除成员的评论

var comment = ... // assume comment from member John Smith
var member = ... // assume member John Smith

member.Comments.Remove(comment);
Run Code Online (Sandbox Code Playgroud)

如果您想Include()发表评论

var member = context.Members
    .Where(m => m.FirstName == "John", m.LastName == "Smith")
    .Include(m => m.Comments);
Run Code Online (Sandbox Code Playgroud)

这一切都像是语法糖,但是如果您愿意进行额外的配置,它确实会给您带来一些好处。无论哪种方式,您似乎都能充分利用这两种方法。