分享关于Entity Framework 进行CRUD操作实验的结果

最新推荐文章于 2021-06-16 05:56:14 发布

转载最新推荐文章于 2021-06-16 05:56:14 发布 · 78 阅读

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原文链接：http://www.cnblogs.com/zuowj/p/4864390.html

本文通过多次实验总结了使用EntityFramework框架进行CRUD操作时可能遇到的问题，特别是针对导航属性和外键属性更新的不同场景进行了详细说明。

我们在使用Entity Framework框架进行CRUD时，经常会出现各种各样的错误，下面请看我的实验结果。

以下是只用一个上下文对象进行操作：

第一次：

            BlogDbContext blog = new BlogDbContext();
            post = blog.Posts.Single(t => t.Id == 2);
            post.AuthorId = 1;
            blog.SaveChanges();

结果：成功

第二次：

            BlogDbContext blog = new BlogDbContext();
            post = blog.Posts.Single(t => t.Id == 2);
            post.Author = blog.Authors.Single(t => t.Id == 3);
            blog.SaveChanges();

结果：成功

第三次：

            BlogDbContext blog = new BlogDbContext();
            post = blog.Posts.Single(t => t.Id == 2);
            post.AuthorId = 1;
            post.Author = blog.Authors.Single(t => t.Id == 4);
            blog.SaveChanges();

结果：失败，报错如下：

Conflicting changes to the role 'Post_Author_Target' of the relationship 'ConsoleApplication1.DDD.Infrastructure.Post_Author' have been detected.

第四次：

            BlogDbContext blog = new BlogDbContext();
            post = blog.Posts.Single(t => t.Id == 2);
            post.Author = new Author() { Id = 4 };
            blog.Entry(post.Author).State = EntityState.Unchanged;
            blog.SaveChanges();

结果：成功

第五次：

            BlogDbContext blog = new BlogDbContext();
            post = blog.Posts.Single(t => t.Id == 2);
            //author = new Author() { Id = 1, Name = "zwj" };//直接实例化或下面查询得出
            author = blog.Authors.AsNoTracking().Single(t => t.Id == 3);
            post.Author = author;
            blog.SaveChanges();

结果：成功，但是Authors表中会新增一笔记录，并将新的ID赋值给Posts表；原因：Author与Post不在同一个上下文中，Post所在的上下文中追踪不到Author的信息，故当成新增。

以下是在两个不同的上下文中进行操作：

第六次：

            BlogDbContext blog = new BlogDbContext();
            post = blog.Posts.AsNoTracking().Single(t => t.Id == 2);
            //blog.Entry(post).State = EntityState.Detached;上面的AsNoTracking可以由这句实现

            BlogDbContext blog2 = new BlogDbContext();
            post.AuthorId = 5;
            blog2.Entry(post).State = EntityState.Modified;
            blog2.SaveChanges();

结果：成功

第七次：

            BlogDbContext blog = new BlogDbContext();
            post = blog.Posts.AsNoTracking().Single(t => t.Id == 2);

            BlogDbContext blog2 = new BlogDbContext();
            post.Author = blog2.Authors.Single(t => t.Id == 5);
            blog2.Entry(post).State = EntityState.Modified;
            blog2.SaveChanges();

结果：失败，报错如下：

A referential integrity constraint violation occurred: The property value(s) of 'Author.Id' on one end of a relationship do not match the property value(s) of 'Post.AuthorId' on the other end.

第八次：

            BlogDbContext blog = new BlogDbContext();
            post = blog.Posts.AsNoTracking().Single(t => t.Id == 2);

            BlogDbContext blog2 = new BlogDbContext();
            author = blog2.Authors.AsNoTracking().Single(t => t.Id == 1);
            post.Author = author;
            blog2.Entry(post).State = EntityState.Modified;
            blog2.SaveChanges();

结果：失败，报错同第七次

第九次：

            BlogDbContext blog = new BlogDbContext();
            post = blog.Posts.AsNoTracking().Single(t => t.Id == 2);
            author = blog.Authors.AsNoTracking().Single(t => t.Id == 3);

            BlogDbContext blog2 = new BlogDbContext();
            post.Author = author;
            blog2.Entry(post).State = EntityState.Modified;
            blog2.SaveChanges();

结果：失败，报错同第七次

第十次：

            BlogDbContext blog = new BlogDbContext();
            post = blog.Posts.AsNoTracking().Single(t => t.Id == 2);

            BlogDbContext blog2 = new BlogDbContext();
            author = new Author() { Id = 1, Name = "zwj" };
            post.Author = author;
            blog2.Entry(post).State = EntityState.Modified;
            blog2.SaveChanges();

结果：失败，报错同第七次

第十一次：

            BlogDbContext blog = new BlogDbContext();
            post = blog.Posts.AsNoTracking().Single(t => t.Id == 2);

            BlogDbContext blog2 = new BlogDbContext();
            author = blog2.Authors.Single(t => t.Id == 1);
            post.Author = author;
            post.AuthorId = author.Id;
            blog2.Entry(post).State = EntityState.Modified;
            blog2.SaveChanges();

结果：成功，但我认为主要是通过赋值AuthorId来完成的，与第六次相同，去掉赋值AuthorId，则与第七次相同的报错

第十二次：

            BlogDbContext blog = new BlogDbContext();
            post = blog.Posts.AsNoTracking().Single(t => t.Id == 2);

            BlogDbContext blog2 = new BlogDbContext();
            author = blog2.Authors.Single(t => t.Id == 3);
            blog2.Entry(post).Reference(t => t.Author).CurrentValue = author;
            blog2.Entry(post).State = EntityState.Modified;
            blog2.SaveChanges();

结果：失败，报错同第七次

第十三次：

            BlogDbContext blog = new BlogDbContext();
            post = blog.Posts.AsNoTracking().Single(t => t.Id == 2);

            BlogDbContext blog2 = new BlogDbContext();
            blog2.Posts.Attach(post);
            post.Author = blog2.Authors.Single(t => t.Id == 1);
            blog2.SaveChanges();

结果：成功

最后得出结论：

1.在同一个上下文中，不论是采用直接赋值导航属性或是直接赋值外键属性，除不能同时对导航属性及外键属性赋不相关联的值外，都可以成功；

2.不在同一个上下文中，若想在完成CUD时，则必需先确保要进行操作的实体处于Detached状态，然后再进行相应的更新，涉及导航属性时，只能采用赋值外键属性，不能通过导航属性直接赋值，否则会报错（这个我觉得是个BUG，不知大家有什么好的解决办法没有），若采用先在新的上下文中Attached，再进行更新操作则与第1条结论相同。

转载于:https://www.cnblogs.com/zuowj/p/4864390.html