Question

我有2个具有一对多关系的模型，如果我保存在DbContext的不同实例中，这会引发异常（违反主键） - 如何避免它？

public class Customer
{
    public Guid Id { get; set; }
    public string Name { get; set; }
}

public class User
{
    public Guid Id { get; set; }
    public string Name { get; set; }
    public Customer Customer { get; set; }
}

public class DbContext : System.Data.Entity.DbContext
{
    public DbContext()
    : base(@"Data Source=(localdb)\MSSQLLocalDB; AttachDBFilename='|DataDirectory|\Sample.mdf'; Integrated Security=True")
    {
        Configuration.ProxyCreationEnabled = false;
        Configuration.AutoDetectChangesEnabled = true;
        Configuration.ValidateOnSaveEnabled = true;
        Configuration.LazyLoadingEnabled = true;
    }

    public IDbSet<Customer> Customers { get; set; }
    public IDbSet<User> Users { get; set; }

    protected override void OnModelCreating(DbModelBuilder modelBuilder)
    {
        modelBuilder.Properties<Guid>()
        .Where(p => p.Name == "Id")
        .Configure(p => { p.IsKey(); p.IsRequired(); });
    }
}

AppDomain.CurrentDomain.SetData(
    "DataDirectory",
    System.Environment.CurrentDirectory);

var customer = new Customer();
customer.Id = Guid.NewGuid();
customer.Name = "customername";

using (var db = new DbContext())
{
    db.Customers.Add(customer);
    db.SaveChanges();
}

var user = new User();
user.Id = Guid.NewGuid();
user.Name = "username";
user.Customer = customer;

using (var db = new DbContext())
{
    db.Users.Add(user);
    db.SaveChanges(); // <- Throw here
}

当然这是一个简化的示例，在编写的内容中，可以只使用DbContext的一个实例，但实际上客户作为参数传递给方法

Answer 1

你是对的 - 第二个实例不会知道你刚刚添加了客户。将它们包装在相同的using语句中，或者您可以告诉第二个实例客户已经存在：

var user = new User();
user.Id = Guid.NewGuid();
user.Name = "username";

using (var db = new DbContext())
{   
    user.Customer = new Customer() { Id = customer.Id };  // only need the id
    db.Customers.Attach(user.Customer);
    db.Users.Add(user);
    db.SaveChanges();
}

Entity Framework: adding existing child POCO to new Parent POCO, creates new child in DB

使用Entity Framework 6违反PRIMARY KEY约束，模型一对多

1 个答案: