Insert语句与外键约束冲突。 实体框架错误

Question

I'm trying to create an one to many relation, where a TypeOfImmobile can have one Immobile or more. Here's my model classes :

public class TypeOfImmobile
{
    public int Id { get; set; }
    [Display(Name = "Type Of Immobile")]
    public string Designation { get; set; }

    public ICollection<Immobile> Immobiles { get; set; }
}

public class Immobile
{
    public int Id { get; set; }
    public int TypeOfImmobileId { get; set; }
    public virtual TypeOfImmobile TypeOfImmobile { get; set; }
}

Error occurred when I create an Immobile:

The INSERT statement conflicted with the FOREIGN KEY constraint "FK_dbo.Immobiles_dbo.TypeOfImmobiles_TypeOfImmobilesId". The conflict occurred in database "aspnet-ARQSI-IT2-20151031052056", table "dbo.TypeOfImmobiles", column 'Id'. The statement has been terminated.

Answer 1

Type of your TypeOfImmobileId property is int , so you can't set it as null . Even you don't assign a value to an integer , it would be equal to 0 not null . If you didn't assign a value for TypeOfImmobileId when your create command was sent to SQL, SQL will take it as 0 in stored procedure that create the record. And probably TypeOfImmobile table doesn't contain a record with a 0 value in its primary key. In other words, value of foreign key of the data you sent is invalid. It doesn't correspond any primary key value in TypeOfImmobile table. Your primary key should not be nullable, but foreign key should be nullable.

You should change type of TypeOfImmobileId from int to Nullable<int> or simply int? .

So TypeOfImmobileId property should be like this:

public int? TypeOfImmobileId { get; set; }

You don't need to add ForeignKey attribute to TypeOfImmobileId property because you follow Code-First conventions and name of Foreign Key matches with name of Primary Key of related entity

You can observe the problem as running SQL Server Profiler. When application runs the command that will create the record, related stored procedure will be listed.

Answer 2

似乎在创建Immobile您可能未将TypeOfImmobileId属性设置为有效值，即现有TypeOfImmobile的Id 。

Answer 3

this causes by one of following reasons:

1-You don't set navigation property (TypeOfImmobile) or TypeOfImmobileId

2-The value of TypeOfImmobileId or TypeOfImmobile is not exist in database

3-Maybe you are saving Immobile before saving TypeOfImmobile

Answer 4

您需要通过[ForeignKey("Your_FK_ImmobileId")]在新的属性ID Your_FK_ImmobileId中设置FK