我有两个实体,当我更新或只添加一个时,一切正常
db.Entry(user1).State = EntityState.Modified;
foreach (var userAudio in user1.Audios)
{
db.Audios.AddOrUpdate(userAudio);
}
db.Users.AddOrUpdate(user1);
db.SaveChanges();
但是如果我尝试添加/更新几个实体:
db.Entry(user1).State = EntityState.Modified;
foreach (var userAudio in user1.Audios)
{
db.Audios.AddOrUpdate(userAudio);
}
db.Users.AddOrUpdate(user1);
db.Entry(user2).State = EntityState.Modified;
foreach (var userAudio in user2.Audios)
{
db.Audios.AddOrUpdate(userAudio);
}
db.Users.AddOrUpdate(user2);
db.SaveChanges();
抛出异常:
附加类型' EfTest.Entities.Audio'失败是因为 另一个相同类型的实体已经拥有相同的主键 值...
也许是因为我在user1和user2中有相同的音频,并且EF无法插入它们......任何人都有想法如何绕过这个?谢谢!
实体
namespace EfTest.Entities
{
public class User
{
[Key]
[DatabaseGenerated(DatabaseGeneratedOption.None)]
public int Id { get; set; }
public string Name { get; set; }
public int Age { get; set; }
public List<Audio> Audios { get; set; }
}
public class Audio
{
[Key]
[DatabaseGenerated(DatabaseGeneratedOption.None)]
public int Id { get; set; }
public string Artist { get; set; }
public string Title { get; set; }
public List<User> Users { get; set; }
}
}
答案 0 :(得分:0)
不幸的是,我只看到一种解决方法。它是独立添加实体然后添加关系:
var userAudios = new List<Audio>();
// Key userId
// Value list of audiosIds
var userAudiosRelations = new List<Relation>();
foreach (var user in users)
{
foreach (var audio in user.Audios)
{
if (!userAudios.Any(x => x.Id == audio.Id))
{
userAudios.Add(audio);
}
userAudiosRelations.Add(new Relation
{
User_Id = user.Id,
Audio_Id = audio.Id
});
}
user.Audios = null;
db.Users.AddOrUpdate(user);
}
foreach (var audio in userAudios)
{
db.Audios.AddOrUpdate(audio);
}
db.SaveChanges();
var existingRelations = db.Database.SqlQuery<Relation>("SELECT * FROM dbo.UserAudios").ToList();
var relationsToAdd =
userAudiosRelations.Where(
x => existingRelations.All(y => x.User_Id != y.User_Id || x.Audio_Id != y.Audio_Id)).ToList();
foreach (var relation in relationsToAdd)
{
db.Database.ExecuteSqlCommand("INSERT INTO dbo.UserAudios (User_Id, Audio_Id) VALUES (@p0, @p1)",
relation.User_Id, relation.Audio_Id);
}
关系模型:
public class Relation
{
public int User_Id { get; set; }
public int Audio_Id { get; set; }
}