如果我想用父子关系绑定两个实体框架对象person1
,person2
。比如说,对象来自不同的ObjectContexts
或者一个是分离的。那么现在,我想写一些类似的东西:
person1.Parent = person2;
SaveChanges()
失败。所以我写道:
person1.ParentReference.EntityKey = person2.EntityKey;
这是解决该问题的正确解决方案,还是应该总是重新加载“坏”对象(当前位于另一个ObjectContext
的对象)?
答案 0 :(得分:0)
直接给你答案
想象你的表
[Tbl_Persons]
PersonId int [PK]
ParentId int [FK]
FirstName nvarchar(100)
LastName nvarchar(100)
CreateUser int
CreateDate datetime
UpdateUser int
UpdateDate datetime
您将有2个选项来分配父
person1.Tbl_Persons = person2;
其中person1和person2都是TblPersons
个对象,或者只是分配Id
person1.ParentId = person2.PersonId;
现在,完整代码就像我通常对在数据库中插入/更新
所做的那样public class MyRepository : IMyRepository
{
MyEntities db = new MyEntities();
DateTime now = DateTime.UTCNow;
public void Save() {
db.SaveChanges();
}
#region Update / Insert Persons
public void UpdatePerson(Tbl_Persons person, int parentId, int userId)
{
// let's check if record is in db
Tbl_Persons p = this.GetPersonById(person.PersonId);
if(p == null)
p = new Person(); // person was not found in db
p.FirstName = person.FirstName;
p.LastName = person.LastName;
// we can now hook up the parent object in 2 ways
// or set the Id, or attach the hole object
// NOTE: only choose one line!
p.MyParent = parentId;
p.TblParent = this.GetPersonById(parentId);
// update info
p.UpdateDate = now;
p.UpdateUser = userId;
if(p.PersonId == 0)
{
// is it new person in db, so we need to INSERT into the db
p.CreateDate = now;
p.CreateUser = userId;
db.Tbl_Persons.AddObject(p);
}
else
{
// It's an existing person, we do need need to do anything
// as internally once we assign a new value to the object that
// come back from the database, EF sets Modified flag on this
}
// let's save
this.Save();
}
#endregion
#region Queries
public Tbl_Persons GetPersonById(int personId)
{
return db.Tbl_Persons.FirstOrDefault(x => x.PersonId == personId);
}
#endregion
}
控制器中的:
[HttpPost]
[ValidateAntiForgeryToken]
public ActionResult Create(Tbl_Persons model, int parentId)
{
if(ModelState.IsValid)
{
db.UpdatePerson(model, parentId, currentUserId);
}
return RedirectToAction("Index");
}
希望它有所帮助。