Entity framework 配置实体框架以加载预定义对象
我有这门课Entity framework 配置实体框架以加载预定义对象,entity-framework,singleton,Entity Framework,Singleton,我有这门课 public class Status { private string status; public string StatusName { get { return status; } } private Status (string _status) { status = _status; } public static rea
public class Status
{
private string status;
public string StatusName
{
get
{
return status;
}
}
private Status (string _status)
{
status = _status;
}
public static readonly Status Open = new Status("Open");
public static readonly Status Closed = new Status("Closed");
}
首先使用代码,我是否可以将EF配置为加载一个预定义对象(打开、关闭),而不是尝试创建一个新对象?或者有更好的方法来实现类似的行为吗?当前的EF版本没有提供任何钩子来替代对象物化(除非您下载源代码并尝试自己实现)。这意味着EF将始终创建自己的状态,并且为了能够做到这一点,它还需要您的
status
类来匹配其规则。我不知道你为什么需要这个,但是如果你真的需要状态总是相同的对象实例,你可以破解它
首先,您需要修改EF的状态
类:
public class Status{
// EF needs access to your property
public string StatusName { get; private set; }
// EF needs parameterless constructor because it will create instances
private Status() {}
private Status (string status) {
StatusName = status;
}
public static readonly Status Open = new Status("Open");
public static readonly Status Closed = new Status("Closed");
}
现在,您需要将EF创建的原始状态
替换为您自己的:
public Context() : base() {
var objectContext = ((IObjectContextAdapter)this).ObjectContext;
objectContext.ObjectMaterialized += OnObjectMaterialized;
}
private void OnObjectMaterialized(object sender, ObjectMaterializedEventArgs args) {
var objectContext = (ObjectContext)sender;
var entity = args.Entity as Entity;
if (entity != null) {
switch (entity.Status.StatusName) {
case "Open":
entity.Status = Status.Open;
break;
case "Closed":
entity.Status = Status.Closed;
break;
}
// This is necessary because previous code made your object modified
objectContext.DetectChanges();
var entry = objectContext.ObjectStateManager.GetObjectStateEntry(entity);
entry.AcceptChanges();
}
}
这是一个丑陋的黑客攻击,但如果你真的需要它,你必须做一些类似的事情。你的状态被用作复杂类型吗?是的,它被用作复杂类型。