My super-powers tell me that you are missing the following on top of your controller file:
using Orchard.ContentManagement;
The generic version of the Query
method is an extension method that is in this namespace.
题
What I'm trying to achieve here is to save the current user instance in my ApiConfigurationRecord table. I already dig around the internet, and most of the example is using UserPartRecord. But the troble I encounter is to get the UserPartRecord object itself.
This is my Entity class look like:
public class ApiConfigurationRecord
{
public virtual int Id { get; set; }
public virtual string Name { get; set; }
public virtual UserPartRecord RegisterBy { get; set; }
}
This is my Migration.cs code look like:
public int Create()
{
SchemaBuilder.CreateTable("ApiConfigurationRecord", table => table
.Column<int>("Id", column => column.PrimaryKey().Identity())
.Column<int>("RegisterBy_id")
.Column<string>("Name", column => column.NotNull())
);
return 1;
}
This is my Action Controller codes:
public ActionResult Test()
{
var userId = this._orchardServices.WorkContext.CurrentUser.Id;
// below code got error: The non-generic method IContentManager.Query() cannot be used with type arguments
this._orchardServices.ContentManager.Query<UserPart, UserPartRecord>().Where(u => u.Id == userId);
return null;
}
For hours I stuck in this problem. Need to know how to save this User relationship object, and most importantly, get the object itself. Please guide me.
解决方案 2
My super-powers tell me that you are missing the following on top of your controller file:
using Orchard.ContentManagement;
The generic version of the Query
method is an extension method that is in this namespace.
其他提示
Or you could just do
_orchardServices.WorkContext.CurrentUser.As<UserPart>().Record;
Though you will probably want to check user is not null there too. And as Bertrand Le Roy says, you will also need
using Orchard.ContentManagement;
to make use of the .As extension method.