Reference built-in Account UserId Assign to Model as Foreign Key - asp.net-mvc-3

Question

after looking for an answer in the already existing questions, I am still a little confused as how I should proceed. I am new to the MVC 3 framework, so if I come off sounding like a dope, I appologize!!
Ok, so I created a MVC 3 internet application, created 3 new Users administrator, user1, and user2. I have created a new model, and controller for my "Posts" I am able to add, edit and delete the items. I currently have a column called UserID in my posts table. I would like this to be automagically populated with the current UsersID. I think I would define this in the controller like this:
[HttpPost]
public ActionResult Create(Post post)
{
if (ModelState.IsValid)
{
public int User = System.Web.Security.Membership.GetUser(id);
db.Posts.Add(post);
db.SaveChanges();
return RedirectToAction("Index");
}
return View(post);
}
Inside the model, this is what I currently have:
public class Post
{
public int PostID { get; set; }
public int UserID { get; set; }
public string PostTitle { get; set; }
public int PostType { get; set; }
public string PostBody { get; set; }
public string PostBlogTitle { get; set; }
public string PostBlogURL { get; set; }
public string PostCategory { get; set; }
public string PostSEO { get; set; }
public int PostStatus { get; set; }
}
public class PostDBContext : DbContext
{
public DbSet<Post> Posts { get; set; }
}
I would like to replace public int UserID { get; set; } with my newly defined variable in the controller, but not sure where/how to add it.

Answer 1

Not sure I'm clear on what you're trying to achieve. Are you just trying to assign the current user to the model, save it and then pass it back to the view?
If so you can just do this:
[HttpPost]
public ActionResult Create(Post post)
{
if (ModelState.IsValid)
{
// Note: I'm assuming this actually works/returns an int
public int User = System.Web.Security.Membership.GetUser(id);
post.UserID = User;
db.Posts.Add(post);
db.SaveChanges();
return RedirectToAction("Index");
}
return View(post);
}