How can I create multiple entities before the SaveChanges? - asp.net-mvc

Question

I have a quality tracking webapp where I need each resolution steps to be signed by a user. To ease the process, I've built my model around following who filled the date at the end of each step.
My simplified Failure model looks like this:
public class failure {
[key]
public string FailureId { get; set; }
...
public int? OpenUserId { get; set; }
public int? DiagUserId { get; set; }
public int? CloseUserId { get; set; }
...
[ForeignKey("OpenUserId")]
public virtual signature OpenUser { get; set; }
[ForeignKey("DiagUserId")]
public virtual signature DiagUser { get; set; }
[ForeignKey("CloseUserId")]
public virtual signature CloseUser { get; set; }
}
and my Signature model:
public class signature {
[key]
[DatabaseGenerated(DatabaseGeneratedOption.Identity)]
public int SignatureId { get; set; }
public string UserName { get; set; }
[Column(TypeName:="Date")]
[DataType(DataType.Date)]
public date DateSign { get; set; }
}
The goal of this model is to minimize the number of signature in the table. So, if a user signs on multiple failure in one day, the code should only need to create one signature and reuse the same Id.
The problem arises when a user fills multiple steps in one save. Two or more signatures are created (which can be a problem in itself but it's not the focus right now) and an error is raised
Multiple added entities may have the same primary key.
because, before the SaveChanges, all Ids are at 0 and the code can't differentiate them.
Here's the POST:
async Task<ActionResult> Edit( FailureVM failure ) {
if (ModelState.IsValid) {
...
failure.OpenUserId = Await TryUpdateSignature(...);
failure.DiagUserId = Await TryUpdateSignature(...);
failure.CloseUserId = Await TryUpdateSignature(...);
...
await db.SaveChangesAsync;
}
}
and my function:
public static async Task<int?> TryUpdateSignature(MyDbContext db, Signature oldSignUser, Date? newDate, string userName)
{
int? SignatureID = null; //Returns null if no date
//Validate if there is a new date
if ((IsNothing(oldSignUser) && newDate != null) || (oldSignUser != null && oldSignUser.DateSign != newDate))
{
Signature recSignature = Await db.Signature.FirstOrDefaultAsync(s => s.UserID == userName && s.DateSign == newDate);
if (IsNothing(recSignature))
{
recSignature = new Signature;
recSignature.UserID = userName;
recSignature.DateSign = newDate;
db.Signature.Add(recSignature);
}
SignatureID = recSignature.SignatureID;
}
else if (oldSignUser != null && newDate != null)
{ //If no change, keep old signature
SignatureID = oldSignUser.SignatureID;
}
return SignatureID;
}
I tried using the object instead of the Ids but it didn't work. I could insert the Signature beforehand but I would prefer having everything saved at once.
At this point, I assume there's a problem with my model or my approach.

Answer 1

All your objects have Id's of 0 after going through the constructor, because 0 is the default int value. Because the PK has to identify the object in the database, EF won't even let you insert multiple objects with the same id values, let alone trying to save those back to the database, because it does not know how to identify each tuple (this even is of relevance to EF, not only DBMS for consistency reasons, because EF uses the PK values as default concurrency tokens).
The way to solve this is to give all your objects distinct PK values. How you do it is up to you - in my project, we've let all objects that are meant to be saved to the database from one common base class, let's call it BOBase:
public abstract class BOBase
{
public long Id { get; set; }
public BOBase()
{
Id = DbTempId.Next();
}
}
internal static class DbTempId
{
private static long _next = -1;
internal static long Next()
{
return _next--;
}
}
Once you let all your BO's inherit from this base class (and don't hide the inherited property), the Id's of newly created objects will be -1, -2 and so on. We even use this to differentiate between new and to be updated objects - new objects will have a negative Id, while already existing objects will have a positive one.