47

I am trying to update an entity with a FK relationship in EntityFramework 4.3 Code First. I try to attach to the related entites by calling: Entry(item).State = EntityState.Unchanged

I get the following exception: An object with the same key already exists in the ObjectStateManager. The ObjectStateManager cannot track multiple objects with the same key.

I do not update these items nor have an id property for them on my main entity. Is it possible to know which entities are attached or not ?

Thanks in advance, Radu

Radu Negrila
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  • You can use ObjectContext.ObjectStateManager.GetObjectStateEntries() method (http://msdn.microsoft.com/en-us/library/bb738497.aspx) to get all entities that are tracked by the context. Note that this is possible that there are entities in the database that are not tracked by ObjectContext. You need to provide a state to let the ObjectStateManager know what entities you need (this enums has Flags attribute so you can combine different values using |) – Pawel Apr 08 '12 at 07:20
  • Does this answer your question? [what is the most reasonable way to find out if entity is attached to dbContext or not?](https://stackoverflow.com/questions/6033390/what-is-the-most-reasonable-way-to-find-out-if-entity-is-attached-to-dbcontext-o) (the accepted answer here literally links to the potential dupe, which, in this case, has the bonus of also being older than this question.) – ruffin Oct 08 '21 at 18:58

4 Answers4

90

You can find the answer here.

public bool Exists<T>(T entity) where T : class
{
    return this.Set<T>().Local.Any(e => e == entity);
}

Place that code into your context or you can turn it into an extension like so.

public static bool Exists<TContext, TEntity>(this TContext context, TEntity entity)
    where TContext : DbContext
    where TEntity : class
{
    return context.Set<TEntity>().Local.Any(e => e == entity);
}
John Bowen
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Tri Q Tran
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    Thanks. This also helped me. Another note with the function above. You need to put where T : class, or else the compiler will complain. – Sylpheed Jan 13 '14 at 18:47
  • @Sylpheed Thanks, I have updated the answer as per your suggestion. – Tri Q Tran Jan 30 '14 at 01:59
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    TContext doesn't seem necessary. Let the first parameter be of type DbContext instead - public static bool Exists(this DbContext context, TEntity entity)... – Peter Hedberg Feb 13 '14 at 12:39
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    @Palpie The answer I provided was for EF 4.5 so DbContext didn't exist at the time. Good suggestion though. – Tri Q Tran Feb 23 '14 at 21:53
  • Sorry, my bad. The question says Entity Framework 4.3! – Peter Hedberg Feb 24 '14 at 10:38
  • The referenced [answer](http://stackoverflow.com/a/6037277/177570) describes a bunch of different methods for different situations--click through! (@TriQ- thanks for linking it!) – Michael Haren Jun 18 '14 at 12:07
  • @Palpie I should have mentioned that this solution applies for EF 4.5 and earlier, so it will work for 4.3 too. – Tri Q Tran Jun 19 '14 at 00:56
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    The "where TContext : DbContext, TEntity : class" did not work on my system and had to be replaced with "where TContext : DbContext where TEntity : class". AKA, replaced the comma with a "where" – Atron Seige Oct 16 '15 at 12:49
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    Just keep in mind that `Local` can also contain new entities that are tracked by the context, but are not in the database yet. If you're updating and inserting entities simultaneously, using this method to determine if an entity is attached before updating it might yield unexpected results. – Rudey Dec 28 '19 at 10:47
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    Not conclusive. If the entity is another *instance* of an attached entity, `Exists(entity)` returns false but it can not be attached. A robust check always requires a check on key values. – Gert Arnold Nov 05 '21 at 13:15
11

You can use this method:

    /// <summary>
    /// Determines whether the specified entity key is attached.
    /// </summary>
    /// <param name="context">The context.</param>
    /// <param name="key">The key.</param>
    /// <returns>
    ///   <c>true</c> if the specified context is attached; otherwise, <c>false</c>.
    /// </returns>
    internal static bool IsAttached(this ObjectContext context, EntityKey key)
    {
        if (key == null)
        {
            throw new ArgumentNullException("key");
        }

        ObjectStateEntry entry;
        if (context.ObjectStateManager.TryGetObjectStateEntry(key, out entry))
        {
            return (entry.State != EntityState.Detached);
        }
        return false;
    }

For example:

     if (!_objectContext.IsAttached(entity.EntityKey))
        {
            _objectContext.Attach(entity);
        }
VictorV
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    I did some performance testing and (surprisingly) found ObjectStateManager.TryGetObjectStateEntry was over 70 times slower than Set().Local.Any( – Brent Jul 09 '14 at 23:34
1

If you have arrived here from an EF Core Lazy Loading scenario in which Navigation properties were filled in a data layer via DbSet<>.Include() clause(s) while the Entity was attached to a DbContext and then that Entity was detached and passed up to a business layer, consider adding something like this to your DbContext.OnConfiguring(DbContextOptionsBuilder optionsBuilder) method: optionsBuilder.ConfigureWarnings(warn => warn.Ignore(CoreEventId.LazyLoadOnDisposedContextWarning)); The error will be ignored and the values that were originally Include()d will be returned.

KramFfud
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0

I use this extension method since I needed to check tracking based on values, not based on instance

internal static class DBExtensions
    {
        internal static bool IsAttached<TEntity>(this DbSet<TEntity> dbSet, Func<TEntity, bool> condition) where TEntity : class
        {
            return dbSet.Local.Any(condition);
        }
    }

Usage:

 if (!context.Items.IsAttached(y => y.ItemId == item.ItemId))
    {
        context.Items.Attach(item);
    }
Beingnin
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  • No, the accepted answer does exactly what is expected. It is important to know that a specific instance is attached, otherwise you may start modifying another unattached instance (that gives `IsAttached` = `true`) and wonder why changes aren't saved. – Gert Arnold Jan 25 '23 at 16:15
  • @GertArnold Feels like that is a point as well. Modified the answer as it is just a solution for specific use case – Beingnin Jan 26 '23 at 00:51