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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T20:11:53+00:00 2026-05-11T20:11:53+00:00

I’m struggling with the Entity Framework when creating a new record in a one

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I’m struggling with the Entity Framework when creating a new record in a one to many relation. I’ve the following tables defined:

Person
PersonInternetAddresses
InternetAddresses

The Entity Framework has created the EntityObjects Person with a navigation property named PersonInternetAddresses and InternetAddresses with a navigation property to persons.

To create a new InternetAddresses record i’ve created a form with a couple of text fields and a bindingsource to InternetAddresses.

Under my “new” button i’ve placed the following code:

internetAddressesBindingSource.AddNew();            
(internetAddressesBindingSource.Current as InternetAddresses).Id = Guid.NewGuid();
(internetAddressesBindingSource.Current as InternetAddresses).Persons = InternetAddresses.Persons;

On the last line i’m getting the following error:

The EntityCollection has already been initialized. The InitializeRelatedCollection method should only be called to initialize a new EntityCollection during deserialization of an object graph.

I’m guessing that i cannot set the person through the navigation property of the InternetAddresses which would mean that the Entity Framework itself would create a PersonInternetAddresses record. Since there isn’t a PersonsInternetAddresses EntityObject in the model i’m pretty stuck!

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  1. Editorial Team
    Editorial Team
    2026-05-11T20:11:53+00:00Added an answer on May 11, 2026 at 8:11 pm

    Found it!

    Persons is of type EntityCollection duh!
    For testing purposes i’ve changed the code to:

    internetAddressesBindingSource.AddNew();            
            (internetAddressesBindingSource.Current as InternetAddresses).Id = Guid.NewGuid();            
    
            foreach(Person p in InternetAddresses.Persons)
            {
                (internetAddressesBindingSource.Current as InternetAddresses).Persons.Add(p);
            }
    
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