Include several references on the second level entity-framework-core


Assume we have this model :

public class Tiers
    public List<Contact> Contacts { get; set; }


public class Contact
    public int Id { get; set; }
    public Tiers Tiers { get; set; }
    public Titre Titre { get; set; }
    public TypeContact TypeContact { get; set; }
    public Langue Langue { get; set; }
    public Fonction Fonction { get; set; }
    public Service Service { get; set; }
    public StatutMail StatutMail { get; set; }

With EF7 I would like to retrieve all data from the Tiers table, with data from the Contact table, from the Titre table, from the TypeContact table and so on ... with one single instruction. With Include/ThenInclude API I can write something like this :

     .Include(tiers => tiers.Contacts)
          .ThenInclude(contact => contact.Titre)

But after Titre property, I can't include others references like TypeContact, Langue, Fonction ... Include method suggests a Tiers objects, and ThenInclude suggests a Titre object, but not a Contact object. How can I include all references from my list of Contact? Can we achieve this with one single instruction?

3/30/2018 6:58:55 AM

Accepted Answer

.ThenInclude() will chain off of either the last .ThenInclude() or the last .Include() (whichever is more recent) to pull in multiple levels. To include multiple siblings at the same level, just use another .Include() chain. Formatting the code right can drastically improve readability.

    .Include(tiers => tiers.Contacts).ThenInclude(contact => contact.Titre)
    .Include(tiers => tiers.Contacts).ThenInclude(contact => contact.TypeContact)
    .Include(tiers => tiers.Contacts).ThenInclude(contact => contact.Langue);
    // etc.
3/30/2018 7:01:23 AM

Popular Answer

For completeness' sake:

It is also possible to include nested properties directly via Include in case they are not collection properties like so:

    .Include(tier => tier.Contact.Titre)
    .Include(tier => tier.Contact.TypeContact)
    .Include(tier => tier.Contact.Langue);

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