How to get a distinct result with nHibernate and QueryOver API?

Question

I have this Repository method

    public IList<Message> ListMessagesBy(string text, IList<Tag> tags, int pageIndex, out int count, out int pageSize)
    {
        pageSize = 10;
        var likeString = string.Format("%{0}%", text);
        var query = session.QueryOver<Message>()
            .Where(Restrictions.On<Message>(m => m.Text).IsLike(likeString) || 
            Restrictions.On<Message>(m => m.Fullname).IsLike(likeString));

        if (tags.Count > 0)
        {
            var tagIds = tags.Select(t => t.Id).ToList();
            query
                .JoinQueryOver<Tag>(m => m.Tags)
                .WhereRestrictionOn(t => t.Id).IsInG(tagIds);
        }            

        count = 0;
        if(pageIndex < 0)
        {
            count = query.ToRowCountQuery().FutureValue<int>().Value;
            pageIndex = 0;
        }
        return query.OrderBy(m => m.Created).Desc.Skip(pageIndex * pageSize).Take(pageSize).List();
    }

You supply a free text search string and a list of Tags. The problem is that if a message has more then one tag it is listed duplicated times. I want a distinct result based on the Message entity. I've looked at

Projections.Distinct

But it requires a list of Properties to to the distinct question on. This Message is my entity root there most be a way of getting this behaviour without supplying all of the entity properties?

Thanks in advance, Anders

Leader

Answer 1

If you're using the ICriteria API, you need:

.SetResultTransformer(new DistinctEntityRootTransformer())

If you're using the QueryOver API, you need:

.TransformUsing(Transformers.DistinctRootEntity)

But beware, this all occurs on client side, so all the duplicate rows are still pulled from SQL database.

Nationwide

The ICriteria API is not very fluent with QUeryOver, but .TransformUsing(Transformers.DistinctRootEntity); works perfect. – Leader

.TransformUsing(Transformers.DistinctRootEntity) Did not work at all with Paging :/ Any other ideas? – Leader

It won't work with paging. For paging queries you will need to use Projections – Nationwide

Does this mean you cannot select to an alias AND do a distinct (as they are both transformers and you only seem to be able to use one transformer)? – Pantograph

I get regular upvotes on this question so I revisit it sometimes. Are there no better solution today? In EF this just works. (They have other problems though) – Leader

Answer 2

Try something like this

public IPagedList<Client> Find(int pageIndex, int pageSize)
{
    Client clientAlias = null;

    var query = Session.QueryOver<Client>(() => clientAlias)

        .Select(
            Projections.Distinct(
                Projections.ProjectionList()
                    .Add(Projections.Property<Client>(x => x.Id).As("Id"))
                    .Add(Projections.Property<Client>(x => x.Name).As("Name"))
                    .Add(Projections.Property<Client>(x => x.Surname).As("Surname"))
                    .Add(Projections.Property<Client>(x => x.GivenName).As("GivenName"))
                    .Add(Projections.Property<Client>(x => x.EmailAddress).As("EmailAddress"))
                    .Add(Projections.Property<Client>(x => x.MobilePhone).As("MobilePhone"))
            )
        )
        .TransformUsing(Transformers.AliasToBean<Client>())

        .OrderBy(() => clientAlias.Surname).Asc
        .ThenBy(() => clientAlias.GivenName).Asc;

    var count = query
        .ToRowCountQuery()
        .FutureValue<int>();

    return query
        .Take(pageSize)
        .Skip(Pagination.FirstResult(pageIndex, pageSize))
        .List<Client>()
        .ToPagedList(pageIndex, pageSize, count.Value);
}

Debutante

Works but... tedious. Instead of As("xxx") you can call the WithAlias extension method instead for a little less magic strings. Thanks – Pantograph

WithAlias extension method example: .WithAlias(() => entity.PropertyName) – Quarterstaff

Answer 3

You can use SelectList and GroupBy, e.g:

tags.SelectList(t => t.SelectGroup(x => x.Id))

Should work and produce the same query plan as distinct.

If you need multiple items in the group, do something like:

tags.SelectList(t => t.SelectGroup(x => x.Id)
                      .SelectGroup(x => x.Name)
               )

Skitter

Answer 4

I have recently created a method to apply select distinct based on a mapped object type. It applies this to an IQueryOver object (property of class). Method also has access to the nhibernate config. You could add these as method parameters. Needs work for production, but method is working great in dev, only used it for one entity so far.

This method was created because I am trying to page my data at the server level and a distinct result transformer would not work.

After you get your object collection (query.List()) you may have to reload the objects to populate one to many child objects. Many to one mappings will be proxied for lazy loads.

 public void DistinctRootProjectionList<E>()
    {
        var classMapping = Context.Config.GetClassMapping(typeof(E));
        var propertyIterator = classMapping.UnjoinedPropertyIterator;
        List<IProjection> projections = new List<IProjection>();
        ProjectionList list = Projections.ProjectionList();

        list.Add(Projections.Property(classMapping.IdentifierProperty.Name), classMapping.IdentifierProperty.Name);

        foreach (var item in propertyIterator)
        {
            if (item.Value.IsSimpleValue || item.Value.Type.IsEntityType)
            {
                list.Add(Projections.Property(item.Name), item.Name);
            }
        }
        query.UnderlyingCriteria.SetProjection(Projections.Distinct(list));
        query.TransformUsing(Transformers.AliasToBean<E>());
    }

Code I used to load one to many relations... T is the entity type.

for (int i = 0; i < resp.Data.Count; i++)
        {
            resp.Data[i] = session.Load<T>(GetInstanceIdValue(resp.Data[i]));
        }

Stairway