In place of parameter "Data" i put the alias i want? i tryed with
"Data" parameter and i got this exception: "Index was outside the
bounds of the array."
I got the SQL generated and executed on SQL Server and it´s ok.
I tryed with the "imagem.Imagem", being "imagem" my alias and
".Imagem" my property and i got an exception from Database saying in
portuguese: "O identificador de várias partes "imagem.Imagem" não pôde
ser associado." ("The multi-part identifier" imagem.Imagem "could not
be associated with.").

Thx for answer.

On 24 jun, 08:56, Kminoru <[email protected]> wrote:
> Hey Diego, the string "dl" is alias of some property or just the name
> of the column from answer?
>
> On 23 jun, 15:50, Diego Mijelshon <[email protected]> wrote:
>
>
>
> > .SetProjection(
> >                Projections.SqlProjection("sum(datalength(Data)) dl",
>
> >                                          new[] {"dl"},
>
> >                                          new[] {NHibernateUtil.Int64}))
>
> >    Diego
>
> > On Wed, Jun 23, 2010 at 14:54, Kminoru <[email protected]> wrote:
> > > i did
> > > this:
> > > .SetProjection(Projections.Alias(Projections.SqlProjection("sum(cast(datale­­ngth(imagem.Imagem)
> > > as bigInt))", New String() {"imagem.Imagem"}, types), "total")) and
> > > got this exception mensage:"O identificador de várias partes
> > > "imagem.Imagem" não pôde ser associado."
>
> > > Translated by google: "The handle multi-part "imagem.Imagem" could not
> > > be associated"
> > >  What´s wrong?
>
> > > On 23 jun, 11:21, Diego Mijelshon <[email protected]> wrote:
> > > > You can use Projections.SqlProjection or .SqlGroupProjection to access
> > > > arbitrary functions.
>
> > > >    Diego
>
> > > > On Wed, Jun 23, 2010 at 09:37, Kminoru <[email protected]> wrote:
> > > > > Hello, in my application, i need select de sum of length of some
> > > > > objects. These objects are varbinary type. I found that SQL SERVER has
> > > > > the function datalength that return it. How can i do this using
> > > > > criteria?
> > > > > Thanks for all.
>
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