Hi there:
I created an ASP.Net project. I added and implemented a component
DataAcess.vb and there are several methods such as

Namespace ABC.Components
Public Class DBAccess
Public Function MakeInParam(...) As SqlParameter

...

End Function

Public Sub Open()


...

End Sub

End Class

End Namespace

In another component (Login.vb) in the same namespace, I have included the
namespace ABC.Components, and the "New" key word is followed with a dropdown
list that contains the class name 'DBAccess', But non of these methods such
as DBAccess.MakeInParam is shown in the other component such as Login.vb

Dim data As New DBAccess 'this line is okay

'this line causes a syntax error: type DBAccess.makeinparam is not defined:

Dim data As New DBAccess.makeinparam



What's wrong? How to fix this problem?

Thanks!

CM

"CM" <(E-Mail Removed)> schrieb
> Hi there:
> I created an ASP.Net project. I added and implemented a component
> DataAcess.vb and there are several methods such as
>
> Namespace ABC.Components
> Public Class DBAccess
> Public Function MakeInParam(...) As SqlParameter
>
> ...
>
> End Function
>
> Public Sub Open()
>
>
> ...
>
> End Sub
>
> End Class
>
> End Namespace
>
> In another component (Login.vb) in the same namespace, I have
> included the namespace ABC.Components, and the "New" key word is
> followed with a dropdown list that contains the class name
> 'DBAccess', But non of these methods such as DBAccess.MakeInParam is
> shown in the other component such as Login.vb
>
> Dim data As New DBAccess 'this line is okay
>
> 'this line causes a syntax error: type DBAccess.makeinparam is not
> defined:
>
> Dim data As New DBAccess.makeinparam
>
>
>
> What's wrong? How to fix this problem?


The new keyword creates a new object. After "New", you write the type name.
DBAccess.makeinparam is not a type, it is a member of a type. You can create
instances of the DBAccess class and use the members of the created object.
See the OOP (object oriented programming) sections in the VB.NET docs.


--
Armin

Hi CM,

What are you trying to do?

MakeInParam is a method, not a class, so you cannot create an instance of
it... Are you trying to call that method?

If so, all you need is:

data.MakeInParam(...)

--
HTH,
-- Tom Spink, Über Geek

Please respond to the newsgroup,
so all can benefit

" System.Reflection Master "

==== Converting to 2002 ====
Remove inline declarations


"CM" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> Hi there:
> I created an ASP.Net project. I added and implemented a component
> DataAcess.vb and there are several methods such as
>
> Namespace ABC.Components
> Public Class DBAccess
> Public Function MakeInParam(...) As SqlParameter
>
> ...
>
> End Function
>
> Public Sub Open()
>
>
> ...
>
> End Sub
>
> End Class
>
> End Namespace
>
> In another component (Login.vb) in the same namespace, I have included the
> namespace ABC.Components, and the "New" key word is followed with a
dropdown
> list that contains the class name 'DBAccess', But non of these methods
such
> as DBAccess.MakeInParam is shown in the other component such as Login.vb
>
> Dim data As New DBAccess 'this line is okay
>
> 'this line causes a syntax error: type DBAccess.makeinparam is not
defined:
>
> Dim data As New DBAccess.makeinparam
>
>
>
> What's wrong? How to fix this problem?
>
> Thanks!
>
> CM
>
>

Hi CM,

Dim data As New DBAccess 'this line is okay

This line <is> ok because 'As' must be followed by a Type and DBAccess
<is> a Type.

Dim data As New DBAccess.makeinparam

But this line will cause a problem because DBAccess.makeinparam is <not> a
Type - it's a method of an instance of a Type.

However,
Dim data As New DBAccess
Dim SquealyPram As SqlParameter
SquealyPram = data.makeinparam (...)
will work, because data is an object, and makeinparam is an object
method.and they like each other just fine.

Regards,
Fergus

"CM" <(E-Mail Removed)> scripsit:
> I created an ASP.Net project. I added and implemented a component
> DataAcess.vb and there are several methods such as
>
> Namespace ABC.Components
> Public Class DBAccess
> Public Function MakeInParam(...) As SqlParameter
>
> ...
>
> End Function
>
> Public Sub Open()
>
>
> ...
>
> End Sub
>
> End Class
>
> End Namespace
>
> In another component (Login.vb) in the same namespace, I have included the
> namespace ABC.Components, and the "New" key word is followed with a dropdown
> list that contains the class name 'DBAccess', But non of these methods such
> as DBAccess.MakeInParam is shown in the other component such as Login.vb
>
> Dim data As New DBAccess 'this line is okay
>
> 'this line causes a syntax error: type DBAccess.makeinparam is not defined:
>
> Dim data As New DBAccess.makeinparam

'MakeInParam' is a method, not a type. You can _call_ the method, but
you cannot instantiate it.

\\\
Dim c As New DBAccess()
Dim p As ... = c.MakeInParam(...)
///

--
Herfried K. Wagner
MVP · VB Classic, VB.NET
<http://www.mvps.org/dotnet>

Thanks for reply.
I think you are all right but this line still not working:

Dim p As ... = c.MakeInParam(...)

Even I type
DBAccess. 'I expect the "open" method will popup in the list, but no.

Nothing follows the dot.

CM


"Herfried K. Wagner [MVP]" <hirf-spam-me-(E-Mail Removed)> wrote in message
news:blvbee$grop3$(E-Mail Removed)...
> "CM" <(E-Mail Removed)> scripsit:
> > I created an ASP.Net project. I added and implemented a component
> > DataAcess.vb and there are several methods such as
> >
> > Namespace ABC.Components
> > Public Class DBAccess
> > Public Function MakeInParam(...) As SqlParameter
> >
> > ...
> >
> > End Function
> >
> > Public Sub Open()
> >
> >
> > ...
> >
> > End Sub
> >
> > End Class
> >
> > End Namespace
> >
> > In another component (Login.vb) in the same namespace, I have included
the
> > namespace ABC.Components, and the "New" key word is followed with a
dropdown
> > list that contains the class name 'DBAccess', But non of these methods
such
> > as DBAccess.MakeInParam is shown in the other component such as Login.vb
> >
> > Dim data As New DBAccess 'this line is okay
> >
> > 'this line causes a syntax error: type DBAccess.makeinparam is not
defined:
> >
> > Dim data As New DBAccess.makeinparam
>
> 'MakeInParam' is a method, not a type. You can _call_ the method, but
> you cannot instantiate it.
>
> \\\
> Dim c As New DBAccess()
> Dim p As ... = c.MakeInParam(...)
> ///
>
> --
> Herfried K. Wagner
> MVP · VB Classic, VB.NET
> <http://www.mvps.org/dotnet>

"CM" <(E-Mail Removed)> schrieb
> Thanks for reply.
> I think you are all right but this line still not working:
>
> Dim p As ... = c.MakeInParam(...)

p As what? What's the declaration of "c"?

> Even I type
> DBAccess.

Where do you type DBAccess?

> 'I expect the "open" method will popup in the list, but
> no.
>
> Nothing follows the dot.



--
Armin

"CM" <(E-Mail Removed)> scripsit:
> I think you are all right but this line still not working:
>
> Dim p As ... = c.MakeInParam(...)
>
> Even I type
> DBAccess. 'I expect the "open" method will popup in the list, but no.
>
> Nothing follows the dot.

Please post more code. Which error/exception do you get?

--
Herfried K. Wagner
MVP · VB Classic, VB.NET
<http://www.mvps.org/dotnet>

Thanks for all everyone's help!

I've checked some error message:

Dim Data1 as new DBAccess 'this line is fine.

This line causes an error( I'll handle it later) : [the value of
'data.MakeInParam("@userID", SqlDbType.Char, 80, userID)' cannot be
converted to one demensional array of
ystem.data.sqlClient.SqlParameter() ]:
SqlParameter() = data.MakeInParam("@userID", SqlDbType.Char, 80, userID)

But for this line (begin at new line) there is nothing following the dot:

DBAccess. 'nothing following the dot

If I put this line:
DBAccess.MakeInParam() 'An error: Reference to a non-shard member requires
an object references.

It is seemed the problem is "non-shard", but sorry, I don't know what this
mean.
Please continue your help!

Thanks!

CM


"Herfried K. Wagner [MVP]" <hirf-spam-me-(E-Mail Removed)> wrote in message
news:blvhdn$gtfrm$(E-Mail Removed)...
> "CM" <(E-Mail Removed)> scripsit:
> > I think you are all right but this line still not working:
> >
> > Dim p As ... = c.MakeInParam(...)
> >
> > Even I type
> > DBAccess. 'I expect the "open" method will popup in the list, but no.
> >
> > Nothing follows the dot.
>
> Please post more code. Which error/exception do you get?
>
> --
> Herfried K. Wagner
> MVP · VB Classic, VB.NET
> <http://www.mvps.org/dotnet>

Hi CM,

SqlParameter() = data.MakeInParam (....)

It's confusing because () can be used with a method which takes no
arguments and with variables that are arrays. In this case the compiler thinks
that SqlParameter() is an array. As MakeInParam only returns a single
SqlParameter you don't need (). But you do need to have defined a variable.

Dim SqlParam As SqlParameter
SqlParam = data.MakeInParam (....)

Have another read of the replies that we've given you. I've already shown
you something very similar to this example.

===========================

DBAccess.

What are you expecting to appear ?? What do you want to do with DBAccess
at this point?

Regards,
Fergus