I see that I can create a List<string> list and bind it to a GridView
control (by setting the DataSource property to the list and calling the
DataBind method).

What's the trick to doing the same thing with my own classes instead of a
simple string? Is there an interface or something I can implement so that
the GridView control can detect the properties of my object and display them
?

Thanks.

--
Jonathan Wood
SoftCircuits Programming
http://www.softcircuits.com

You can just set the DataSource to be a List<MyObject> and DataBind. You
won't get any help with setting up the columns at design time if you do it
that way - you'll have type in all the details. If you want design time
help consider using an ObjectDataSource that points to a method that returns
the list.

Nick



"Jonathan Wood" <(E-Mail Removed)> wrote in message
news:%(E-Mail Removed)...
>I see that I can create a List<string> list and bind it to a GridView
>control (by setting the DataSource property to the list and calling the
>DataBind method).
>
> What's the trick to doing the same thing with my own classes instead of a
> simple string? Is there an interface or something I can implement so that
> the GridView control can detect the properties of my object and display
> them ?
>
> Thanks.
>
> --
> Jonathan Wood
> SoftCircuits Programming
> http://www.softcircuits.com
>

Nick,

> You can just set the DataSource to be a List<MyObject> and DataBind. You
> won't get any help with setting up the columns at design time if you do it
> that way - you'll have type in all the details. If you want design time
> help consider using an ObjectDataSource that points to a method that
> returns the list.

Yeah, that's how it's done with a List<string> so I thought that might work.
But, with my class defined like this:

public class ClientMenuItem
{
public int MealNum;
public float Substitutions;
public string Group;
public float Units;
public string Measure;
public string Description;
public float Calories;
public float Protein;
public float Carbohydrate;
public float Fat;
}

I tried the following:

List<ClientMenuItem> items = ClientUsers.GetMenuItems(menus[0].MenuID);
GridView1.DataSource = items;
GridView1.DataBind();

And I get the following error at runtime on the last line above:

"The data source for GridView with id 'GridView1' did not have any
properties or attributes from which to generate columns. Ensure that your
data source has content."

--
Jonathan Wood
SoftCircuits Programming
http://www.softcircuits.com

The message is correct: your class doesn't have any properties - it has
public fields, which isn't a good thing. Try replacing your fields with
properties - e.g. convert

public int MealNum;

to

public int MealNum { get; set; }

if you're in VS2008, or this

private int _mealNum;

public int MealNum
{
get { return _mealNum; }
set { _mealNum = value; }
}

if in VS2005 or VS2003.





"Jonathan Wood" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> Nick,
>
>> You can just set the DataSource to be a List<MyObject> and DataBind. You
>> won't get any help with setting up the columns at design time if you do
>> it that way - you'll have type in all the details. If you want design
>> time help consider using an ObjectDataSource that points to a method that
>> returns the list.
>
> Yeah, that's how it's done with a List<string> so I thought that might
> work. But, with my class defined like this:
>
> public class ClientMenuItem
> {
> public int MealNum;
> public float Substitutions;
> public string Group;
> public float Units;
> public string Measure;
> public string Description;
> public float Calories;
> public float Protein;
> public float Carbohydrate;
> public float Fat;
> }
>
> I tried the following:
>
> List<ClientMenuItem> items = ClientUsers.GetMenuItems(menus[0].MenuID);
> GridView1.DataSource = items;
> GridView1.DataBind();
>
> And I get the following error at runtime on the last line above:
>
> "The data source for GridView with id 'GridView1' did not have any
> properties or attributes from which to generate columns. Ensure that your
> data source has content."
>
> --
> Jonathan Wood
> SoftCircuits Programming
> http://www.softcircuits.com
>

That's the deal? I thought about that. Actually, I was putting all
properties in my classes until I decided it was WAAAAAAAAAAAAY too much
typing and just seemed unnecessary.

Yup, that's it! Nice. Cool feature.

Thanks.

--
Jonathan Wood
SoftCircuits Programming
http://www.softcircuits.com

"Nick Bennett" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> The message is correct: your class doesn't have any properties - it has
> public fields, which isn't a good thing. Try replacing your fields with
> properties - e.g. convert
>
> public int MealNum;
>
> to
>
> public int MealNum { get; set; }
>
> if you're in VS2008, or this
>
> private int _mealNum;
>
> public int MealNum
> {
> get { return _mealNum; }
> set { _mealNum = value; }
> }
>
> if in VS2005 or VS2003.
>
>
>
>
>
> "Jonathan Wood" <(E-Mail Removed)> wrote in message
> news:(E-Mail Removed)...
>> Nick,
>>
>>> You can just set the DataSource to be a List<MyObject> and DataBind. You
>>> won't get any help with setting up the columns at design time if you do
>>> it that way - you'll have type in all the details. If you want design
>>> time help consider using an ObjectDataSource that points to a method
>>> that returns the list.
>>
>> Yeah, that's how it's done with a List<string> so I thought that might
>> work. But, with my class defined like this:
>>
>> public class ClientMenuItem
>> {
>> public int MealNum;
>> public float Substitutions;
>> public string Group;
>> public float Units;
>> public string Measure;
>> public string Description;
>> public float Calories;
>> public float Protein;
>> public float Carbohydrate;
>> public float Fat;
>> }
>>
>> I tried the following:
>>
>> List<ClientMenuItem> items = ClientUsers.GetMenuItems(menus[0].MenuID);
>> GridView1.DataSource = items;
>> GridView1.DataBind();
>>
>> And I get the following error at runtime on the last line above:
>>
>> "The data source for GridView with id 'GridView1' did not have any
>> properties or attributes from which to generate columns. Ensure that
>> your data source has content."
>>
>> --
>> Jonathan Wood
>> SoftCircuits Programming
>> http://www.softcircuits.com
>>
>
>

Howdy,

You don't need to type property declaration yourself. VS >= 2005 supports
refactoring and includes very useful commands to make your life easier:
Three ways to quickly implement a property:
1. Press Ctrl+K,X or go to main menu Edit->Intelli Sense->Insert Snippet
select "prop" or Visual C#\prop from drop down list, and change property's
type and name.
2. in the code type
private [AnyType] fieldName;
(Ctrl+R,E) or right click -> Refactor ->Encapsulate Field.
3. Create class diagram, and design your classes through a GUI.

Hope this helps
--
Milosz


"Jonathan Wood" wrote:

> That's the deal? I thought about that. Actually, I was putting all
> properties in my classes until I decided it was WAAAAAAAAAAAAY too much
> typing and just seemed unnecessary.
>
> Yup, that's it! Nice. Cool feature.
>
> Thanks.
>
> --
> Jonathan Wood
> SoftCircuits Programming
> http://www.softcircuits.com
>
> "Nick Bennett" <(E-Mail Removed)> wrote in message
> news:(E-Mail Removed)...
> > The message is correct: your class doesn't have any properties - it has
> > public fields, which isn't a good thing. Try replacing your fields with
> > properties - e.g. convert
> >
> > public int MealNum;
> >
> > to
> >
> > public int MealNum { get; set; }
> >
> > if you're in VS2008, or this
> >
> > private int _mealNum;
> >
> > public int MealNum
> > {
> > get { return _mealNum; }
> > set { _mealNum = value; }
> > }
> >
> > if in VS2005 or VS2003.
> >
> >
> >
> >
> >
> > "Jonathan Wood" <(E-Mail Removed)> wrote in message
> > news:(E-Mail Removed)...
> >> Nick,
> >>
> >>> You can just set the DataSource to be a List<MyObject> and DataBind. You
> >>> won't get any help with setting up the columns at design time if you do
> >>> it that way - you'll have type in all the details. If you want design
> >>> time help consider using an ObjectDataSource that points to a method
> >>> that returns the list.
> >>
> >> Yeah, that's how it's done with a List<string> so I thought that might
> >> work. But, with my class defined like this:
> >>
> >> public class ClientMenuItem
> >> {
> >> public int MealNum;
> >> public float Substitutions;
> >> public string Group;
> >> public float Units;
> >> public string Measure;
> >> public string Description;
> >> public float Calories;
> >> public float Protein;
> >> public float Carbohydrate;
> >> public float Fat;
> >> }
> >>
> >> I tried the following:
> >>
> >> List<ClientMenuItem> items = ClientUsers.GetMenuItems(menus[0].MenuID);
> >> GridView1.DataSource = items;
> >> GridView1.DataBind();
> >>
> >> And I get the following error at runtime on the last line above:
> >>
> >> "The data source for GridView with id 'GridView1' did not have any
> >> properties or attributes from which to generate columns. Ensure that
> >> your data source has content."
> >>
> >> --
> >> Jonathan Wood
> >> SoftCircuits Programming
> >> http://www.softcircuits.com
> >>
> >
> >
>
>

Sorry for the slow delay but the Encapsulate Field command looks like the
one I was trying to find.

Thanks!

--
Jonathan Wood
SoftCircuits Programming
http://www.softcircuits.com

"Milosz Skalecki [MCAD]" <(E-Mail Removed)> wrote in message
news:EB31650A-EEBC-4A5D-B56B-(E-Mail Removed)...
> Howdy,
>
> You don't need to type property declaration yourself. VS >= 2005 supports
> refactoring and includes very useful commands to make your life easier:
> Three ways to quickly implement a property:
> 1. Press Ctrl+K,X or go to main menu Edit->Intelli Sense->Insert Snippet
> select "prop" or Visual C#\prop from drop down list, and change property's
> type and name.
> 2. in the code type
> private [AnyType] fieldName;
> (Ctrl+R,E) or right click -> Refactor ->Encapsulate Field.
> 3. Create class diagram, and design your classes through a GUI.
>
> Hope this helps
> --
> Milosz
>
>
> "Jonathan Wood" wrote:
>
>> That's the deal? I thought about that. Actually, I was putting all
>> properties in my classes until I decided it was WAAAAAAAAAAAAY too much
>> typing and just seemed unnecessary.
>>
>> Yup, that's it! Nice. Cool feature.
>>
>> Thanks.
>>
>> --
>> Jonathan Wood
>> SoftCircuits Programming
>> http://www.softcircuits.com
>>
>> "Nick Bennett" <(E-Mail Removed)> wrote in message
>> news:(E-Mail Removed)...
>> > The message is correct: your class doesn't have any properties - it has
>> > public fields, which isn't a good thing. Try replacing your fields
>> > with
>> > properties - e.g. convert
>> >
>> > public int MealNum;
>> >
>> > to
>> >
>> > public int MealNum { get; set; }
>> >
>> > if you're in VS2008, or this
>> >
>> > private int _mealNum;
>> >
>> > public int MealNum
>> > {
>> > get { return _mealNum; }
>> > set { _mealNum = value; }
>> > }
>> >
>> > if in VS2005 or VS2003.
>> >
>> >
>> >
>> >
>> >
>> > "Jonathan Wood" <(E-Mail Removed)> wrote in message
>> > news:(E-Mail Removed)...
>> >> Nick,
>> >>
>> >>> You can just set the DataSource to be a List<MyObject> and DataBind.
>> >>> You
>> >>> won't get any help with setting up the columns at design time if you
>> >>> do
>> >>> it that way - you'll have type in all the details. If you want
>> >>> design
>> >>> time help consider using an ObjectDataSource that points to a method
>> >>> that returns the list.
>> >>
>> >> Yeah, that's how it's done with a List<string> so I thought that might
>> >> work. But, with my class defined like this:
>> >>
>> >> public class ClientMenuItem
>> >> {
>> >> public int MealNum;
>> >> public float Substitutions;
>> >> public string Group;
>> >> public float Units;
>> >> public string Measure;
>> >> public string Description;
>> >> public float Calories;
>> >> public float Protein;
>> >> public float Carbohydrate;
>> >> public float Fat;
>> >> }
>> >>
>> >> I tried the following:
>> >>
>> >> List<ClientMenuItem> items =
>> >> ClientUsers.GetMenuItems(menus[0].MenuID);
>> >> GridView1.DataSource = items;
>> >> GridView1.DataBind();
>> >>
>> >> And I get the following error at runtime on the last line above:
>> >>
>> >> "The data source for GridView with id 'GridView1' did not have any
>> >> properties or attributes from which to generate columns. Ensure that
>> >> your data source has content."
>> >>
>> >> --
>> >> Jonathan Wood
>> >> SoftCircuits Programming
>> >> http://www.softcircuits.com
>> >>
>> >
>> >
>>
>>