Class object from XML.

Question

Hi.
I want to create DataGrid with itemReneder for each column.
This is inline generated itemRenderer:
DataGridColumn.itemRenderer = new
ClassFactory(MyCustomClass);
// MyCustomClass - it manually generated class which contains
some form items(TextInput, Combox..e.t.c.)
I want to consturct ItemRenderer dynamicly. For this i need
to create Class object and send it to itemRenderer:
var component:DynamicFormClass = new DynamicFormClass();
var newComponent:Class =
component.someMethodWhichWillCreateForm(configData); // - this will
return Class which will generate form items that described in
configData
DatagridColumn.itemRenderer = new ClassFactory(newComponent);
This code is how i see solvetion of this problem, but this
doesnt work.
I really don't know how it to create and will be very
thankful for any dicision and answers how to make it.
Thank you.

Answer 1

"Dmitry Sergeev" <[email protected]> wrote
in message
news:g8eghj$881$[email protected]..
> Hi.
> I want to create DataGrid with itemReneder for each
column.
>
> This is inline generated itemRenderer:
> DataGridColumn.itemRenderer = new
ClassFactory(MyCustomClass);
> // MyCustomClass - it manually generated class which
contains some form
> items(TextInput, Combox..e.t.c.)
>
> I want to consturct ItemRenderer dynamicly. For this i
need to create
> Class
> object and send it to itemRenderer:
> var component:DynamicFormClass = new DynamicFormClass();
> var newComponent:Class =
> component.someMethodWhichWillCreateForm(configData);
> // - this will return Class which will generate form
items that described
> in
> configData
> DatagridColumn.itemRenderer = new
ClassFactory(newComponent);
Use getDefinitionByName and ClassFactory to dynamically
create your class.
Just be sure that you actually have at least one "hard"
reference to each
class you intend to use, or the class might not get compiled
into your swf.
Here's a full write-up of how to do that:
http://www.paulofierro.com/archives/520/
HTH;
Amy