Using StructureMap to create classes by a name?

Posted by Bevan on Stack Overflow See other posts from Stack Overflow or by Bevan
Published on 2010-03-31T22:21:18Z Indexed on 2010/03/31 22:23 UTC
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How can I use StructureMap to resolve to an appropriate implementation of an interface based on a name stored in an attribute?

In my project, I have many different kinds of widgets, each descending from IWidget, and each decorated with an attribute specifying the kind of associated element.

To illustrate:

[Configuration("header")]
public class HeaderWidget : IWidget { }

[Configuration("linegraph")]
public class LineGraphWidget : IWidget { }

When processing my (XML) configuration file, I want to obtain an instance of the appropriate concrete class based on the name of the element I'm processing.

public IWidget CreateWidget(XElement definition)
{
    var kind = definition.Name.LocalName;
    var widget = // What goes here?
    widget.Configure(definition);
    return widget;
}

Each definition should result in a different widget being created - I don't need or want the instances to be shared.

In the past I've written plenty of code to do this kind of thing manually, including writing a custom "roll-your-own" IoC container for one project. However, one of my goals with this project is to become proficient with StructureMap instead of reinventing the wheel.

I think I've already managed to set up automatic scanning of assemblies so that StructureMap knows about all my IWidget implementations:

public class WidgetRegistration : Registry
{
    public WidgetRegistration()
    {
        Scan(
            scanner =>
            {
                scanner.AssembliesFromApplicationBaseDirectory();
                scanner.AddAllTypesOf<IWidget>();
            });
    }
}

However, this isn't registering the names of my widgets with StructureMap. What do I need to add to make my scenario work?

(While I am trying to use StructureMap in this project, an answer showing me how to solve this problem with a different DI/IoC tool would still be valuable.)

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