Question

我编写了以下代码，我正在寻求改进它。使用Unity DI，我需要注册具体的类以及它们实现的接口。该方法的优点是无需在添加时继续注册新类。

我已将代码缩减为


    IUnityContainer container = new UnityContainer();
            // from
            // Factories
            // container.RegisterType();

            // Services
            // container.RegisterType();
            // container.RegisterType();
            //......
            // to

    var assemblies = AppDomain.CurrentDomain.GetAssemblies()
        .Where(x => x.GetName().Name.Contains("Romabravo.ApplicationManager"));

    foreach (var assembly in assemblies)
    {
        var types = assembly.GetExportedTypes();
        foreach (Type t in types)
        {
            var interfaces = t.GetInterfaces().Where(x => x.FullName != null &&
            x.FullName.Contains("Romabravo.ApplicationManager")).ToList();

                foreach (var item in interfaces)
                {
                    var interfaceImplementers = AppDomain.CurrentDomain.GetAssemblies()
                    .SelectMany(s => s.GetTypes())
                    .Where(p => item.GetTypeInfo().IsAssignableFrom(p) && 
                    item.FullName.Contains("Romabravo.ApplicationManager"))
                    .Select(x => x).Where(x => !x.IsInterface && !x.IsAbstract).ToList();

                    if (interfaceImplementers != null)
                    {
                        foreach (var implementer in interfaceImplementers)
                        {
                            container.RegisterType(item, implementer);
                        }
                    }
                }
            }
        }

    return container;

只是想方设法进一步优化它。

Answer 1

如果您使用的是Unity 3，那么您可以使用registration by convention来简化代码：

container.RegisterTypes(
    AllClasses.FromLoadedAssemblies()
      .Where(a => a.FullName.Contains("Romabravo.ApplicationManager")),
    WithMappings.FromMatchingInterface);

将接口映射到具体类

1 个答案: