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POIs it ok to add static reference to a service layer in Global.asax?
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  1. POIs it ok to add static reference to a service layer in Global.asax?
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    copied!<p>I am assessing the <a href="http://nuget.org/packages/Highway.Data.EntityFramework.Unity/3.0.3.2" rel="nofollow">Highway.Data.EntityFramework.Unity</a> package in the context of a single project ASP.NET Web Forms application.</p> <p>I want easy access to my service layer. Here is the relevant part of Global:</p> <pre><code>public class Global : HttpApplication { private static readonly IUnityContainer Container = new UnityContainer(); public static IEmployeeService Service { get; private set; } protected void Application_Start(object sender, EventArgs e) { Container.BuildHighway(); Container.RegisterType&lt;IMappingConfiguration, EntityMapping&gt;(); Container.RegisterType&lt;IEmployeeService, EmployeeService&gt;( new PerRequestLifetimeManager()); Container.RegisterType&lt;IRepository, Repository&gt;( new PerRequestLifetimeManager()); Container.RegisterType&lt;IDataContext, DataContext&gt;( new PerRequestLifetimeManager(), new InjectionConstructor("DataContext", new EntityMapping())); Service = Container.Resolve&lt;IEmployeeService&gt;(); } } </code></pre> <p>In my client, I can access like this:</p> <pre><code>this.Employees.DataSource = this.service.GetEmployees(); this.Employees.DataBind(); </code></pre> <p>Works fine but I haven't taken this approach before and just because it appears to be ok... well, is it? If not, what do I do?</p> <p>[Editited] For requested clarity.</p> <p>Service:</p> <pre><code>public class EmployeeService : IEmployeeService { private readonly IRepository repository; public EmployeeService(IRepository repository) { this.repository = repository; } public IEnumerable&lt;Employee&gt; GetEmployees() { return this.repository.Find( new AllEmployeesQuery()) .ToList() .Select(ObjectMapper.Map&lt;EmployeeEntity, Employee&gt;); } } </code></pre> <p>AllEmployeesQuery is a specification. Business object is mapped to EF entity by AutoMapper and vice-versa.</p> <p>Thanks, Richard</p>
 

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