can I use it because the dependency Injections
As soon as you reference the DLL / project you can use all classes the same ways as if they were in the project.
To use a class as a service:
- Provide the service
- Inject the service
There's a lot of documentation available, so I'll keep this short:
// provide in startup.cs
services.AddTransient<IUserService, UserService>();
// Inject where you need it
MyConstructor(IUserService userService) {}
See https://learn.microsoft.com/en-us/aspnet/core/fundamentals/dependency-injection?view=aspnetcore-5.0
Provide Extension Method
If we take a look at other libs, most of them provide a method to setup the services.
Example: Entity framework core
public void ConfigureServices(IServiceCollection services)
{
services.AddDbContext<MyDbContext>(options => options.UseSqlServer(...));
}
So you could:
- In your lib, create an extension method for IServicesCollection that adds all services of your lib.
- In the consuming project, call
services.AddMyLibServices()
.
This could look like so:
public static class ServicesConfiguration
{
public static void AddDataLayer(this IServiceCollection services)
{
services.AddTransient<IUserService, UserService>();
// ... same for all services of your lib
}
}
Here's a tutorial with more details:
https://dotnetcoretutorials.com/2017/01/24/servicecollection-extension-pattern/
Lamar service registries
An optional and alternative approach are service registries. It's very similar to the extension methods but uses a class to do the setup. See https://jasperfx.github.io/lamar/documentation/ioc/registration/registry-dsl/
Composition Root
You may want to read about the composition root pattern, e.g. What is a composition root in the context of dependency injection?
In a simple app, your startup.cs is your composition root. In more complex apps, you could create a separate project to have a single place to configure your apps services.
Create the DLL
There are two ways to create the DLL:
- As a project in your solution (so your solution has multiple projects, each will result in a separate DLL)
- As a separate solution and as nuget package