在Ninject中注入相同类型的多个参数

Question

Lets say we have a class X like this:

class X
{
   X(Z a, Z b)
   { }
}

And the Z class would have a Y dependency:

class Z
{
   Z(Y c)
   { }
}

What's the proper way to bind these classes such that two instances of Z, each of them with a different instance of Y, get injected into X?

I know these have to do with Context Binding, but I'm not sure how to go about it.

EDIT:

The Y class would be:

class Y
{
    Y(string someString)
    { }
}

I want the two instances of Y with a different string as well.

Thanks

Answer 1

You could used Named Bindings . Other types of contextual bindings can be found here .

Bind<X>().To<XA>().Named("A");
Bind<X>().To<XB>().Named("B");

public class Z {
    public Z([Named("A")] X x1, [Named("B")] X x2) {}
}

Answer 2

If your bindings for X and Y are not scoped, then each time an X object is requested from the kernel, it will get constructed with two unique Z objects (each having a unique instance of Y). This will happen by default.

Answer 3

pass your dependencies that implement the same interface as an IEnumerable. Otherwise, Ninject will yell at you if you have multiple bindings for one interface. Use the .WithConstructorArgument() method to define your string. If the string is not constant, you can use .ToMethod() instead and have the string determined at runtime.

public class Z
{
    public Z(IEnumerable<Y> dependencies)
    {
        if (dependencies == null) throw new ArgumentNullException("dependencies");
        _dependencies = dependencies;
    }
}

static void Main() // or other entry point
{
    var kernel = new StandardKernel();
    kernel.Bind<Z>().To<ZImplementation>();
    kernel.Bind<Y>().To<YImplementation1>().WithConstructorArgument("c", "string to inject 1");
    kernel.Bind<Y>().To<YImplementation2>().WithConstructorArgument("c", "string to inject 2");
}