使用泛型减少重复代码 - 接受两个泛型参数的方法答案

【问题标题】：Use generic to reduce repeated code - Method accepting two generic parameters使用泛型减少重复代码 - 接受两个泛型参数的方法
【发布时间】：2021-06-21 21:56:39
【问题描述】：

我有一个方法，它有一组重复的语句，它们要求重构，但我不知道如何泛型应该执行重复任务的方法的参数。

这是我的代码：

private void RegisterServices(IServiceCollection services)
{
    services.AddTransient<CommonApiMessageHandler>();

    var baseAddress = Configuration["ApiSetup:BaseAddress"];

    services.AddHttpClient<ILookupService, LookupService>("Api.Lookup", client =>
    {
        client.BaseAddress = new Uri(baseAddress);
    }).AddHttpMessageHandler<CommonApiMessageHandler>();

    services.AddHttpClient<IBusinessLookupService, BusinessLookupService>("Api.Lookup", client =>
    {
        client.BaseAddress = new Uri(baseAddress);
    }).AddHttpMessageHandler<CommonApiMessageHandler>();

    services.AddHttpClient<IApiUserService, ApiUserService>("Api.ApiUser", client =>
    {
        client.BaseAddress = new Uri(baseAddress);
    }).AddHttpMessageHandler<CommonApiMessageHandler>();

    services.AddHttpClient<IRequestService, RequestService>("Api.Request", client =>
    {
        client.BaseAddress = new Uri(baseAddress);
    }).AddHttpMessageHandler<CommonApiMessageHandler>();

    services.AddHttpClient<IResourceService, ResourceService>("Api.Resource", client =>
    {
        client.BaseAddress = new Uri(baseAddress);
    }).AddHttpMessageHandler<CommonApiMessageHandler>();

    services.AddHttpClient<IApproverTemplateService, ApproverTemplateService>("Api.ApproverTemplate", client =>
    {
        client.BaseAddress = new Uri(baseAddress);
    }).AddHttpMessageHandler<CommonApiMessageHandler>();

    services.AddHttpClient<IAttachmentService, AttachmentService>("Api.Attachment", client =>
    {
        client.BaseAddress = new Uri(baseAddress);
    }).AddHttpMessageHandler<CommonApiMessageHandler>();

    services.AddHttpClient<IDacBudgetService, DacBudgetServiceClient>("Api.DaCBudget", client =>
    {
        client.BaseAddress = new Uri(baseAddress);
    }).AddHttpMessageHandler<CommonApiMessageHandler>();
}

我觉得应该是这样的：

private static void NewMethod(IServiceCollection services, string baseAddress, Generic T, Generic U)
{
    services.AddHttpClient<T, U>("Api.Lookup", client =>
    {
        client.BaseAddress = new Uri(baseAddress);
    }).AddHttpMessageHandler<CommonApiMessageHandler>();
}

这可能吗？

【问题讨论】：

是的。应该是private static void NewMethod<T, U>(IServiceCollection services, string baseAddress, T serviceInterface, U serviceImpl) where U : T。您在尖括号中定义泛型参数，然后可以像使用类型一样使用它们。 where U : T 是一个约束，意味着 U（您的服务类）必须实现 T（您的服务接口），这样可以避免在您不小心尝试输入不兼容的类型时出错。
@ckuri 可能想将其作为您的答案，而不是评论。
@ckuri 我在使用serviceInterface & serviceImpl services.AddHttpClient<serviceInterface, serviceImpl>(name, client => { ... } 时出错：serviceInterface 是一个变量，但用作一个类型

标签： c# generics dependency-injection

【解决方案1】：

您不仅可以使用泛型，还可以将其定义为extension method，因此您可以像调用其他服务一样调用它。

public static class MyExtensions
{
    public static IServiceCollection AddMyService<TInterface,TImplementation>(this IServiceCollection services, string serviceName, string baseAddress)
        where TInterface: class 
        where TImplementation : class, TInterface
    {
        services.AddHttpClient<TInterface, TImplementation>(serviceName, client =>
        {
            client.BaseAddress = new Uri(baseAddress);
        }).AddHttpMessageHandler<CommonApiMessageHandler>();

        return services;
    }
}

然后这样称呼它：

services.AddMyService<IApiUserService, ApiUserService>("Api.ApiUser", baseAddress);
services.AddMyService<IRequestService, RequestService>("Api.Request", baseAddress);

//etc....

或者，由于我们返回 IServiceCollection，您可以将它们链接起来：

services
    .AddMyService<IApiUserService, ApiUserService>("Api.ApiUser", baseAddress)
    .AddMyService<IRequestService, RequestService>("Api.Request", baseAddress);

【讨论】：

扩展方法中需要有return语句吗？
已编辑。很高兴返回 IServiceCollection 以便调用者可以链接调用。
这是完美的。