c# - ActionFilterAttribute 中的 GetService 返回 null

Question

我创建了一个 Action 过滤器来根据传入的 id 检查 id 是否正确。
过滤器的构造方法接受一个服务类型参数。
在 ActionMethod 运行之前，我需要服务来检查此 ID。但是 GetService()方法返回空值。
Action 过滤器

 public class ContainsFilterAttribute : ActionFilterAttribute
{
    private Type _service;
    private Type _entity;
    private IdSections _idSections;
    public ContainsFilterAttribute(Type service, Type entity, IdSections idSections)
    {
        if (!typeof(ICommonService).IsAssignableFrom(service) || !typeof(IEntity).IsAssignableFrom(entity))
        {
            throw new System.Exception("Service or entity undefined.");
        }
        _service = service;
        _entity = entity;
        _idSections = idSections;
    }
    public override void OnActionExecuting(ActionExecutingContext context)
    {
        //returns null service
        var service = (ICommonService)context.HttpContext.RequestServices.GetService(_service);
        var entity = (IEntity)Activator.CreateInstance(_entity);
        
        if (IdSections.FromRoute == _idSections)
        {
            entity.Id = (int)context.ActionArguments.FirstOrDefault(pair => pair.Key.ToUpper().Contains("ID")).Value;
        }
        var res = service.Contains(entity);
    }
}

操作方法

    [HttpGet]
    [Route("{userId}/user-channels")]
    [ContainsFilter(typeof(UserManager), typeof(User), IdSections.FromRoute)]
    public IActionResult GetUserChannels(int userId)
    {
        var result = _channelService.GetUserChannels(userId);
        if (result.IsSuccessful)
        {
            var mapResult = _mapper.Map<List<ChannelForListDto>>(result.Data);
            return Ok(mapResult);
        }
        return this.ServerError(result.Message);
    }

注塑

services.AddScoped<IUserService, UserManager>();

Answer 1

IUserService注册为服务

services.AddScoped<IUserService, UserManager>();

但您正在尝试解决实现 UserManager在显示的属性中

 [ContainsFilter(typeof(UserManager), typeof(User), IdSections.FromRoute)]

称呼

[ContainsFilter(typeof(IUserManager), typeof(User), IdSections.FromRoute)] 

反而