asp.net-mvc - 我如何同时绑定(bind) FromQuery 和 FromRoute 参数？

Question

我需要同时支持基于查询参数的路由 (/api/models?id=1) 和基于路由的路由 (/api/models/1 >) 同时仍然允许明确访问模型集合 (/api/models)？

我的 Controller 看起来(有点)像这样:

[Route("/api/{controller}")]
public class ModelsController : Controller
{
    [HttpGet]
    public Models[] GetModels([FromQuery]QueryOptions queryOptions)
    {
        //...
    }    

    [HttpGet("{id:int}")]
    public Model Get([FromRoute] int id)
    {
        //...
    }

    [HttpGet("?{id:int}")]
    public Model Get2Try1([FromQuery] int id)
    {
       //Fails with ": The literal section '?' is invalid. 
       //Literal sections cannot contain the '?' character."
       //Which makes sense after some reading...
    }

    [HttpGet]
    public Model Get2Try2([FromQuery] int id)
    {
       //Fails with "AmbiguousActionException: Multiple actions matched. 
       //The following actions matched route data and had all constraints satisfied: 
       //GetModels and Get2Try2"
       //Which I think I understand as well...the absence of optional params
       //means ambiguous routing...
    }

    [HttpGet] //What here?
    public Model Get2Try3([FromQuery] int id) //and/or here?
    {

    }
}

我觉得应该有某种方法(通过声明式路由)来实现这一点。有没有人按照这些思路做过任何事情？

此外，当前代码库是 ASP.NET Core (RC1)，不久将升级到 RTM/1.0。任何一方的细节可能相似，但我对任何一方/两者都感兴趣。

Answer 1

我发现以下方法有效:

[HttpGet, Route("{id?}")]

...关键是'?'。您不需要方法签名中的任何 [FromX]，这可以解决问题并满足查询字符串和路由参数传递的需要。

不幸的是，Swagger UI 不喜欢它并期望一些显式参数开箱即用(https://github.com/domaindrivendev/Ahoy/issues/47 或 https://github.com/domaindrivendev/Ahoy/issues/182)，但这是另一回事了:)