Url.Link始终返回在asp..net web api中使用的null

问题描述 投票:2回答:2

我想为REST Api返回的DTO创建一些超链接。

变量url总是为空,我不知道为什么?!

为什么网址没有创建,我还缺少什么?

路线的名称是GetStatusFeedback,这是正确的,我也使用动作和控制器作为路线参数+ leadId参数!

public class ValuesController : ApiController
{
    [Route("")]
    [HttpGet]
    public IHttpActionResult Get()
    {
        var leadsFromDataBase = new List<Lead> { new Lead { Id = 1 }, new Lead { Id = 2 } };
        var leadDtos = new List<LeadDto>();
        foreach (var lead in leadsFromDataBase)
        {
            var leadDto = new LeadDto();
            string url = Url.Link("GetStatusFeedback", new { LeadId = lead.Id, Action = "Accept", Controller = "values"});
            leadDto.AcceptLink = new Link { Url = url, Verb = "Get" };


            leadDtos.Add(leadDto);
        }
        return Ok(leadDtos);
    }

    [Route("leads/{id:int}/statusfeedback", Name = "GetStatusFeedback")]
    [HttpPost]
    public void Accept(int leadId)
    {

    }
c# asp.net-web-api asp.net-web-api-routing
2个回答
4
投票

路由参数需要匹配路由模板中的预期参数

string url = Url.Link("GetStatusFeedback", new { id = lead.Id });

应该基于"leads/{id:int}/statusfeedback"路线模板匹配

路由模板占位符也需要匹配操作中的参数名称

[Route("leads/{id:int}/statusfeedback", Name = "GetStatusFeedback")]
[HttpPost]
public void Accept(int id) {

}

最后,您应该使用有助于描述路线的正确路线名称。


0
投票

我花了一段时间从接受的答案中得到我需要的东西(我的错误与OP不同),这是一个基于Nkosi答案和OP评论的简化版本:

public class ValuesController : ApiController
{
  [HttpGet]
  public IHttpActionResult Get()
  {
    string url = Url.Link("GetStatusFeedback", new { Id = 1, Action = "Accept", Controller = "values"});
    /*or*/ url = Url.Link(nameof(Accept), new { Id = 1,      Action = "Accept", Controller = "values"});
    //not  url = Url.Link(nameof(Accept), new { LeadId = 1,  Action = "Accept", Controller = "values"});

    return Ok(url);
  }

// Controller we want to get the URL of:
[HttpPost("leads/{id:int}/statusfeedback", Name = "GetStatusFeedback")]
public void Accept(int id) //param name matches the line above {id} (in OP it mismatched)
{

}
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