Question

我有以下内容将JSON发布到C＃Web API：

submitForm(data: any): Observable<Response> {
    return this.http.post(
        'https://localhost:44396/api/PostNewComputer/AddItem/?=', data,
        { headers: new Headers({ 'Content-Type': 'application/x-www-form-urlencoded' }) }).subscribe(data => { return "test"; });
}

API控制器：

public class PostNewComputerController : ApiController
    {
        [HttpPost]
        public IHttpActionResult AddItem(HttpRequestMessage request)
        {

            //var dynamicObject = Json.Decode(jsonString);
            var jsonString = request.GetQueryNameValuePairs();

            string message;

            using (SqlConnection con = new SqlConnection("data source = MYSERVER; initial catalog = AssetDB; integrated security = SSPI; persist security info = True; Trusted_Connection = Yes"))
            {
                using (SqlCommand cmd = new SqlCommand("POST_NEW_COMPUTER", con) { CommandType = CommandType.StoredProcedure })
                {
                    try
                    {
                        cmd.Parameters.Add(new SqlParameter("@JSON_TEXT", SqlDbType.NVarChar)).Value = jsonString;

                        con.Open();
                        cmd.ExecuteNonQuery();

                        con.Close();

                        message = "Item Successfully Added";
                    }
                    catch (Exception e)
                    {
                        //throw e;
                        message = e.Message;
                    }

                }

                return Ok(message);
            }
        }
    }

我正在尝试从POST调用创建一个响应消息，然后当用户提交表单以显示帖子已经工作时，该消息被反馈到网页。

然而，这个当前的设置不起作用，我正在努力解决这个问题。

有正确的方法可以做到这一点吗？

Answer 1

使用HttpResponseMessage将讯息回发为：

[HttpPost]
[ResponseType(typeof(string))]
public HttpResponseMessage AddItem(HttpRequestMessage request)
{
    HttpResponseMessage response = null;
    string message = string.Empty;

    try 
    {
        .....

        message = "Item Successfully Added";
    }
    catch (Exception e)
    {
        .....

        message = e.Message;
    }

    response = Request.CreateResponse<string>(HttpStatusCode.OK, message);
    return response;
}

Answer 2

问题出在你的Angular服务中。您的API是正确的

在将映射回映射到subscribe之前，您尝试json()来自API的响应，您需要执行以下操作，请参阅以下内容：

首先，这应该是您的组件中的内容：

submitForm(data: any): Observable<Response> {
  //this url can't be correct?
  let url:string = 'https://localhost:44396/api/PostNewComputer/AddItem/?=';

  return this.http.post(url, data, this._headers())
    .map((res:Response) => res.json());
}

//Created a method to do the headers for you
private _headers():RequestOptionsArgs {   
  let headers:Headers = new headers();    
  headers.append('Content-Type', 'application/json');

  let options:RequestOptionsArgs = new RequestOptions();
  options.headers = headers;

  return options;
}

然后，您可以按如下方式从您的组件中调用您的服务：

public functionName():void {
  this.service.submitForm(data)
    .subscribe((res:any) => {
      //this will contain your response
    },
    (error:any) => {
      //this will contain your error
    });
}

您需要确保您发布的模型与预期的模型匹配。关于您要发布的网址，这似乎不正确，是不是https://localhost:44396/api/PostNewComputer/AddItem/？

从JSON帖子

2 个答案: