Question

我试图跟踪一家公司但没有成功。

我为所有范围生成了一个访问令牌，它也适用于API的其他功能：

范围= w_messages + rw_company_admin + rw_nus + r_emailaddress + r_basicprofile + rw_groups + r_fullprofile + r_network + r_contactinfo

以下代码到达结尾并显示带有消息的MessageBox： “BadRequest，无法解析JSON公司文档。\ n请求正文：\ n \ n错误：\ nnull”

我很乐意帮忙。我已经完成了以下文档，但它没有准确地说明如何关注该公司，所以它给我留下了这个问题和示例如下： https://developer.linkedin.com/documents/company-follow-and-suggestions

        String companyID = "9288340";
        requestUrl = "https://api.linkedin.com/v1/people/~/following/companies/id=" + companyID + "?oauth2_access_token=MYTOKENGOESHERE";

        RestSharp.RestClient rc = new RestSharp.RestClient();
        RestSharp.RestRequest request = new RestSharp.RestRequest(requestUrl, RestSharp.Method.POST); //POST = Follow
        request.AddHeader("Content-Type", "application/json");
        request.AddHeader("x-li-format", "json");

        request.RequestFormat = RestSharp.DataFormat.Json;

        RestSharp.RestResponse restResponse = (RestSharp.RestResponse)rc.Execute(request);
        RestSharp.ResponseStatus responseStatus = restResponse.ResponseStatus;

        MessageBox.Show(restResponse.StatusCode.ToString() + "," + restResponse.Content.ToString());

Answer 1

经过大量测试后，我才能成功跟进公司。所以下面的代码正在运行：

    requestUrl = "https://api.linkedin.com/v1/people/~/following/companies?oauth2_access_token=MYTOKENGOESHERE";

                        var BODY = new
                        {
                            id = "9288340" //companyID 
                        };


    RestSharp.RestClient rc = new RestSharp.RestClient();
    RestSharp.RestRequest request = new RestSharp.RestRequest(requestUrl, RestSharp.Method.POST); //POST = Follow
    request.AddHeader("Content-Type", "application/json");
    request.AddHeader("x-li-format", "json");

                        request.RequestFormat = RestSharp.DataFormat.Json;
                        request.AddBody(BODY);


    RestSharp.RestResponse restResponse = (RestSharp.RestResponse)rc.Execute(request);
    RestSharp.ResponseStatus responseStatus = restResponse.ResponseStatus;

    MessageBox.Show(restResponse.StatusCode.ToString() + "," + restResponse.Content.ToString());

关注公司问题（LinkedIn API）

1 个答案: