编写了一个控制台应用程序,以使用RestSharp推送数据,但是当我尝试创建Windows Service时,它不起作用

时间:2018-09-17 18:43:35

标签: c# windows-services restsharp

我的任务是创建一个Windows服务,该服务使用RestSharp将数据推送到API。我已经创建了满足我需要的控制台应用程序,但是当我尝试将其转换为Windows Service时,该方法无法运行。我什至无法提交错误。我已附上以下代码:

    protected override void OnStart(string[] args)
    {
        timer1 = new Timer();
        this.timer1.Interval = 150000;
        this.timer1.Elapsed += new System.Timers.ElapsedEventHandler(this.timer1_Tick);
        timer1.Enabled = true;

        Log("Service Started successfully with Grovo-Test with Grovo in Try catch2");

        try
        {
            Grovo();
        }
        catch (Exception e)
        {
            Log(e.Message);
        }
    }

    public void Grovo()
    {
        var seconds = DateTime.Now.Second.ToString();
        var grovolist = new List<GrovoModel>();
        GrovoModel model = new GrovoModel()
        {
            employeeId = "testing",
            firstName = "testing" + seconds,
        };

        var jsonobject = "";

        foreach (var json in grovolist)
        {
            var post = JsonConvert.SerializeObject(json);
            jsonobject += post;
            jsonobject += "\n";
        }

        var clientrest = new RestClient("http://public-api.grovo.com/users/batch-sync");
        var request = new RestRequest(Method.POST);
        request.AddHeader("Postman-Token", "5fd35576-9ee6-4a65-8389-40a3b7eb1e8c");
        request.AddHeader("Cache-Control", "no-cache");
        request.AddHeader("content-type", "application/x-ndjson");
        request.AddHeader("x-grovo-onboarding-option", "email");
        request.AddHeader("x-api-key", "5Uss0T9T7b3Cdy04sGGkf7DFF7RYhPXV8mau11wh");
        //request.AddParameter("application/x-ndjson", postBody, ParameterType.RequestBody);
        request.AddParameter("application/x-ndjson", jsonobject, ParameterType.RequestBody);

        IRestResponse response = clientrest.Execute(request);
    }

Log()方法有效,每次服务循环通过时,该方法都会添加到本地文本文件中,但Grovo()则不会。我目前正在尝试在本地计算机上运行此程序,因此我在本地系统下有服务ProcessInstaller帐户。

1 个答案:

答案 0 :(得分:0)

您的json对象为空字符串。

var grovolist = new List<GrovoModel>();

foreach (var json in grovolist)
        {
            var post = JsonConvert.SerializeObject(json);
            jsonobject += post;
            jsonobject += "\n";
        }

您永远不会在grovolist中加载任何内容。您创建了模型,但没有将其添加到列表中。