【问题标题】:The operation has timed out in windows formsWindows 窗体中的操作已超时
【发布时间】:2015-09-26 14:12:47
【问题描述】:

我在windows服务和windows应用程序中使用了这个代码,我强制执行“操作已超时”,但是当我在网站中使用这个代码时,我从来没有强制异常,有什么问题?!

string con = System.Configuration.ConfigurationManager.ConnectionStrings["TelegramServiceConnectionString"].ConnectionString;
        SqlConnection cn = new SqlConnection(con);
        SqlDataAdapter da = new SqlDataAdapter("SELECT  SH_Message_Sent.phone, SH_Message_Sent.text, SH_Message_Sent.update_id, FaranegarApiUsers.chat_id, FaranegarApiUsers.phone AS Expr1 FROM         SH_Message_Sent INNER JOIN FaranegarApiUsers ON SH_Message_Sent.phone = FaranegarApiUsers.phone  where SH_Message_Sent.phone='09127218155' order by update_id asc ", cn);
        DataSet ds = new DataSet();
        da.FillSchema(ds, SchemaType.Source);
        da.Fill(ds);
        DataTable t = ds.Tables[0];
        foreach (DataRow row in t.Rows)
        {
            WebRequest request2 = WebRequest.Create("https://api.telegram.org/bot99452812:AAE2MntQnStPr_J2KmrOsp_gvGZLZNsy3mE/sendMessage?chat_id=" + row[3] + "&text=" + row[1]);
            request2.Timeout = 20000;
            WebResponse response2 = request2.GetResponse();

        }

【问题讨论】:

    标签: asp.net service ado.net httprequest telegram-bot


    【解决方案1】:

    我修改了我的代码,它的工作正常 WebRequest request2 = WebRequest.Create("@987654321@" + row[3] + "&text=" + row[1]); request2.Timeout = 20000; WebResponse response2 = request2.GetResponse(); response2.Close();

    【讨论】:

      猜你喜欢
      • 1970-01-01
      • 2023-03-30
      • 2013-01-26
      • 1970-01-01
      • 1970-01-01
      • 2012-09-21
      • 2011-08-27
      • 1970-01-01
      • 1970-01-01
      相关资源
      最近更新 更多