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我在 Windows 服务和 Windows 应用程序中使用了此代码,并且我强制执行“操作已超时”,但是当我在网站中使用此代码时,我从未强制执行异常,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();

        }
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1 回答 1

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我修改了我的代码并且它的工作正常 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(); response2.Close();

于 2015-09-28T11:03:15.390 回答