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Pasted by Ivan Jensen

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SqlParameter paramTid = new SqlParameter();
                paramTid.ParameterName = "@Tid";
                paramTid.Value = System.DateTime.Now.ToLongTimeString();

                SqlParameter paramTEMP1 = new SqlParameter();
                paramTEMP1.ParameterName = "@Temp1";
                paramTEMP1.Value = temp1;

                SqlParameter paramTEMP2 = new SqlParameter();
                paramTEMP2.ParameterName = "@Temp2";
                paramTEMP2.Value = temp2;

                SqlParameter paramAlarm = new SqlParameter();
                paramAlarm.ParameterName = "@Alarm";
                paramAlarm.Value = a;

                SqlParameter paramBatteri = new SqlParameter();
                paramBatteri.ParameterName = "@Batteri";
                paramBatteri.Value = b;

                SqlParameter paramMax = new SqlParameter();
                paramMax.ParameterName = "@MaxTemp";
                paramMax.Value = H;

                SqlParameter paramMin = new SqlParameter();
                paramMin.ParameterName = "@MinTemp";
                paramMin.Value = L;

                SqlParameter paramDigi = new SqlParameter();
                paramDigi.ParameterName = "@Digi";
                paramDigi.Value = digi;

                SqlConnection conn = new SqlConnection(ConnString);

                string InsertStr = @"Insert into Home (Tid, Temp1, Temp2, Alarm, Batteri, MaxTemp, MinTemp, Digi) Values(@Tid, @Temp1, @Temp2, @Alarm, @Batteri, @MaxTemp, @MinTemp, @Digi)";

                SqlCommand cmdInsert = new SqlCommand(InsertStr, conn);

                conn.Open();
                cmdInsert.ExecuteNonQuery();
                conn.Close();


Følgende fejl bliver skrevet i logmessage:
10:05:53: Must declare the scalar variable "@Tid".
10:05:56: Must declare the scalar variable "@Tid".

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