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Pasted by Ivan Jensen
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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