Must declare the scalar variable “@Format”

Question

if (CheckBox1.Checked==false)
       {
           String strConnString = ConfigurationManager.ConnectionStrings["CallcenterConnectionString"].ConnectionString;
           SqlConnection con = new SqlConnection(strConnString);
           SqlCommand cmd = new SqlCommand("insert into CallCenter..Loy_DispMstr (CallType, SUBFormat, Disposition, SubDisposition) values (@CallType, @Format, @Disposition, @SubDisposition)", con);
           cmd.Parameters.AddWithValue("CallType" , SqlDbType.VarChar).Value = ddlCalltype.SelectedItem.Value;
           cmd.Parameters.AddWithValue("SUBFormat", SqlDbType.VarChar).Value = ddlFormat.SelectedItem.Value;
           cmd.Parameters.AddWithValue("Disposition", SqlDbType.VarChar).Value = ddlDisp.SelectedItem.Value;
           cmd.Parameters.AddWithValue("SubDisposition", SqlDbType.VarChar).Value = ddlSubdisp.SelectedItem.Value;
           con.Open();
           int i = cmd.ExecuteNonQuery();
           con.Close();  
           Label2.Text = " Your data is been saved in the database";
           Label2.ForeColor = System.Drawing.Color.ForestGreen;

       }
       else if(flag==0 && CheckBox1.Checked==true)
       {
           String strConnString = ConfigurationManager.ConnectionStrings["CallcenterConnectionString"].ConnectionString;
           SqlConnection con = new SqlConnection(strConnString);
           SqlCommand cmd = new SqlCommand("insert into CallCenter..Loy_DispMstr (CallType, SUBFormat,Disposition, SubDisposition) values (@CallType, @Format,@Disposition, @SubDisposition)", con);
           cmd.Parameters.AddWithValue("CallType", SqlDbType.VarChar).Value= ddlCalltype.Text;
           cmd.Parameters.AddWithValue("SUBFormat", SqlDbType.VarChar).Value= ddlFormat.Text;
           cmd.Parameters.AddWithValue("Disposition", SqlDbType.VarChar).Value= TextBox1.Text;
           cmd.Parameters.AddWithValue("SubDisposition", SqlDbType.VarChar).Value= TextBox2.Text;

           con.Open();
           int i = cmd.ExecuteNonQuery();//error on this
           con.Close();  


       }   
   }

the error is on the executenonquery the error is on the executenonquery

the error is on the executenonquery

the error is on the executenonquery

1. Change AddWithValue to Add. 2. add @ to the start of the parameters names. 3. You declared a parameter called @Format but added a parameter called SubFormat to your parameters collection. — Zohar Peled
@ZoharPeled Please do not provide answers as comments. There are several negative impacts that this behavior creates. — mickmackusa

RAJNIK PATEL RAJNIK PATEL · Accepted Answer · 2017-04-18T12:01:22

Try this it will work for you :

if (CheckBox1.Checked == false)
{
    String strConnString = ConfigurationManager.ConnectionStrings["CallcenterConnectionString"].ConnectionString;
    SqlConnection con = new SqlConnection(strConnString);
    SqlCommand cmd = new SqlCommand("insert into CallCenter..Loy_DispMstr (CallType, SUBFormat, Disposition, SubDisposition) values (@CallType, @Format, @Disposition, @SubDisposition)", con);
    cmd.Parameters.Add("@CallType", ddlCalltype.SelectedItem.Value);
    cmd.Parameters.Add("@Format", ddlFormat.SelectedItem.Value);
    cmd.Parameters.Add("@Disposition", ddlDisp.SelectedItem.Value);
    cmd.Parameters.Add("@SubDisposition", ddlSubdisp.SelectedItem.Value);
    con.Open();
    int i = cmd.ExecuteNonQuery();
    con.Close();
    Label2.Text = " Your data is been saved in the database";
    Label2.ForeColor = System.Drawing.Color.ForestGreen;

}
else if (flag == 0 && CheckBox1.Checked == true)
{
    String strConnString = ConfigurationManager.ConnectionStrings["CallcenterConnectionString"].ConnectionString;
    SqlConnection con = new SqlConnection(strConnString);
    SqlCommand cmd = new SqlCommand("insert into CallCenter..Loy_DispMstr (CallType, SUBFormat,Disposition, SubDisposition) values (@CallType, @Format,@Disposition, @SubDisposition)", con);
    cmd.Parameters.Add("@CallType", ddlCalltype.Text);
    cmd.Parameters.Add("@Format", ddlFormat.Text);
    cmd.Parameters.Add("@Disposition", TextBox1.Text);
    cmd.Parameters.Add("@SubDisposition", TextBox2.Text);

    con.Open();
    int i = cmd.ExecuteNonQuery();//error on this
    con.Close();
}

Must declare the scalar variable “@Format”

1 Answers