How to pass session value to next page?

Question

I am using .net and I am changing my .aspx.cs file to add the user choice in radio button to a row in database.

 private DataSet CreateData()
    {
        DataTable table = new DataTable();
        table.Columns.Add(new DataColumn("OptionID", System.Type.GetType("System.String")));
        table.Columns.Add(new DataColumn("Option", System.Type.GetType("System.String")));

        DataRow row1 = table.NewRow();
        row1["OptionID"] = "1";
        row1["Option"] = "option 1";

        DataRow row2 = table.NewRow();
        row2["OptionID"] = "2";
        row2["Option"] = "two";

        DataRow row3 = table.NewRow();
        row3["OptionID"] = "3";
        row3["Option"] = "three";

        table.Rows.Add(row1);
        table.Rows.Add(row2);
        table.Rows.Add(row3);

        DataSet ds = new DataSet();
        ds.Tables.Add(table);

        return ds;

    }

    private void bindRadioList()
    {
        DataSet ds = CreateData();
        RadioButtonList1.DataSource = ds;
        RadioButtonList1.DataTextField = "Option";
        RadioButtonList1.DataValueField = "OptionID";
        RadioButtonList1.DataBind();

        if (RadioButtonList1.Items.Count > 0)
            RadioButtonList1.Items[0].Selected = true; //you can set a selected items you want.


    }

    public void Button1_Click(object sender, System.EventArgs e)
    {
        Session["choice"] = RadioButtonList1.SelectedValue;

    }

But I do not know how could I get this value at my last step when people click submit at the next page. I tried:

  string sqlIns = "INSERT INTO Choice (Email, Choice) VALUES (@Email, @Choice)";
             SqlConnection conn = new SqlConnection(ConfigurationSettings.AppSettings["connectionString"]);
                conn.Open();
            try
            {
                SqlCommand cmdIns = new SqlCommand(sqlIns, conn);
                cmdIns.Parameters.Add("@Email", userRecord.email);
                cmdIns.Parameters.Add("@Choice", (string)Session["choice"]);
                cmdIns.ExecuteNonQuery();

But I got this error:

Exception Details: System.Data.SqlClient.SqlException: The parameterized query '(@Email nvarchar(19),@Choice nvarchar(4000))INSERT INTO Choice (' expects the parameter '@Choice', which was not supplied.

Answer 1

I found out the problem and solution for this: The problem is Session["choice"] = Null and it is not acceptable under MSN standard. So I need to add this to prevent Null value :

   if ((string)Session["choice"] == null)
                {
                    cmdIns.Parameters.Add("@Choice", SqlDbType.NVarChar).Value = DBNull.Value;
                }
                else
                {
                   // cmdIns.Parameters.AddWithValue("@Choice", (string)Session["choice"]);
                    cmdIns.Parameters.Add("@Choice", SqlDbType.NVarChar).Value = (string)Session["choice"];
                }

To fix the null value, in .aspx file(I still can't figure out why the old code give null, but this one can avoid it)

   <div>                                                                                                                               
              <asp:RadioButton  ID="RadioButton1" GroupName="Group1" Text="Choice 1" OnCheckedChanged="Button1_Click" AutoPostBack="true" runat="server" />
              <asp:RadioButton  ID="RadioButton2" GroupName="Group1" Text="Choice 2" OnCheckedChanged="Button1_Click" AutoPostBack="true" runat="server" /> 
              <asp:RadioButton  ID="RadioButton3" GroupName="Group1" Text="Choice 3" OnCheckedChanged="Button1_Click" AutoPostBack="true" runat="server" />
        </div>

and .cs file:

protected void Button1_Click(object sender, System.EventArgs e)
        {
           // Session["choice"] = RadioButtonList1.SelectedItem.Text;
            if (RadioButton1.Checked)
            {
                Session["choice"] = "Choice 1";
            }

            if (RadioButton2.Checked)
            {
                Session["choice"] = "Choice 2";
            }
            if (RadioButton3.Checked)
            {
                Session["choice"] = "Choice 3";
            }

        }

Answer 2

Use AddWithValue instead of Add like;

cmdIns.Parameters.AddWithValue("@Email", userRecord.email);
cmdIns.Parameters.AddWithValue("@Choice", (string)Session["choice"]);

From it's documentation;

SqlParameterCollection.Add Method (String, Object) overload has been deprecated. Use AddWithValue(String parameterName, Object value) instead. http://go.microsoft.com/fwlink/?linkid=14202"

Or you can use (String, SqlDbType) overload (which I almost always prefer instead of AddWithValue) of Add method like (I assume both your column is NVarChar type);

cmdIns.Parameters.Add("@Email", SqlDbType.NVarChar).Value = userRecord.email;
cmdIns.Parameters.Add("@Choice", SqlDbType.NVarChar).Value = (string)Session["choice"];

Also use using statement to dispose your SqlConnection adn SqlCommand like;

using(SqlConnection conn = new SqlConnection(ConfigurationSettings.AppSettings["connectionString"]))
using(SqlCommand cmdIns = conn.CreateCommand())
{
   cmdIns.CommandText = sqlIns;
   ...
   ...
}

Answer 3

please check your store procedure where you are passing paramerters there you are not passing parameters @Choice