Web-Form not accepting responses(Asp.net) - html

Question

Recently i am working on a web-form written in asp.net framework. I have kept in mind to include !IsPostBack() but yet my form is not accepting the responses and updating the database accordingly.
The body of the .aspx code which basically contains 4 RadioButtonList and one textbox. Here is the code for one of the RadioButtonList and the textBox, the submit button and and a link button:
<li class="form-line jf-required" data-type="control_scale" id="id_24">
<label class="form-label form-label-top form-label-auto" id="label_24" for="input_24">
Do you feel the training provided value for money/time?
<span class="form-required">*
</span>
</label>
<asp:RadioButtonList ID="RadioButtonList4" runat="server" AutoPostBack="True" CellPadding="10" CellSpacing="10" RepeatDirection="Horizontal" Font-Names="Comic Sans MS" Font-Size="Large" Height="50px">
<asp:ListItem Value="1" class="radiobuttonlist1">Worse</asp:ListItem>
<asp:ListItem Value="2" class="radiobuttonlist2">Boring</asp:ListItem>
<asp:ListItem Value="3" class="radiobuttonlist3">Satisfactory</asp:ListItem>
<asp:ListItem Value="4" class="radiobuttonlist4">good</asp:ListItem>
<asp:ListItem Value="5" class="radiobuttonlist5">Fantastic</asp:ListItem>
</asp:RadioButtonList>
</li>
<li id="cid_27" class="form-input-wide" data-type="control_head">
<div class="form-header-group header-default">
<div class="header-text httal htvam">
<h2 id="header_27" class="form-header" data-component="header">Final Thoughts?
</h2>
</div>
</div>
</li>
<li class="form-line" data-type="control_textarea" id="id_45">
<label class="form-label form-label-top form-label-auto" id="label_45" for="input_45">REMARKS / SUGGESTIONS</label>
<div id="cid_45" class="form-input-wide" data-layout="full">
<asp:TextBox ID="TextBox1" runat="server" style="width: 40px; height: 6px"></asp:TextBox>
</div>
</li>
<li id="cid_28" class="form-input-wide" data-type="control_head">
<div align="center">
<asp:Button ID="Button1" runat="server" Text="Submit" Text-align="center" CssClass="btn btn-primary" OnClick="Button1_Click" BackColor="#66FFCC" Height="40px" Width="80px" />
</div>
<div>
<asp:LinkButton ID="LinkButton1" runat="server" CssClass="btn btn-link" OnClick="LinkButton1_Click">Go back to Home Page</asp:LinkButton>
<br />
<br />
</div>
Here is the .aspx.cs of Page_Load:
public MySqlConnection mycon;
public MySqlCommand MyCm, cmd1;
public MySqlDataAdapter myad;
string userId;
protected void Page_Load(object sender, EventArgs e)
{
userId= Session["newFeedback"].ToString();
try
{
mycon = new MySqlConnection("server=localhost;database=feedbackform;user id=abcd;password=abcdabcd;");
}
catch (Exception ex)
{
Response.Write(ex.ToString());
}
}
Here is the .aspx.cs code for Button_Click:
protected void Button1_Click(object sender, EventArgs e)
{
if (!IsPostBack)
{
if (Check_feedback_Exists())
{
Response.Write("<script>alert('You have already filled feedback form')</script>");
}
else
{
try
{
if (mycon.State == ConnectionState.Closed)
{
mycon.Open();
}
MySqlCommand cmd = new MySqlCommand("insert into feedback (Reference_Number, Feed1,Feed2,Feed3,Feed4,Remarks_Suggestion)values(userId, RadioButtonList1.SelectedItem.Text,RadioButtonList2.SelectedItem.Text,RadioButtonList3.SelectedItem.Text,RadioButtonList4.SelectedItem.Text,Textbox1.Text);", mycon);
cmd.ExecuteNonQuery();
mycon.Close();
Response.Write("<script> alert ('hello success'); </Script>");
}
catch (Exception ex)
{
Response.Write("<script> alert ('" + ex.Message + "'); </Script>");
}
}
}
}
Here is the code for the Check_feedback_exists:
bool Check_feedback_Exists()
{
try
{
if (mycon.State == ConnectionState.Closed) //checking that the connection between database and our project is open or not if not then open it
{
mycon.Open();
}
MySqlCommand cmd = new MySqlCommand("select * from feedback where Reference_Number = '" + userId + "' ; ", mycon);
MySqlDataReader dr = cmd.ExecuteReader();
if (dr.HasRows)
{
return true;
}
else
{
return false;
}
}
catch (Exception ex)
{
Response.Write("<script> alert ('" + ex.Message + "'); </script>");
return false;
}
}
The check_feedback_exists function is working well since if i submit for an already existing feedback, it gives the alert message: 'You have already filled feedback form' as coded in the Button1_click function. But if i enter a new feedback for a new user and click on submit, it doesn't enter the database, neither shows the success alert. Here is the database description in which I am inserting these values:

Answer 1

Makes no sense to check !IsPostBack in a button click. Such a button click will ALWAYS be a post-back. So, that's confusing.
And same goes for your sql insert. Your trying to insert the "text" of the control.
eg this:
string strSQL =
"insert into feedback (Reference_Number, Feed1, Feed2, Feed3, Feed4, Remarks_Suggestion) " +
"VALUES (#userId, #RB1,#RB2,#RB3,#RB4,#TBOX1"
SqlCommand cmd = new SqlCommand(strSQL, conn);
cmd.Parameters.Add("#usewrId", SqlDbType.Int).Value = userid;
cmd.Parameters.Add("#RB1", SqlDbType.NVarChar).Value = RadioButtonList1.SelectedItem.Text;
cmd.Parameters.Add("#RB2", SqlDbType.NVarChar).Value = RadioButtonList2.SelectedItem.Text;
etc. etc. etc.
Now, I used the sql provider, - it looks like your using MySQL. But, the sytnax should be simular to above. So, for your insert, you can't use a string of
string strSQL = "textBox1.Text, TextBox2.Text"
Since that string will have the above textbox names as string.
You would need
string strSQL = "VALUES ('" + TextBox1.Text + "','" + TextBox2.Text + "')"
But, the above is dangerous - due to sql injection. So use the above parameters approach.
And remove the !IsPostBack from your button stubb - not required, and a button click is ALWAYS a post-back anyway.