Solved

How to insert a new row to gridview by adding footer row columns in the code behind.

Posted on 2013-01-04
3
2,021 Views
Last Modified: 2013-01-05
Hi,
I am adding the columns to GridView in code behind and binding it to Data Table.
private void BindGridView(DataTable SQLDt)
        {

            grd1.Columns.Clear();
            grd1.DataSource = SQLDt;

            BoundField bnd = new BoundField();
            bnd.DataField = "Number";
            bnd.SortExpression = "Number";
            bnd.ShowHeader = true;
            bnd.HeaderText = "Number";
            grd1.Columns.Add(bnd);
           
            ButtonField btn = new ButtonField();
            btn.ButtonType = ButtonType.Link;
            btn.DataTextField = "AMOUNT";
            btn.SortExpression = "AMOUNT";
            btn.CommandName = "Select";
            btn.ShowHeader = true;
            btn.HeaderText = "Amount";
            grd1.Columns.Add(btn);

            bnd = new BoundField();
            bnd.DataField = "DESCRIPTION";
            bnd.SortExpression = "DESCRIPTION";
            bnd.ShowHeader = true;
            bnd.HeaderText = "Type";
            grd1.Columns.Add(bnd);
         

            bnd = new BoundField();
            bnd.DataField = "Date";
            bnd.SortExpression = "Date";
            bnd.ShowHeader = true;
            bnd.HeaderText = "Date";
            grd1.Columns.Add(bnd);

            bnd = new BoundField();
            bnd.DataField = "TIME";
            bnd.SortExpression = "TIME";
            bnd.ShowHeader = true;
            bnd.HeaderText = "Time";
            grd1.Columns.Add(bnd);

           
           
            grd1.DataBind();
            grd1.Columns[0].Visible = false;
        }

There is a Add button in the page. When Add button is clicked I want to show the footer row and add the columns to the footer row in the code behind.
Basically I want to insert a new row to gridview by editing footer row columns.  I want to add the footer row  it in the code behind.
 
 protected void AddButton_Click(object sender, EventArgs e)
        {
            try
            {
                grd1.FooterRow.Visible = true;
            }
            catch
            {

            }

        }
   adding columns to footer row in code behind is possible.

Please give me suggestions.
Thanks
0
Comment
Question by:KavyaVS
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
3 Comments
 
LVL 19

Assisted Solution

by:Manoj Patil
Manoj Patil earned 250 total points
ID: 38744077
0
 
LVL 18

Accepted Solution

by:
Rajar Ahmed earned 250 total points
ID: 38746097
Example :
http://www.dotnetfunda.com/articles/article180.aspx
How it Works .
Gridview Design Part :
 1.Enable footer on gridview showfooter="true".
 2.Add footer template to each column on your gridview .
 3.Create event for the add new button in the footer template
<asp:GridView ID="GridView1" runat="Server" AutoGenerateColumns="False" BackColor="White"
            BorderColor="#DEDFDE" BorderStyle="None" BorderWidth="1px" CellPadding="4" ForeColor="Black"
            GridLines="Vertical" Width="100%" OnRowCommand="GridView1_OnRowCommand1" ShowFooter="true"
            DataKeyNames="AutoID">
            <Columns>
                <asp:BoundField DataField="AutoID" HeaderText="AutoID" ReadOnly="True" />
                <asp:TemplateField HeaderText="UserName">
                    <ItemTemplate>
                        <%# Eval("UserName") %>
                    </ItemTemplate>
                    <EditItemTemplate>
                        <asp:TextBox ID="txtUserName" runat="Server" Text='<%# Eval("UserName") %>'></asp:TextBox>
                    </EditItemTemplate>
                    <FooterTemplate>
                        <asp:TextBox ID="txtUserName" runat="Server"></asp:TextBox>
                    </FooterTemplate>
                </asp:TemplateField>
                <asp:TemplateField HeaderText="Password">
                    <ItemTemplate>
                        <%# Eval("Password") %>
                    </ItemTemplate>
                    <EditItemTemplate>
                        <asp:TextBox ID="txtPassword" runat="Server" Text='<%# Eval("Password") %>'></asp:TextBox>
                    </EditItemTemplate>
                    <FooterTemplate>
                        <asp:TextBox ID="txtPassword" runat="Server"></asp:TextBox>
                    </FooterTemplate>
                </asp:TemplateField>
                <asp:TemplateField HeaderText="Active">
                    <ItemTemplate>
                        <%# Eval("Active") %>
                    </ItemTemplate>
                    <EditItemTemplate>
                        <asp:DropDownList ID="dropActive" runat="server" SelectedValue='<%# Eval("Active").ToString().ToLower().Equals("true") ? "True" : "False" %>'>
                            <asp:ListItem Text="Yes" Value="True"></asp:ListItem>
                            <asp:ListItem Text="No" Value="False"></asp:ListItem>
                        </asp:DropDownList>
                    </EditItemTemplate>
                    <FooterTemplate>
                        <asp:DropDownList ID="dropActive" runat="server">
                            <asp:ListItem Text="Yes" Value="True"></asp:ListItem>
                            <asp:ListItem Text="No" Value="False"></asp:ListItem>
                        </asp:DropDownList>
                        <br />
                        <asp:Button ID="btnInsert" runat="Server" Text="Insert" CommandName="Insert" UseSubmitBehavior="False" />
                    </FooterTemplate>
                </asp:TemplateField>
            </Columns>
        </asp:GridView>

Open in new window

Codebehind :Event of the add new button
1.Find control of footer template.
2.Pass it to your insert procedure.
3.Bind Gridview Again.

  protected void GridView1_OnRowCommand1(object sender, GridViewCommandEventArgs e)
    {

        if (e.CommandName.Equals("Insert"))
        {

            TextBox u = GridView1.FooterRow.FindControl("txtUserName") as TextBox;

            TextBox p = GridView1.FooterRow.FindControl("txtPassword") as TextBox;

            new AdminBAL().Insert(u.Text, p.Text, sp.Text, dT.SelectedValue, ad.Text, bool.Parse(dA.SelectedValue));

            lblError.Text = "<br />Record inserted successfully<br />";

            BindCurrentUsers(); // rebind the data

        }

    }

Open in new window

Meeran03
0
 

Author Closing Comment

by:KavyaVS
ID: 38746839
Thanks.
0

Featured Post

Industry Leaders: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

We all know that functional code is the leg that any good program stands on when it comes right down to it, however, if your program lacks a good user interface your product may not have the appeal needed to keep your customers happy. This issue can…
This article is for Object-Oriented Programming (OOP) beginners. An Interface contains declarations of events, indexers, methods and/or properties. Any class which implements the Interface should provide the concrete implementation for each Inter…
In this video, viewers will be given step by step instructions on adjusting mouse, pointer and cursor visibility in Microsoft Windows 10. The video seeks to educate those who are struggling with the new Windows 10 Graphical User Interface. Change Cu…
Monitoring a network: why having a policy is the best policy? Michael Kulchisky, MCSE, MCSA, MCP, VTSP, VSP, CCSP outlines the enormous benefits of having a policy-based approach when monitoring medium and large networks. Software utilized in this v…

622 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question