Add Columns dynamically to Grid-view with itemtemplate

Question

i want to know how to add column dynmically to gridview. the grid view suppose to get user input. i know how to use itemtemplate for specific no of columns, but i\'m not awa

Answer 1

You need to create a class implementing ITemplate , Full code as following:

public class DynamicTemplateField : ITemplate
{

    public void InstantiateIn(Control container)
    {
        //define the control to be added , i take text box as your need
        TextBox txt1 = new TextBox();
        txt1.ID = "txt1";
        container.Controls.Add(txt1);
    }
}

//Method to bind the Grid View
public void BindData()
{
    TemplateField temp1  = new TemplateField();  //Create instance of Template field
    temp1.HeaderText = "New Dynamic Temp Field"; //Give the header text

    temp1.ItemTemplate = new DynamicTemplateField(); //Set the properties **ItemTemplate** as the instance of DynamicTemplateField class.


    gv.Columns.Add(temp1); //add the instance if template field in columns of grid view

    //Bind the grid  view
    gv.DataSource = [your data source];
    gv.DataBind();

 }

RowDataBound

protected void gv_RowDataBound(object sender, System.Web.UI.WebControls.GridViewRowEventArgs e)
{
  if(e.Row.RowType == DataControlRowType.DataRow)
   {
      TextBox txt1 = e.Row.FindControl("txt1") as TextBox;
      txt1.Text = e.Row.DataItem["Name"]; //Assign any column value of your datasource
    }

}

.aspx page

You can manipulate DynamicTemplateField class to add different types of control