Displaying buttons in dynamicaly populatet GridView

I developed a wizard using ASP.net with which the user will be able to register requests of a certain type. Each request may have one or more contact persons, which the user will be able to add on a given step of the wizard. This step contains a form to add a new contact person and a table (GridView Control) which displays all contact persons the user already provided.

 

At the end of the whole wizard, all data will be saved in a database. Meanwhile I store the data (e.g. the contact persons) in a session variable. This session variable is a typed list of contact persons (List<ContactPerson>). On each page load all contact persons available in this session variable will be displayed in the table mentioned.

 

The requirement is to display two buttons in each row enabling the user to either edit or delete a contact person. Here is my problem: I am not able to create a table (currently a GridView Control, but may be anything else) from the data in the session variable and add the buttons (I now try to do so programmatically, but any solution is appreciated).


//Currently I use the following code in the *.aspx-file: 
 
 
 
<asp:GridView ID="GV_ContactPersons" runat="server" Width="100%">
 
</asp:GridView>
 
 
 
//& and the following code in the *.aspx.cs-file where you can see how I currently am converting the data in the session variable to the table (a DataTable which I then bind to the GridView): 
 
 
 
    private void FillDataGrid()
 
    {
 
      DataTable L_Table = new DataTable();
 
 
 
      DataColumn L_ColumnName = new DataColumn("Name");
 
      L_Table.Columns.Add(L_ColumnName);
 
 
 
      DataColumn L_ColumnAddress = new DataColumn("Adresse");
 
      L_Table.Columns.Add(L_ColumnAddress);
 
 
 
      DataColumn L_ColumnCity = new DataColumn("Ortschaft");
 
      L_Table.Columns.Add(L_ColumnCity);
 
 
 
      DataColumn L_ColumnPhone = new DataColumn("Telefon");
 
      L_Table.Columns.Add(L_ColumnPhone);
 
 
 
      DataColumn L_ColumnEMail = new DataColumn("E-Mail");
 
      L_Table.Columns.Add(L_ColumnEMail);
 
 
 
      //ButtonColumn L_ButtonColumn = new ButtonColumn();
 
      //ButtonField L_ButtonField = new ButtonField();
 
      //GridViewRow L_GridViewRow = new GridViewRow(0, 0, DataControlRowType.DataRow, DataControlRowState.Normal);
 
 
 
      try
 
      {
 
        foreach (ContactPerson L_ContactPerson in (List<ContactPerson>)Session["ContactPersons"])
 
        {
 
          //LinkButton L_EditButton = new LinkButton();
 
          //L_EditButton.Text = "Bearbeiten";
 
          //L_EditButton.CommandArgument = L_ContactPerson.ContactPerson_PK.ToString();
 
          //L_EditButton.Command += new CommandEventHandler(L_EditButton_Command);
 
 
 
          DataRow L_Row = L_Table.NewRow();
 
          //GridViewRow L_Row = new GridViewRow(0, 0, DataControlRowType.DataRow, DataControlRowState.Normal);
 
 
 
          L_Row[L_ColumnName] = L_ContactPerson.ContactPerson_Title + " " + L_ContactPerson.ContactPerson_FirstName + " " + L_ContactPerson.ContactPerson_LastName;
 
          L_Row[L_ColumnAddress] = L_ContactPerson.ContactPerson_Address + " " + L_ContactPerson.ContactPerson_Address2;
 
          L_Row[L_ColumnCity] = L_ContactPerson.ContactPerson_PostalCode + " " + L_ContactPerson.ContactPerson_City;
 
          L_Row[L_ColumnPhone] = L_ContactPerson.ContactPerson_Phone;
 
          L_Row[L_ColumnEMail] = L_ContactPerson.ContactPerson_EMail;
 
 
 
          L_Table.Rows.Add(L_Row);
 
        }
 
 
 
        GV_ContactPersons.DataSource = L_Table;
 
        GV_ContactPersons.DataBind();
 
      }
 
      catch (Exception L_Ex)
 
      {
 
        Session.Add("ErrorMessage", "Die Tabelle kann nicht angezeigt werden");
 
        Session.Add("ErrorDescription", L_Ex.Message);
 
 
 
        Response.Redirect("ErrorPage.aspx");
 
      }
 
    }

Open in new window

wizard-step-contactpersons.png
Martin KreisAsked:
Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

RamuncikasCommented:
You can define your gridview control in aspx page as usual. I mean there is no real need to create columns in code (see code snippet). Also in same aspx you can add a template filed with buttons in it for editing/deleting.

Gridview control can bind not only to datatable or dataset. If binds to typed list (List) as well. So when it comes to showing data you just assign your list to gridview as datasource and call DataBind method (see snippet). You just have to make sure that DataField property of a BoundField in aspx matches a property of ContactPerson class. In a given sample in a snippet it is assumed that ContactPerson class has a public property Name.

HTH
R
// aspx
<asp:GridView ID="GV_ContactPersons" runat="server" Width="100%">
  <Columns>
    <asp:BoundField DataField="Name" HeaderText="Name" />
    ...
  </Columns>
</asp:GridView>
 
// cs
this.GV_ContactPersons.DataSource = (List<ContactPerson>)Session["myData"];
this.GV_ContactPersons.DataBind();

Open in new window

0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
Martin KreisAuthor Commented:
Thank you very much!!
0
Martin KreisAuthor Commented:
Thank you very much for your help!
With your input, I was now able to solve my problem (with TemplatesFields).

Best regards
0
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
ASP.NET

From novice to tech pro — start learning today.

Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.