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Displaying buttons in dynamicaly populatet GridView

Posted on 2008-10-29
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Last Modified: 2011-10-19
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");
 
      }
 
    }

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wizard-step-contactpersons.png
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Question by:Martin Kreis
  • 2
3 Comments
 
LVL 14

Accepted Solution

by:
Ramuncikas earned 500 total points
ID: 22838603
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();

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0
 

Author Closing Comment

by:Martin Kreis
ID: 31511139
Thank you very much!!
0
 

Author Comment

by:Martin Kreis
ID: 22839420
Thank you very much for your help!
With your input, I was now able to solve my problem (with TemplatesFields).

Best regards
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