Dynamical link Button in ASP.NET

Question

0.00/5 (No votes)

See more:

I have created link button in my application now i have binded the event handler to it but its not Firing any suggestion.

Posted 18-Oct-12 0:55am

rohit24c

Add a Solution

Comments

bbirajdar 18-Oct-12 7:05am

"not Firing any suggestion" ..How can it fire a suggestion ?

adriancs 18-Oct-12 8:10am

I think that simply a grammar mistake and the sentences sounds to be like this:
......but its not Firing. Any suggestion?

AshishChaudha 18-Oct-12 7:06am

share your code please..

adriancs 18-Oct-12 8:11am

how do you "bind the event handler"?

bbirajdar 18-Oct-12 9:51am

using += operator...

2 solutions

Add a Solution

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Maksud Saifullah Pulak · Answer 1 · 2013-02-23T03:28:00

First Add Your link Button and Div message in .ASPX

<asp:LinkButton ID="ancDelete" CommandName="delete" OnCommand="delete_EventCategories" runat="server">Delete </asp:LinkButton>

XML

<div id="divMessage" runat="server" class="grid_7 alpha" style="display: none;">
        <asp:Literal runat="server" ID="ltlMessage" Text="test"></asp:Literal>
    </div>

Then write a Show method.

C#

public static void ShowMessage(string _class, string msg, System.Web.UI.HtmlControls.HtmlGenericControl divMessage, Literal ltlMessage)
       {
           divMessage.Style.Add("display", "block");
           divMessage.Attributes.Add("class", _class);
           ltlMessage.Text = msg;
       }

And finally Event Hadelers.

C#

protected void delete_EventCategories(object Sender, CommandEventArgs e)
        {
            if (e.CommandName == "delete")
            {
                try
                {

             ShowMessage("alert alert-success", " Successfully!",divMessage, ltlMessage);
                }
                catch (Exception ex)
                {

                  ShowMessage("alert alert-error", ex.InnerException.Message, divMessage,ltlMessage);
                }
            }
        }

S. M. Ahasan Habib · Answer 2 · 2013-02-23T19:38:00

If it is matter of dynamic link control then just mind it that its state not persist my ASP.net framework with the help of view state. For example if you create a link button like as follows

ASP.NET

<div id="divLink"  runat="server">
</div>

C#

if (!base.IsPostBack)//comment out this condition if you want to fire Click event 
{
    var control = new LinkButton();
    control.Text = "Clik ME";           
    control.Click += control_Click;
    divLink.Controls.Add(control);
}

C#

void control_Click(object sender, EventArgs e)
{
    Response.Write("............................");
}

You see that next postback event the LinkButton not showing and event not firing.
If you want to fire that event you need to create that dynamic control every request (either postback or not postback.

Dynamical link Button in ASP.NET

2 solutions

Solution 1

Solution 2

Add your solution here

Preview 0

Dynamical link Button in ASP.NET

2 solutions

Solution 1

Solution 2

Add your solution here

Preview 0

Existing Members

...or Join us