How I can display Image on mouseover event on asp:linkButton
I am using asp.net framework 3.5 and c# 2.0. I am able to bind data in to gridview.
Actually I want to display Image on mouseover event on asp:linkButton, which is used in grid view
<asp:GridView ID="gvTemplate" runat="server" AllowPaging="True" AutoGenerateColumns="False"
CellPadding="4" ForeColor="#333333" GridLines="None" ShowFooter="True" Width="100%"
OnRowCommand="gvTemplate_RowCommand" OnRowDataBound="RowDataBound" OnPageIndexChanging="gvTemplate_PageIndexChanging">
<RowStyle BackColor="#F7F6F3" ForeColor="#333333" />
<Columns>
<asp:TemplateField HeaderStyle-HorizontalAlign="Left" HeaderText="Sr. No">
<ItemTemplate>
<%# Container.DataItemIndex + 1 %>
ItemTemplate>
<HeaderStyle HorizontalAlign="Left" />
asp:TemplateField>
<asp:TemplateField HeaderText="Template Name" ItemStyle-HorizontalAlign="Left">
<ItemTemplate>
<asp:LinkButton ID="lBtTempID" runat="server" Text='<%# DataBinder.Eval(Container.DataItem,"tempName") %>'
CommandName='<%# DataBinder.Eval(Container.DataItem,"tempID") %>' CommandArgument="tempID" onmouseover="javascript:showImage(this.id)"/>
ItemTemplate>
asp:TemplateField>
<asp:TemplateField HeaderStyle-HorizontalAlign="Left" HeaderText="Template Type">
<ItemTemplate>
<asp:Label ID="lbltempType" runat="server" Text='<%# DataBinder.Eval(Container.DataItem,"tempType") %>' />
ItemTemplate>
<HeaderStyle HorizontalAlign="Left" />
asp:TemplateField>
Columns>
<FooterStyle BackColor="#5D7B9D" Font-Bold="True" ForeColor="White" />
<PagerStyle BackColor="#284775" ForeColor="White" HorizontalAlign="Center" />
<SelectedRowStyle BackColor="#E2DED6" Font-Bold="True" ForeColor="#333333" />
<HeaderStyle BackColor="#5D7B9D" Font-Bold="True" ForeColor="White" />
<EditRowStyle BackColor="#999999" />
<AlternatingRowStyle BackColor="White" ForeColor="#284775" />
asp:GridView>
Dadu DaPosted Aug 29, 2012, 5:48 AM
I it definitely help full in my future also.
Thanks Dear It is really work nice.
I want to ask another question regarding url rewriting.
I need to make dynamic website like google provide services of develop and design on line website,
I have completed all things of database and design, but how could i get the url at loading time
of the web application
if page not found then 404 error acre
in google
the website url would be www.website.google.com/ourlwebsitename
here ourlwebsitename is as a example of user created name.
same thing I want to do.
www.mywebsite.com/userchiceaname
here www.mywebsite.com is my website url
and userchiceaname is provided and selected by user in the time of developing a webbsite
Krishna GaradPosted Aug 28, 2012, 8:59 AM