<asp:TemplateField HeaderText="命令">
                         <ItemTemplate>
                             <asp:HyperLink ID="HyperLink1" runat="server" Target="_blank" NavigateUrl='<%# "InfoEdit.aspx?id="+Eval("Info_Id")+"&mid="+Eval("Menu_ID")+"&pid="+Eval("Parent_Code")+"&sign=1" %>'>添加</asp:HyperLink>
                             |<asp:LinkButton ID="LinkButton2" runat="server" CommandName="del"
                                
                                CommandArgument='<%# Bind("Info_Id") %>'>删除</asp:LinkButton>
                             |<asp:HyperLink ID="HyperLink2" runat="server" Target="_blank" NavigateUrl='<%# "InfoEdit.aspx?id="+Eval("Info_Id")+"&mid="+Eval("Menu_ID")+"&pid="+Eval("Parent_Code")+"&sign=2" %>'>修改</asp:HyperLink>                           
                            |<asp:HyperLink ID="HyperLink4" runat="server" Target="_blank" NavigateUrl='<%# "AddImage.aspx?id="+Eval("Info_Id")+"&mid="+Eval("Menu_ID")+"&pid="+Eval("Parent_Code") %>'>上传图片</asp:HyperLink>
                         </ItemTemplate>
                         <FooterStyle CssClass="GirdViewFootRight" />
                         <HeaderStyle CssClass="GirdViewHeadRight" HorizontalAlign="Center" />
                         <ItemStyle HorizontalAlign="Center" />
                     </asp:TemplateField>
 
我想对上面的“上传图片”进行一个判断，比如当mid为9时就显示“上传图片”，否则就不显示，请问我的代码该怎么写呢，请各位大大们帮帮忙啊，我在线等，谢谢。

既然HyperLink4是个服务器控件应该有可见属性Visible
后台
判断一下
if(Request["mid"]==9)
   HyperLink4.Visble=true;
else
   HyperLink4.Visble=false;
这样不就可以了？

谢谢指导