如何根据条件设置链接按钮的状态？

Question

我有 2 个链接按钮，它们链接到一个多视图，并且根据我按下的按钮将更改活动视图。我希望其各自视图的链接按钮显示为活动状态。

<asp:Panel runat="server" >
        <div>
            <asp:LinkButton ID="linkDeviceList"  CommandName="SwitchViewByID" CommandArgument="viewDeviceList" runat="server" OnClick="linkDeviceList_Click" CssClass="button-link">Device List</asp:LinkButton>
            <asp:LinkButton ID="linkFTPFolders"  CommandName="SwitchViewByID" CommandArgument="viewFTPFolders" runat="server" OnClick="linkFTPFolders_Click" CssClass="button-link">FTP Folders</asp:LinkButton>           
        </div>
    </asp:Panel>

事件处理程序。我以为我会在一段时间内更改按钮的状态，但不知道如何应用样式更改。

protected void linkFTPFolders_Click(object sender, EventArgs e)
    {
        MultiView1.SetActiveView(viewFTPFolders);

        while (MultiView1.GetActiveView() == viewFTPFolders)
        {
            
        }
    }

    protected void linkDeviceList_Click(object sender, EventArgs e)
    {
        MultiView1.SetActiveView(viewDeviceList);
        while (MultiView1.GetActiveView() == viewDeviceList)
        {
            
        }
    }

Answer 1

我有一个类似的控件。这就是我所做的。

1. 我在单击的按钮上删除/添加了“活动”类。

2. 我禁用了被点击的按钮，使其无法被点击 再次。

       protected void lbListView_Click(object sender, EventArgs e)
       {
           lbGridView.CssClass = "btn btn-default btn-sm pull-right dt-margin-left-5";
           lbGridView.Enabled = true;
   
           lbListView.CssClass = "btn btn-default btn-sm pull-right dt-margin-left-5 active";
           lbListView.Enabled = false;
   
           repGridResults.Visible = false;
           repListResults.Visible = true;
       }
   
       protected void lbGridView_Click(object sender, EventArgs e)
       {
           lbListView.CssClass = "btn btn-default btn-sm pull-right dt-margin-left-5";
           lbListView.Enabled = true;
   
           lbGridView.CssClass = "btn btn-default btn-sm pull-right dt-margin-left-5 active";
           lbGridView.Enabled = false;
   
           repListResults.Visible = false;
           repGridResults.Visible = true;
       }