Question

我有两个表单，每个表单都有按钮，我用两个面板抽象它们所以当按下每个面板中的键时，会发出正确的按钮。在一个面板中我有登录表单

 <asp:Panel ID="pnlRegistered" runat="server" DefaultButton="btnLoginButton">
    <LayoutTemplate>
       <asp:Button ID="btnLoginButton" runat="server" CommandName="Login" Text="Login" ValidationGroup="Login1" OnClick="btnLoginButton_Click"  ClientIDMode="Static" />
    </LayoutTemplate>
 <asp:Panel>

但我得到例外

System.InvalidOperationException

The DefaultButton of 'pnlRegistered' must be the ID of a control of type IButtonControl.

Answer 1

您缺少DefaultButton的值：

<asp:Panel ID="pnlRegistered" runat="server" DefaultButton="btnLoginButton">
  <asp:Button ID="btnLoginButton" runat="server" CommandName="Login" Text="Login" ValidationGroup="Login1" OnClick="btnLoginButton_Click"  ClientIDMode="Static" />
<asp:Panel>

编辑：你也可以在代码隐藏中设置DefaultButton，如下所示：

pnlRegistered.DefaultButton = btnLoginButton.UniqueID;

Answer 2

将您的面板放在登录控件的layouttemplate

中

<asp:Login ID="Login1" runat="server">
    <LayoutTemplate>
        <asp:Panel ID="pnlRegistered" runat="server" DefaultButton="btnLoginButton">
            <asp:Button ID="btnLoginButton" runat="server" CommandName="Login" Text="Login" ValidationGroup="Login1"
                ClientIDMode="Static" />
        </asp:Panel>
    </LayoutTemplate>
</asp:Login>

Panel.DefaultButton和登录按钮

2 个答案: