我有两个表单,每个表单都有按钮,我将它们抽象为两个面板,因此当按下每个面板中的enter键时,适当的按钮将发送。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>但我得到了例外
System.InvalidOperationException
The DefaultButton of 'pnlRegistered' must be the ID of a control of type IButtonControl.发布于 2013-06-15 04:01:11
将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>发布于 2013-06-15 03:53:33
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:您也可以在代码隐藏中设置DefaultButton,如下所示:
pnlRegistered.DefaultButton = btnLoginButton.UniqueID;https://stackoverflow.com/questions/17116228
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