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我想顯示模式彈出我點擊Save按鈕保存在DB值之後,雖然顯示模式彈出,應該繼續保存,無需關閉模式彈出 我使用這個代碼顯示模式彈出在按鈕單擊
protected void btsave_Click(object sender, EventArgs e)
{
ModalPopupExtender1.Show();
//My Code
}
和我的aspx代碼
<input type="button" runat="server" id="btmodel" style="display: none" />
<ajaxToolkit:ModalPopupExtender ID="ModalPopupExtender1" runat="server" TargetControlID="btmodel"
PopupControlID="Panel1" PopupDragHandleControlID="PopupHeader" Drag="true" DropShadow="true"
OkControlID="OkButton" CancelControlID="CancelButton" BackgroundCssClass="ModalPopupBG">
</ajaxToolkit:ModalPopupExtender>
<asp:Panel ID="Panel1" Style="display: none" runat="server">
<div class="progress-popup">
<div class="potit">
Saving
</div>
<img src="/images/prog.png" alt="" style="margin: 0 auto; display: block;" />
<div class="potit-cancel">
<span class="cncl">
<asp:Button ID="btnOkay" runat="server" Text="Ok" OnClick="btnOkay_Click" ValidationGroup="vg"
CssClass="popupcancl" />
</span>
</div>
</div>
</asp:Panel>
有了這個代碼,它首先將數據保存到數據庫,然後顯示Modalpopup
嘗試使用AJAX更新面板 –
我應該將兩個控件放在ajax面板中嗎?我的意思是按鈕和莫代爾彈出? –
是的,你可以使用 –