我有一个模态,我想从模态后面的代码中运行它:
<asp:ScriptManager ID="ScriptManager1" runat="server"></asp:ScriptManager>
<asp:UpdatePanel ID="UpdatePanel1" runat="server">
<ContentTemplate>
<div dir="rtl" id="mymodal" class="modal hide fade" tabindex="-1" role="dialog" aria-labelledby="myModalRapidAddLabel" aria-hidden="true">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-hidden="true" style="float: left">×</button>
<h3 id="myModalRapidAddLabel">Edit Brand</h3>
</div>
<div class="modal-body">
<p>Enter New Name</p>
<div class="control-group">
<div class="controls">
<asp:TextBox ID="txtnewname" runat="server"></asp:TextBox>
</div>
</div>
</div>
<div class="modal-footer">
<button class="btn" data-dismiss="modal" aria-hidden="true">cancel</button>
</div>
</div></ContentTemplate>
</asp:UpdatePanel>
而我的代码背后:
ScriptManager.RegisterClientScriptBlock(this, this.GetType(), "none","<script>$('#mymodal').modal('show');</script>", false);
ScriptManager.RegisterStartupScript(Page, Page.GetType(), "mymodal", "$('#mymodal').modal();", true);
我剪切代码的每个地方都必须运行模式,但这对我不起作用,为什么?
试试这个,创建一个asp button
:
<asp:UpdatePanel runat="server">
<ContentTemplate>
<asp:Button runat="server" CssClass="btn btn-info" Text="Search" ID="cmdSearch" OnClick="cmdSearch_Click" />
</ContentTemplate>
</asp:UpdatePanel>
在后面的代码中:
protected void cmdSearch_Click(object sender, EventArgs e)
{
ScriptManager.RegisterStartupScript(this, this.GetType(), "",
"$('#mymodal').modal('show');", true);
}
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