asp.net – DropDownList的SelectedIndexChanged()如何在没有PostBack的情况下工作？

DropDownList的SelectedindexChanged()事件填充页面上的ListBox.显然,这会将页面发布回服务器.有没有办法在没有完全回发的情况下实现它？

protected void ddlTablo_SelectedindexChanged(object sender,EventArgs e)
{
    List<string> list = new List<string>();
    ListBox1.Items.Clear();
    var columnNames= from t in typeof(Person).GetProperties() select t.Name;
    foreach (var item in columnNames)
    {
         list.Add(item);
    }
    ListBox1.DataSource = list;
    ListBox.DataBind();
}

解决方法

您可以将DropDownList放入< asp：UpdatePanel>并在DropDownList上设置AutopostBack =“true”.您必须将触发器设置为SelectedindexChanged事件.

像这样的东西(不要忘记脚本管理器)

<asp:ScriptManager ID="ScriptManager1" runat="server" />

<asp:UpdatePanel ID="UpdatePanel1" runat="server">
   <ContentTemplate>
      <asp:DropDownList ID="drop1" runat="server" OnSelectedindexChanged="ddlTablo_SelectedindexChanged" />
   </ContentTemplate>
   <Triggers>
      <asp:AsyncPostbackTrigger ControlID="drop1" EventName="SelectedindexChanged" />
   </Triggers>
</asp:UpdatePanel>