DropDownList的SelectedIndexChanged（）如何使用PostBack？ [英] How DropDownList's SelectedIndexChanged() works without PostBack?

问题描述

DropDownList的 SelectedIndexChanged（）事件填充页面上的ListBox。显然，这个页面将该页面发送回服务器。有没有什么办法可以实现没有完整的回发？

DropDownList's SelectedIndexChanged() Event fills the ListBox on the page. Obviously this posts the page back to the server. Is there any way to make it happen without full postback?

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();
}

推荐答案

您可以将DropDownList在DropDownList上设置< asp：UpdatePanel> 并设置 AutoPostBack =true您必须将触发器设置为 SelectedIndexChanged 事件。

You could put the DropDownList into an <asp:UpdatePanel> and set AutoPostBack="true" on the DropDownList. You have to set the trigger to the SelectedIndexChanged event.

这样的事情（不要忘记脚本管理器）

Something like this (don't forget the script manager)

<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>

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