听起来您可能未正确使用UpdatePanel功能。如果您将UpdatePanel设置为在子事件触发事件时进行更新，则仅UpdatePanel应该刷新，而不刷新整个页面。下面的代码的行为似乎与您想要的类似。更改下拉菜单时，只有更新面板会发回到服务器上，刷新页面时，您可以从会话中获取值。

ASPX代码

<form id="form1" runat="server">
    <asp:ScriptManager ID="ScriptManager1" runat="server">
    </asp:ScriptManager>
    <div>
        Current Time: <asp:Label ID="lblTime" runat="server" /><br />
        Session Value: <asp:Label ID="lblSessionValue" runat="server" /><br />
        <br />
        <asp:UpdatePanel ID="upSetSession" runat="server">
            <ContentTemplate>
                <asp:DropDownList ID="ddlMyList" runat="server" 
                    onselectedindexchanged="ddlMyList_SelectedIndexChanged"
                    AutoPostBack="true">
                    <asp:ListItem>Select One</asp:ListItem>
                    <asp:ListItem>Maybe</asp:ListItem>
                    <asp:ListItem>Yes</asp:ListItem>
                </asp:DropDownList>
            </ContentTemplate>
            <Triggers>
                <asp:AsyncPostBackTrigger ControlID="ddlMyList" 
                    EventName="SelectedIndexChanged" />
            </Triggers>
        </asp:UpdatePanel>
    </div>
</form>

后面的代码

    protected void Page_Load(object sender, EventArgs e)
    {
        this.lblTime.Text = DateTime.Now.ToShortTimeString();
        if (Session["MyValue"] != null) 
            this.lblSessionValue.Text = Session["MyValue"].ToString();
    }

    protected void ddlMyList_SelectedIndexChanged(object sender, EventArgs e)
    {
        Session.Remove("MyValue");
        Session.Add("MyValue", this.ddlMyList.SelectedValue);
    }