post 后翻页

Scroll page after post back

我想在 post 返回后向下滚动页面。该代码不适用于/没有更新面板。

浏览器控制台错误:

Uncaught SyntaxError: Unexpected token < VM465:1

点击VM465:1时,指向这一行:

<script type="text/javascript">$('html, body').animate({ scrollTop: 1600 }, 'slow');</script>

ASPX:

<asp:UpdatePanel ID="up1" runat="server">
    <Triggers>
        <asp:AsyncPostBackTrigger ControlID="btnMore" EventName="Click" />
    </Triggers>
    <ContentTemplate>
<asp:LinkButton ID="btnMore" runat="server" OnClick="btnMore_Click">ShowMore</asp:LinkButton>
</ContentTemplate>
</asp:UpdatePanel>

C#:

protected void btnMore_Click(object sender, EventArgs e)
    {
        //string message = "alert('Hello!')";
        //ScriptManager.RegisterClientScriptBlock(sender as Control, this.GetType(), "alert", message, true);
        ScriptManager.RegisterClientScriptBlock(sender as Control, this.GetType(), "ScrollPage", GetPageScrollScript(900), true);
    }

private string GetPageScrollScript(int heightToScroll)
    {
        string ScrollPage = "<script type=\"text/javascript\">$('html, body').animate({ scrollTop: " + heightToScroll + " }, 'slow');</script>";
        return ScrollPage;
    }

当我取消注释(有和没有更新面板)时警报有效。
应该怎么做才能使卷轴起作用。

更新: 它以这种方式对我有用:

ScriptManager.RegisterClientScriptBlock(sender as System.Web.UI.Control, this.GetType(), "ScrollPage", "$('html, body').animate({ scrollTop: " + 1000 + " }, 'slow')", true);

您正在使用像

这样的警报
alert('Hello!')

这是可行的,但是对于您要执行的脚本,您需要将其包裹在标签周围

<script type=\"text/javascript\">$('html, body').animate({ scrollTop: " + heightToScroll + " }, 'slow');</script>

没有他们试试?

$('html, body').animate({ scrollTop: " + heightToScroll + " }, 'slow')