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我对WebControl / CompositeControl世界真的很陌生,而且我正在玩一个小型测试类。它只是一个在单击时更新的LinkBut​​ton当我将它排除在UpdatePanel之外时,一切都很好 。但是当我尝试在里面运行它时,我仍然得到一个完整的页面 POST 响应。如何使此类在UpdatePanel中工作?

这是课程:

public class Test2 : CompositeControl 
{
    private static readonly object testButtonEvent = new object();

    public event EventHandler OnTestClick
    {
        add { Events.AddHandler(testButtonEvent, value); }
        remove { Events.RemoveHandler(testButtonEvent, value); }
    }

    private LinkButton testLinkButton;

    public virtual string testLinkButtonText
    {
        get
        {
            object o = ViewState["testLinkButtonText"];
            return (o == null) ? String.Empty : (string)o;
        }
        set
        {
            if (value == null)
                ViewState.Remove("testLinkButtonText");
            else
                ViewState["testLinkButtonText"] = value;
        }
    }   

    protected override void OnInit(EventArgs e)
    {
        /* This stuff makes it ajax friendly but stops the text rendering
        EnsureChildControls();

        ScriptManager ScMan = ScriptManager.GetCurrent(Page);
        if (ScMan != null)
        {
            ScMan.RegisterAsyncPostBackControl(testLinkButton);
        }            */

        base.OnInit(e);
    }

    protected override void CreateChildControls()
    {
        Controls.Clear();

        testLinkButton = new LinkButton();
        testLinkButton.Command += new CommandEventHandler(testClick);
        testLinkButtonText = "Test ViewState Text";

        Controls.Add(testLinkButton);
    }

    void testClick(object sender, CommandEventArgs e)
    {
        testLinkButtonText = "Updated Text On " + DateTime.Now.ToLongTimeString();
    }

    protected override void Render(HtmlTextWriter writer)
    {
        RenderContents(writer);
    }

    protected override void RenderContents(HtmlTextWriter writer)
    {
        EnsureChildControls();

        testLinkButton.Text = testLinkButtonText;
        testLinkButton.RenderControl(writer);            
    }

}

中的代码OnInit()导致控件正确发布,但我没有得到LinkBut​​ton的更新文本。它仍在触发事件 - 当我调试时,我可以看到它被调用。设置此控件以在UpdatePanel中使用的正确方法是什么?

用法,以防万一:

<asp:UpdatePanel runat="server" UpdateMode="Conditional">
    <ContentTemplate>
        <cc:Test2 ID="jqTest02" runat="server" />
    </ContentTemplate>
</asp:UpdatePanel>
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1 回答 1

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您必须为按钮提供一个 ID 属性...这用于驱动UpdatePanel. 更具体地说,它列在要拦截和执行异步回发的控件列表中。

testLinkButton.ID = "btn";
于 2010-01-28T21:23:51.237 回答