【发布时间】:2016-08-17 15:10:33
【问题描述】:
我有一个要添加到服务器控件的复选框。此复选框也没有添加通常的 onclick...__doPostBack() 调用。这导致它不会像我希望的那样导致回发。
private void CreateGrid()
{
StringWriter sWriter = new StringWriter();
HtmlTextWriter writer = new HtmlTextWriter(sWriter);
GridItem.ID = "gridItem";
GridHeader.ID = "gridHeader";
GridHeader.Attributes["class"] += " no-select";
GridCount.ID = "GridCount";
GridDescription.ID = "GridDescription";
if (cBoxID == null) tBox.Visible = false;
else
{
tBox.ID = cBoxID;
tBox.AutoPostBack = true;
tBox.EnableViewState = true;
ScriptManager.GetCurrent(this.Page).RegisterPostBackControl(tBox);
//tBox.CheckedChanged += new EventHandler(Force_Post_Back);
}
QuickFilter.Attributes["class"] = "quick-filter";
QuickFilter.Attributes["title"] = "Quick Filter";
IconMagnifier.Attributes["class"] = "icon-magnifier";
GridResults.ID = "ltlGridResults";
GridResults.ClientIDMode = ClientIDMode.Static;
//GridResults.EnableViewState = false;
QuickFilter.Controls.Add(IconMagnifier);
GridHeader.Controls.AddAt(0, tBox);
GridHeader.Controls.Add(QuickFilter);
GridHeader.Controls.Add(GridCount);
GridHeader.Controls.Add(GridDescription);
GridItem.Controls.Add(GridHeader);
GridItem.Controls.Add(GridResults);
}
protected void Page_PreRender(object sender, EventArgs e)
{
CreateGrid();
}
protected override void Render(HtmlTextWriter writer)
{
GridItem.RenderControl(writer);
}
【问题讨论】:
-
请显示您的 HTML(CheckBox 控件)并显示您的服务器端事件。
-
我附上了一张图片,显示了从代码隐藏中生成的 html。
标签: c# asp.net .net custom-server-controls