我的代碼如下任何替代方案在asp.net菜單控制的XmlDataSource
DataSet ds = new DataSet();
string connStr = "Data Source=PARITAS00024;Initial Catalog=MenuDb;Persist Security Info=True;User ID=sa;Password=paritas123";
using (SqlConnection conn = new SqlConnection(connStr))
{
string sql = "Select MenuId, MenuTitle, MenuDesc, MenuURL, ParentMenuId from tblMenus where Status=1 and RecordStatus=1 order by ParentMenuId, DisplayOrder";
SqlDataAdapter da = new SqlDataAdapter(sql, conn);
da.Fill(ds);
da.Dispose();
}
ds.DataSetName = "Menus";
ds.Tables[0].TableName = "Menu";
DataRelation relation = new DataRelation("ParentChild", ds.Tables["Menu"].Columns["MenuId"], ds.Tables["Menu"].Columns["ParentMenuId"], true);
relation.Nested = true;
ds.Relations.Add(relation);
System.Web.UI.WebControls.XmlDataSource xds = new System.Web.UI.WebControls.XmlDataSource();
xds.TransformFile = "~/TransformXSLT.xsl";
xds.XPath = "MenuItems/MenuItem";
xds.Data = ds.GetXml();
xds.ID = "xmlDataSourceMenu";
Menu1.DataSource = xds;
Menu1.DataBind();
使用的XmlDataSource的這種正確的方法?
依賴於XML和XSL文件的內容..你的問題到底是什麼? – 2011-02-11 08:48:48
這裏我不使用xmldatasource控件,而是創建了我自己的xmldatasource對象,並使用它進行即時通訊。我想知道的是,這是正確的方式嗎? – Harsha 2011-02-11 09:02:05