获取XML节点

Question

一下代码为XML格式<Record Uin="2051" Pwd="" Nick="陈国桢" Name="陈国桢" Face="19" Age="0" Gender="1" Birthday="0" BloodType="5" Constellation="0" Mobile="" HomePage="" Email="" Phone="" Fax="" Address="" PostCode="" Country="" Province="" City="" Memo="" Collage="" MobileType="0" AuthType="0" Dept="" Position="" OpenGsmInfo="0" OpenContactInfo="0" PubOutUin="0" PubOutNick="" PubOutName="" PubOutDept="" PubOutPosition="" PubOutInfo="" OuterPublish="0" LastModifyTime="1253498968" LastLogonTime="1268185124" Deny="0" Allow="2014" LdapUser="0"/>Record 有1K条左右，想获取每个Record下的各个参数，哪位指点一下，ASP语法最好，谢谢各位

Answer 1

XmlDocument xmldoc = new XmlDocument();        xmldoc.Load(@"E:\WorkProject\WebSite21\XMLFile.xml");        XmlNodeList xmlNodeList = xmldoc.SelectNodes("Record");        DataTable dt = new DataTable();        dt.Columns.Add("Uin", typeof(string));
        //完整Column...        for (int i = 0; i < xmlNodeList.Count; i++)
        {
            DataRow dr = dt.NewRow();
            for (int j = 0; j < xmlNodeList[i].Attributes.Count; j++)
            {
                dr[j] = xmlNodeList[i].Attributes[j].Value;
            }
            dt.Rows.Add(dr);     
        }        //dt为参数集表