给指定的ELEMENT追加一个属性节点如何操作（在线等待，谢谢！）

Question

需求如下：
<?xml version="1.0">
  <ListSection>
  <ListPart>
  <ListNo>1</ListNo>
  <ListSymbol>test1</ListSymbol>
  <ListComment>test1Comment</ListComment>
  </ListPart>
  <ListPart>
  <ListNo>2</ListNo>
  <ListSymbol>test2</ListSymbol>
  <ListComment>test2Comment</ListComment>
  </ListPart></ListSection>xXDocument xmlSource = XDocument.Load(new StringReader(strXml));
想通过xmlSource 给ListSymbol=test2的节点追加一个属性并附上值，即<ListSymbol attr="hello">test2</ListSymbol>
如何做呢，请教大家帮帮忙。谢谢

Answer 1

xmlSource.Elements().Skip(1).First().Elements("ListSymbol").First().Add(new XAttribute("attr", "hello")); 

Answer 2

// 设定根节点ItemsCount属性值
    // Parameter：
    //  RootNodeName:根节点名称
    //  ds:数据集
    // Return:
    //      XmlDataDocument
    private XmlDataDocument SetItemsCountAttribute(string strRootNodeName, DataSet ds)
    {
        try
        {
            XmlDataDocument xmlDoc;
            int ItemCount = 0;            ds.DataSetName = strRootNodeName;
            ds.EnforceConstraints = false;
            if (ds.Tables.Count == 0  )
            {
                xmlDoc = new XmlDataDocument();                string xml = "<" + strRootNodeName + "></" + strRootNodeName + ">";
                xmlDoc.LoadXml(xml);
            }
            else
            {
                ds.Tables[0].TableName = "Item";                ItemCount = ds.Tables[0].Rows.Count;
                if (ItemCount == 0)
                {
                    xmlDoc = new XmlDataDocument();                    string xml = "<" + strRootNodeName + "></" + strRootNodeName + ">";
                    xmlDoc.LoadXml(xml);
                }
                else
                {
                    xmlDoc = new XmlDataDocument(ds);
                }
            }            XmlNode root = xmlDoc.FirstChild;
            // 创建节点
            XmlNode attrCount = xmlDoc.CreateNode(XmlNodeType.Attribute, "ItemsCount", null);
            attrCount.Value = ItemCount.ToString();
            // 添加节点属性
            root.Attributes.SetNamedItem(attrCount);            return xmlDoc;
        }
        catch (Exception e)
        {
            string strMsg = e.Message;
            return null;
        }
    }

Answer 3

代码已经经过测试:
var document = XDocument.Load("XMLFile1.xml");            var query =  from ele in document.Descendants(XName.Get("ListSymbol"))
                         where ele.Value == "test2"
                         select ele;
            var node = query.SingleOrDefault();
            if (node != null)
            {
                node.Add(new XAttribute("attr", "hello"));
            }            document.Save("XmlResult.xml");
XmlResult.xml:<?xml version="1.0" encoding="utf-8"?>
<ListSection>
  <ListPart>
    <ListNo>1</ListNo>
    <ListSymbol>test1</ListSymbol>
    <ListComment>test1Comment</ListComment>
  </ListPart>
  <ListPart>
    <ListNo>2</ListNo>
    <ListSymbol attr="hello">test2</ListSymbol>
    <ListComment>test2Comment</ListComment>
  </ListPart>
</ListSection>

Answer 4

+1
用linq to xml 爽.