用C#在Winform中怎么判断XML文件中某个节点是否有属性

XmlDocument xmlDoc = new XmlDocument();
xmlDoc.LoadXml("");
XmlNode myNode = xmlDoc.SelectSingleNode("root/aaly");
if (myNode.Attributes.Count > 0)
{
//...
}
else
{
//...
}
设置个事件呗，触发亩前后加散斗载Xml文件并获得节点名。然迅掘清后就像上面代码一样了

XmlDocument xmlDoc = new XmlDocument();

xmlDoc.Load("Demo.xml");
XmlNamespaceManager nsmanager = new XmlNamespaceManager(xmlDoc.NameTable);
nsmanager.AddNamespace("xx", "http://www.springframework.net");
XmlNodeList nodeList = xmlDoc.SelectSingleNode("/xx:objects/xx:object[@id=\"BaseRemoteServiceConfig\"]", nsmanager).ChildNodes;

for (int i = 0; i < nodeList.Count; i++)
{
//将子宽颤颤节点类型转换为XmlElement类型慎败
XmlElement xe = (XmlElement)nodeList[i];
if (xe.Name == "property")
{
string str = "textBox"洞态 + (i+1).ToString();
((TextBox)(this.Controls.Find(str, false)[0])).Text = xe.GetAttribute("value");
}
}