C# 读取XML中的数据信息内容的方法

internal class ReadXML

{

/// <summary>

/// 获取XML文件person节点信息

/// </summary>

/// <param name="filePath">xml文件路径</param>

/// <returns></returns>

internal object Get(string filePath)

{

XmlDocument doc = new XmlDocument();

doc.Load(filePath);

StringBuilder sb = new StringBuilder();

foreach (XmlNode element in doc.DocumentElement.getElementsByTagName_r("person1"))

{

//1 获取person节点->name节点->address属性值

sb.Append(element["name"].GetAttribute("address"));

//2 获取person节点->name值

//sb.Append(element["name"].InnerText);

//3 获取person节点->name/sex/age属性值

//sb.Append(element["name"].InnerText)

//.Append(element["sex"].InnerText)

//.Append(element["age"].InnerText);

}

return sb;

}

}

static void Main(string[] args)

{

ReadXML read = new ReadXML();

Console.WriteLine(read.Get("person.xml").ToString());

Console.ReadKey();

}

person.xml文件内容

<?xml version="1.0" encoding="utf-8" ?>

<ss>

<person name="张三" sex="男" age="21">

</person>

<person name="张三2" sex="男" age="22">

</person>

<person name="张三3" sex="男" age="23">

</person>

<person1>

<name address="四川成都">李四</name>

<sex>女</sex>

<age>23</age>

</person1>

简单的一句：select * from sysobjects for xml path('root') 以xml格式显示数据 path里面的参数是跟节点名称

</ss>