Vin*_*alo 4 c# linq-to-xml xml-parsing
在这个xml文件中(http://www.studiovincent.net/list.xml):
<list version="1.0">
<meta>
<type>resource-list</type>
</meta>
<resources start="0" count="4">
<resource classname="Quote">
<field name="name">Vincent</field>
<field name="username">Hill</field>
<field name="age">31</field>
<field name="hair">black</field>
</resource>
<resource classname="Quote">
<field name="name">John</field>
<field name="username">Tedelon</field>
<field name="age">27</field>
<field name="hair">brown</field>
</resource>
<resource classname="Quote">
<field name="name">Michael</field>
<field name="username">Lopez</field>
<field name="age">20</field>
<field name="hair">red</field>
</resource>
<resource classname="Quote">
<field name="name">Frank</field>
<field name="username">Lopez</field>
<field name="age">25</field>
<field name="hair">black</field>
</resource>
</resources>
</list>
Run Code Online (Sandbox Code Playgroud)
使用此代码:
using System.Xml;
using.System.Xml.Linq;
XmlReader reader = XmlReader.Create("http://www.studiovincent.net/list.xml");
XElement el = XElement.Load(reader);
reader.Close();
var items = el.Elements("resources").Elements("resource").Descendants().DescendantNodes();
var items = from item in el.Elements("resources").Elements("resource").Descendants()
where item.Attribute("name").Value == "name" select item.FirstNode;
foreach (XNode node in items)
{
Console.WriteLine(node.ToString());
}
Run Code Online (Sandbox Code Playgroud)
我有这个输出:
Vincent
John
Michael
Frank
Run Code Online (Sandbox Code Playgroud)
代码工作得非常好,但我需要获得值31,其对应字段名称="年龄",其中字段名称="名称"是Vincent.我怎样才能得到这个结果?
我建议您自然地阅读XML.
在您的代码中,尝试查找name属性设置为的所有字段"name".
此过程不能用于将名称与年龄相关联.读取XML并检查所有resource元素更自然.然后向元素添加元素中描述的一些信息field.
// Using LINQ to SQL
XDocument document = XDocument.Load("http://www.studiovincent.net/list.xml"); // Loads the XML document.
XElement resourcesElement = document.Root.Element("resources"); // Gets the "resources" element that is in the root "list" of the document.
XElement resourceElementVincent = (from resourceElement in resourcesElement.Elements("resource")// Gets all the "resource" elements in the "resources" element
let fieldNameElement = resourceElement.Elements("field").Single(fieldElement => fieldElement.Attribute("name").Value == "name") // Gets the field that contains the name (there one and only one "name" field in the "resource" element -> use of Single())
where fieldNameElement.Value == "Vincent" // To get only resources called "Vincent"
select resourceElement).Single(); // We suppose there is one and only 1 resource called "Vincent" -> Use of Single()
XElement fieldAgeElement = resourceElementVincent.Elements("field").Single(fieldElement => fieldElement.Attribute("name").Value == "age"); // Gets the corresponding "age" field
int age = int.Parse(fieldAgeElement.Value, CultureInfo.InvariantCulture); // Gets the age by Parse it as an integer
Console.WriteLine(age);
Run Code Online (Sandbox Code Playgroud)
它能做你想要的吗?
| 归档时间: |
|
| 查看次数: |
14731 次 |
| 最近记录: |