如何解析XML文件

Question

我有一個看起來像一個XML文件：

<results> 
    <result> 
     <title>Welcome+to+The+JASON+Project%21</title> 
     <url>http%3A%2F%2Fwww.jason.org%2F</url> 
     <domain /> 
     <inside_links> 
     <inside_link> 
      <description>News</description> 
      <url>http%3A%2F%2Fwww.jason.org%2FPublic%2FNews%2FNews.aspx</url> 
     </inside_link> 
     <inside_link> 
      <description>register</description> 
      <url>http%3A%2F%2Fwww.jason.org%2Fpublic%2Fregistration%2Fregistration.aspx</url> 
     </inside_link> 
     <inside_link> 
      <description>Argonauts</description> 
      <url>http%3A%2F%2Fwww.jason.org%2FPublic%2FArgonauts%2FArgonauts.aspx</url> 
     </inside_link> 
     <inside_link> 
      <description>Curriculum</description> 
      <url>http%3A%2F%2Fwww.jason.org%2FPublic%2FCurriculum%2FCurriculum.aspx</url> 
     </inside_link> 
     <inside_link> 
      <description>Credits</description> 
      <url>http%3A%2F%2Fwww.jason.org%2Fpublic%2FMisc%2FCredits.aspx</url> 
     </inside_link> 
     </inside_links> 
     <inside_keywords>National+Science+Education+Standards, National+Geographic+Society, Physical+Science, Professional+Development, Earth+Science</inside_keywords> 
    </result> 
    </results> 

...而且我對如何讀它很迷​​茫。我只想將標題，說明和URL分爲不同的字符串。喜歡的東西：

foreach line in lines 
string title = gettitle; 
string description = getdescription; 
string url = geturl; 

...我已經看了這麼多教程，但所有的人似乎並不相對於什麼，我需要做的..有人可以幫我這個？

Answer 1

到LINQ擴展到XML的建議，您可以使用SELECT子句來創建對象代表解析鏈接：

XDocument doc = XDocument.Load(filename); 
var links = from link in doc.Descendants("inside_link") 
      select new 
      { 
       Description = (string)link.Element("description"), 
       Url = HttpUtility.UrlDecode((string)link.Element("url")) 
      }; 

foreach(var l in links) 
    Console.WriteLine("<a href=\"{0}\">{1}</a>", l.Url, l.Description); 

在這種情況下，links將有一個匿名類型對象的可序列Description和Url屬性，其中Url已解碼。這foreach會顯示這樣的事情：

<a href="http://www.jason.org/Public/News/News.aspx">News</a> 
<a href="http://www.jason.org/public/registration/registration.aspx">register</a> 
... 

Answer 2

如果您使用的是.NET 3.5，我建議使用LINQ to XML ...

XDocument doc = XDocument.Load(filename); 
XElement insideLinks = doc.Root.Element("result").Element("inside_links"); 
foreach (XElement insideLink in insideLinks.Elements()) 
{ 
    string description = (string)insideLink.Element("description"); 
    string url = (string)insideLink.Element("url"); 
} 

這也使您可以使用內置的「查詢」語法，所以你可以做這樣的事情...

XDocument doc = XDocument.Load(filename); 
XElement insideLinks = doc.Root.Element("result").Element("inside_links"); 
var allTitles = from XElement insideLink 
       in insideLinks.Elements("inside_link") 
       select (string)insideLink.Element("title"); 

（根據註釋編輯）

Answer 3

試試這個：

XmlDocument xmlDoc = new XmlDocument(); 
xmlDoc.Load("yourfile.xml"); 
foreach (XmlNode result in xmlDoc.SelectNodes("/results/result")) 
{ 
    string title = result.SelectSingleNode("title").InnerText; 
    string url = result.SelectSingleNode("url").InnerText; 
    foreach (XmlNode insideLink in result.SelectNodes("inside_links/inside_link")) 
    { 
     string description = insideLink.SelectSingleNode("description").InnerText; 
    } 
}