C# 在 XmlReader 中解析 XML 元素
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Parsing through XML elements in XmlReader
提问by Ross
I'm building an application that needs to run through an XML feed but I'm having a little trouble with getting certain elements.
我正在构建一个需要通过 XML 提要运行的应用程序，但在获取某些元素时遇到了一些麻烦。
I'm using the Twitter feedand want to run through all the <item>elements. I can connect fine and get the content from the feed but I can't figure out how to select only the itemelements when I'm loopuing through reader.Read();.
我正在使用Twitter 提要并希望浏览所有<item>元素。我可以正常连接并从提要中获取内容，但我无法弄清楚如何item在循环时只选择元素reader.Read();。
Thanks for your help!
谢谢你的帮助！
采纳答案by Vinko Vrsalovic
The easiest way to do that is to use XPath. Example to follow.
最简单的方法是使用 XPath。要遵循的例子。
string xml = @"<?xml version=""1.0"" encoding=""UTF-8""?>
<rss version=""2.0"">
<channel>
<title>Twitter public timeline</title>
<link>http://twitter.com/public_timeline</link>
<description>Twitter updates from everyone!</description>
<language>en-us</language>
<ttl>40</ttl>
<item>
<title>yasu_kobayashi: rTwT: @junm : yayaya</title>
<description>yasu_kobayashi: rTwT: @junm : yayaya</description>
<pubDate>Tue, 28 Oct 2008 12:04:48 +0000</pubDate>
<guid>http://twitter.com/yasu_kobayashi/statuses/978829930</guid>
<link>http://twitter.com/yasu_kobayashi/statuses/978829930</link>
</item><item>
<title>FreeGroup: WikiFortio - foobar http://tinyurl.com/5gvttf</title>
<description>FreeGroup: WikiFortio - foobar
http://tinyurl.com/5gvttf</description>
<pubDate>Tue, 28 Oct 2008 12:04:47 +0000</pubDate>
<guid>http://twitter.com/FreeGroup/statuses/978829929</guid>
<link>http://twitter.com/FreeGroup/statuses/978829929</link>
</item></channel></rss>
";
XPathDocument doc = new XPathDocument(new StringReader(xml));
XPathNavigator nav = doc.CreateNavigator();
// Compile a standard XPath expression
XPathExpression expr;
expr = nav.Compile("/rss/channel/item");
XPathNodeIterator iterator = nav.Select(expr);
// Iterate on the node set
try
{
while (iterator.MoveNext())
{
XPathNavigator nav2 = iterator.Current.Clone();
nav2.MoveToChild("title","");
Console.WriteLine(nav2.Value);
nav2.MoveToParent();
nav2.MoveToChild("pubDate","");
Console.WriteLine(nav2.Value);
}
}
catch (Exception ex)
{
Console.WriteLine(ex.Message);
}
Console.ReadKey();
And this is jan's approach working
这是 jan 的工作方法
XmlDocument doc2 = new XmlDocument();
doc2.LoadXml(xml);
XmlNode root = doc2.DocumentElement;
foreach (XmlNode item in root.SelectNodes(@"/rss/channel/item"))
{
Console.WriteLine(item.SelectSingleNode("title").FirstChild.Value);
Console.WriteLine(item.SelectSingleNode("pubDate").FirstChild.Value);
}
回答by jan
An alternative:
替代：
// starts as in Vinko Vrsalovic 's answer
// and not including decent eror handling
XmlDocument doc = new XmlDocument(new StringReader(xml));
foreach (XmlNode item in doc.SelectNodes(@"/rss/channel/item"))
{
Console.WriteLine(item.SelectSingleNode("title").Value);
Console.WriteLine(item.SelectSingleNode("pubDate").Value);
}
I don't know if this code is slower or bad practice. Please do comment.
我不知道这段代码是较慢还是不好的做法。请做评论。
I find it more readable than the other one using Navigator and Iterator.
我发现它比使用 Navigator 和 Iterator 的另一个更具可读性。
Edit:I use an XmlDocument. An XPathDocument as in Vinko Vrsalovic 's answer doesn't support this way of working, but is a lot faster: (MSDN)
编辑：我使用Xml文档。Vinko Vrsalovic 的回答中的XPath文档不支持这种工作方式，但速度要快得多： （MSDN）
