您的位置:首页 > 其它

.net中使用XPath语言在xml中判断是否存在节点值的方法

2015-06-25 16:00 741 查看
book.xml
<?xml version="1.0" encoding="utf-8" ?> <bookstore>
<book category="COOKING">
<title lang="en">Everyday Italian</title>
<author>Giada De Laurentiis</author>
<year>2005</year>
<price>30.00</price>
</book>
<book category="CHILDREN">
<title lang="en">Harry Potter</title>
<author>J K. Rowling</author>
<year>2005</year>
<price>29.99</price>
</book>
<book category="WEB">
<title lang="en">XQuery Kick Start</title>
<author>James McGovern</author>
<author>Per Bothner</author>
<author>Kurt Cagle</author>
<author>James Linn</author>
<author>Vaidyanathan Nagarajan</author>
<year>2003</year>
<price>49.99</price>
</book>
<book category="WEB">
<title lang="en">Learning XML</title>
<author>Erik T. Ray</author>
<year>2003</year>
<price>39.95</price>
</book>
</bookstore>
c#代码:
//判断是否存在title的取值为:'Harry Potter'的节点存在
Stopwatch watch = Stopwatch.StartNew();
var xmlDoc2 = new XmlDocument();
xmlDoc2.Load(@"book.xml");
var titleTextExpr4 = "/bookstore/book[title='Harry Potter']/title";
var titleTextNodes4 = xmlDoc2.SelectNodes(titleTextExpr4);
Console.WriteLine("XPath表达式为 /bookstore/book[title='Harry Potter']/title,节点数为:" + titleTextNodes4.Count);
if(titleTextNodes4.Count>0)
{
Console.WriteLine("title='Harry Potter'的节点存在");
}
else
{
Console.WriteLine("title='Harry Potter'的节点不存在");
}
watch.Stop();
Console.WriteLine("take times(ms)="+watch.ElapsedMilliseconds);
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: