Question

<?xml version="1.0"?>

-<bookstore>            
        <book > 
            <title>aaaa</title> 
            -<author > 
                <first-name>firts</first-name> 
                <last-name>last</last-name> 
            </author> 
            <price>8.23</price> 
            <otherbooks>
                    <book > 
                        <title>bbb</title>      
                        <price>18.23</price> 
                    </book>     
                    <book > 
                        <title>ccc</title>      
                        <price>11.22</price> 
                    </book>     
            </otherbooks>
        </book> 
</bookstore>

我从xml文件中选择了所有书籍。如何使用XPath为每本书选择标题，作者（名字和姓氏）和价格？

xPathDoc = new XPathDocument(filePath);
xPathNavigator = xPathDoc.CreateNavigator();
XPathNodeIterator xPathIterator = xPathNavigator.Select("/bookstore//book");
foreach (XPathNavigator book in xPathIterator)
{
    ??
}

Answer 1

使用SelectSingleNode()和Value：

  XPathDocument xPathDoc = new XPathDocument(filePath); 
  XPathNavigator xPathNavigator = xPathDoc.CreateNavigator(); 
  XPathNodeIterator xPathIterator = xPathNavigator.Select("/bookstore//book"); 
  foreach (XPathNavigator book in xPathIterator) 
  {
    XPathNavigator nav = book.SelectSingleNode("title");
    string title = nav==null ? string.Empty : nav.Value;
    nav = book.SelectSingleNode("author/first-name");
    string authorFirstName = nav==null ? string.Empty : nav.Value;
    nav = book.SelectSingleNode("author/last-name");
    string authorLastName = nav==null ? string.Empty : nav.Value;
    nav = book.SelectSingleNode("price");
    string price = nav==null ? string.Empty : nav.Value;;
    Console.WriteLine("{0} {1} {2} {3}", title, authorFirstName, authorLastName, price);
  }

Answer 2

您可以使用LINQ2XML

XElement doc=XElement.Load("yourXML.xml");//loads your xml
var bookList=doc.Descendants().Elements("book").Select(
x=>//your book node
    new{
           title=x.Element("title").Value,
           author=new //accessing your author node
           {
               firstName=x.Element("author").Element("first-name").Value,
               lastName=x.Element("author").Element("last-name").Value
           },
           price=x.Element("price").Value
       }
);

bookList现在拥有您想要的所有元素

所以，你现在可以这样做

foreach(var book in bookList)
{
book.title;//contains title of the book
book.author.firstName;//contains firstname of that book's author
book.author.lastName;
}

Answer 3

我喜欢Mimo提供的解决方案，但只需稍加改动，创建一个扩展方法来重用部分功能：

public static class XPathNavigatorExtensions
{
    public static string GetChildNodeValue(this XPathNavigator navigator, string nodePath)
    {
        XPathNavigator nav = navigator.SelectSingleNode(nodePath);
        return nav == null ? string.Empty : nav.Value;
    }
}

结果代码看起来更干净：

        ICollection<Book> books = new List<Book>();
        foreach (XPathNavigator node in iterator)
        {
            Book book = new Book() { Author = new Author() };
            book.Title = node.GetChildNodeValue("title");
            book.Author.FirstName = node.GetChildNodeValue("author/first-name");
            book.Author.LastName = node.GetChildNodeValue("author/last-name");
            book.Price = node.GetChildNodeValue("price");
            books.Add(book);
        }

XPathNodeIterator从迭代数据访问子节点

3 个答案: