0

说，我有这样的xml文件：

<?xml version="1.0"?>
<catalog>
   <title>My book catalog</title>
   <link>http://example.com/catalog</link>
   <book id="bk101">
      <author>Gambardella, Matthew</author>
      <title>XML Developer's Guide</title>
      <genre>Computer</genre>
      <price>44.95</price>
      <publish_date>2000-10-01</publish_date>
      <description>An in-depth look at creating applications 
      with XML.</description>
   </book>
   <book id="bk102">
      <author>Ralls, Kim</author>
      <title>Midnight Rain</title>
      <genre>Fantasy</genre>
      <price>5.95</price>
      <publish_date>2000-12-16</publish_date>
      <description>A former architect battles corporate zombies, 
      an evil sorceress, and her own childhood to become queen 
      of the world.</description>
   </book>
</catalog>

最后需要获取对象，我可以使用如下：

xml.title    //must return "My book catalog"
xml.link     //must return "http://example.com/catalog"
xml.book[0]  //is an object with following properties:
             //author, title, genre etc...
             //i.e., xml.book[0].author must return Gambardella, Matthew

希望没有类似的问题，如果有的话，我很抱歉，我没有找到。另外，如果有任何关于此的文档，特殊情况，请指出它，因为我找不到它。有很多关于以各种方式解析 xml 的文档，但没有关于这种情况的信息。

提前致谢。

4

2 回答 2

1

这是使用这些扩展方法的工作 LinqToXml 代码：

[DebuggerDisplay("{Title}")]
public class Catalog
{
    XElement self;
    public Catalog(XElement catalog) { self = catalog; }
    public string Title { get { return self.Get("title", string.Empty); } }
    public Uri Link { get { return self.Get<Uri>("link", null); } }
    public Book[] Books
    {
        get { return _Books ?? (_Books = self.GetEnumerable("book", x => new Book(x)).ToArray()); }
    }
    Book[] _Books;
    [DebuggerDisplay("{Title} by {Author}")]
    public class Book
    {
        XElement self;
        public Book(XElement book) { self = book; }
        public string Id { get { return self.Get("id", string.Empty); } }
        public string Author { get { return self.Get("author", string.Empty); } }
        public string Title { get { return self.Get("title", string.Empty); } }
        public string Genre { get { return self.Get("genre", string.Empty); } }
        public decimal Price { get { return self.Get<decimal>("price", 0); } }
        public DateTime PublishDate { get { return self.Get("publish_date", DateTime.MinValue); } }
        public string Description { get { return self.Get("description", string.Empty); } }
    }
}

并使用它：

Catalog catalog = new Catalog(XElement.Load(file)); // or .Parse(string)
于 2012-04-15T00:30:35.800 回答
0 
doc = XDocument.Load("path/to/file.xml");
List<book> items = doc.Root.Element("catalog").Elements("book").
        Select(e => new book(
            (string)e.Element("author").Value,
            (string)e.Element("title").Value,
            (string)e.Element("price").Value,
            (string)e.Element("publish_date").Value,
            (string)e.Element("description").Value,
            (string)e.Element("genre").Value
        )).ToList();
于 2012-04-18T12:54:52.383 回答