解析XML文件，并抢值

  • Z时代
  • 2024-01-10
  • 分类：问答

XML文件是这样的：解析XML文件，并抢值

<?xml version="1.0"?> 
<catalog> 
    <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>

我需要的所有数据读取我的课：

public class Book 
{ 
    public string Id { get; set; } 
    public string Author { get; set; } 
    public string Title { get; set; } 
    public Genre Genre { get; set; } 
    public decimal Price { get; set; } 
    public DateTime PublishDate { get; set; } 
    public string Description { get; set; } 
}

下面的代码不工作，我可以从节点抓取bookId，但不能标题，作者.. 如何实现该结果？这是我到目前为止有：

const string filePath = @"C:\Users\Michał\Desktop\books.xml"; 
XDocument xmlDoc = XDocument.Load(filePath); 
var dupa = xmlDoc 
    .Descendants("book") 
    .Select(x => new Book() 
    { 
     Id = (string) x.Attribute("bookid"), 
     Title = (string) x.Attribute("title") // Title is empty after that code runs 
    }).ToList();

回答：

id是<book>的attribute但其他节点是孩子XML elements所以你需要使用XContainer.Element(XName name)访问它们：

var dupa = xmlDoc 
    .Descendants("book") 
    .Select(x => new Book 
    { 
     Id = (string)x.Attribute("bookid"), 
     Author = (string)x.Element("author"), 
     Title = (string)x.Element("title"), 
     Genre = (Genre)Enum.Parse(typeof(Genre), (string)x.Element("genre")), 
     Price = (decimal)x.Element("price"), 
     PublishDate = (DateTime)x.Element("publish_date"), 
     Description = (string)x.Element("description"), 
    }).ToList();

在这里，我假设Genre是一个枚举类似于：

public enum Genre 
{ 
    Computer, 
    Fantasy, 
};

样例fiddle #1。

或者，您可以创建一个包含您的书籍列表中Catalog类型，与attributes that control XML serialization注释，并Book，并使用XmlSerializer反序列化的一切。

定义你的类型，像这样：

[XmlRoot(ElementName = "book")] 
public class Book 
{ 
    [XmlElement(ElementName = "author")] 
    public string Author { get; set; } 
    [XmlElement(ElementName = "title")] 
    public string Title { get; set; } 
    [XmlElement(ElementName = "genre")] 
    public Genre Genre { get; set; } 
    [XmlElement(ElementName = "price")] 
    public decimal Price { get; set; } 
    [XmlElement(ElementName = "publish_date")] 
    public string PublishDate { get; set; } 
    [XmlElement(ElementName = "description")] 
    public string Description { get; set; } 
    [XmlAttribute(AttributeName = "id")] 
    public string Id { get; set; } 
} 
[XmlRoot(ElementName = "catalog")] 
public class Catalog 
{ 
    [XmlElement(ElementName = "book")] 
    public List<Book> Books { get; set; } 
} 
public enum Genre 
{ 
    Computer, 
    Fantasy, 
};

然后反序列化如下：

List<Book> dupa; 
using (var reader = XmlReader.Create(filePath)) 
{ 
    var serializer = new XmlSerializer(typeof(Catalog)); 
    dupa = ((Catalog)serializer.Deserialize(reader)).Books; 
}

样品fiddle #2。

注：

  • 考虑从enum改变Genre到string。使用当前的模型，如果新的流派稍后添加到XML中，反序列化将失败。

  • 有多种工具可以自动生成与XmlSerializer兼容的XML类型。我使用http://xmltocsharp.azurewebsites.net/，然后根据需要修改属性名称和类型。另一种选择是xsd.exe。欲了解更多信息，请参阅Generate C# class from XML。

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