XmlSerializer C# 类数组，不包装集合元素

Question

我正在尝试使用 C# 中的 XmlSerializer 解析外部 XML 文件，源 XML 看起来像这样：

<product>
  <id>71464280-6d57-49de-ba26-35c9919d808d</id>
  <name>Random Product</name>
  <property name="color">Blue</property>
  <property name="country">SE</property>
  <property name="brand">Fancy Brand</property>
</product>

我试图让它与 C# 类匹配，问题是“属性”元素，首先我将它们视为列表/数组，但我没有设法在没有包装的情况下解析它们像..这样的元素在源代码中，然后我开始认为它们可能是类上的属性，如下所示：

 public class ProductXml
{

    public Guid Id { get; set; }
    public string Name { get; set; }

    [XmlElement(ElementName = "Property")]
    public string Color { get; set; }

    [XmlElement(ElementName = "Property")]
    public string Country { get; set; }

    [XmlElement(ElementName = "Property")]
    public string Brand { get; set; }
}

这样的类是要序列化的，它不会呈现，因为有多个“属性”元素。

有人知道我应该如何处理这个问题吗？

谢谢！

Answer 1

原来这很简单，只需要确保列表属性的元素名称与“属性”类的元素名称匹配。

我使用这个很棒的实用程序https://xmltocsharp.azurewebsites.net/发现了这一点

[XmlRoot(ElementName = "product")]
public class Product
{
    [XmlElement(ElementName = "id")]
    public string Id { get; set; }

    [XmlElement(ElementName = "name")]
    public string Name { get; set; }

    [XmlElement(ElementName = "property")]
    public List<Property> Property { get; set; }
}

[XmlRoot(ElementName = "property")]
public class Property
{
    [XmlAttribute(AttributeName = "name")]
    public string Name { get; set; }
    [XmlText]
    public string Text { get; set; }
}

Answer 2

你可以这样做：

<?xml version="1.0" encoding="UTF-8"?>
<product>
  <id>71464280-6d57-49de-ba26-35c9919d808d</id>
  <name>Random Product</name>
  <property name="color">Blue</property>
  <property name="country">SE</property>
  <property name="brand">Fancy Brand</property>
</product>

public class Product
{
    public string ID { get; set; }
    public string Name { get; set; }
    public string Color { get; set; }
    public string Country { get; set; }
    public string Brand { get; set; }

    public Product(XElement element)
    {
        ID = element.Element("id").Value;
        Name = element.Element("name").Value;
        List<XElement> properties = element.Elements("property").ToList();
        Color = properties.Where(x => x.Attribute("name").Value == "color").First().Value;
        Country = properties.Where(x => x.Attribute("name").Value == "country").First().Value;
        Brand = properties.Where(x => x.Attribute("name").Value == "brand").First().Value;
    }
}

class Program
{
    static void Main(string[] args)
    {
        XDocument doc = XDocument.Load("xml.xml");
        XElement productElement = doc.Descendants("product").First();
        Product product = new Product(productElement);
    }
}

Answer 3

这是另一种获取单个属性值的方法

    public class ProductXml
    {

        public Guid Id { get; set; }
        public string Name { get; set; }
        private string Color { get; set; }
        private string Country { get; set; }
        private string Brand { get; set; }

        [XmlElement(ElementName = "Property")]
        public List<Property> property
        {
            get {
                return new List<Property>() {
                    new Property() { Name = "color", Text = Color},
                    new Property() { Name = "country", Text = Country},
                    new Property() { Name = "brand", Text = Brand},
                };          
            }
            set {
                foreach (Property prop in value)
                {
                    switch (prop.Name)
                    {
                        case "color" :
                            Color = prop.Text;
                            break;
                        case "country":
                            Country = prop.Text;
                            break;
                        case "brand":
                            Brand = prop.Text;
                            break;
                    }
                }
            }
        }
    }
    [XmlRoot(ElementName = "property")]
    public class Property
    {
        [XmlAttribute(AttributeName = "name")]
        public string Name { get; set; }
        [XmlText]
        public string Text { get; set; }
    }