Question

我有一系列水果对象，其中一些是橘子，其中一些是苹果我想将它们序列化为一个看起来像的列表：

<Fruits>
    <AppleFruit>
         <IsRotten>true</IsRotten>
         <FellFarFromTree>false</FellFarFromTree>
    </AppleFruit>
    <OrangeFruit>
         <IsRotten>false</IsRotten>
         <NumberOfSegments>6</NumberOfSegments>
    </OrangeFruit>
</Fruits>

所以我尝试以下方法：

[Serializable]
[XmlInclude(typeof(Apple))]
[XmlInclude(typeof(Orange))]
public abstract class Fruit {
    public bool IsRotten { get; set; }
}

[Serializable]
[XmlRoot("AppleFruit")]
public class Apple : Fruit {
    public bool FellFarFromTree { get; set; }
}

[Serializable]
[XmlRoot("OrangeFruit")]
public class Orange : Fruit {
    public int NumberOfSegments { get; set; }
}

public class Blender {
    public void XmlBlend(params Fruit[] fruits) {
        using (var writer = new XmlTextWriter(@"c:\test\blended_fruits.xml", Encoding.UTF8)) {
            writer.Formatting = Formatting.Indented;
            writer.WriteStartDocument();
            writer.WriteStartElement("Fruits");

            var serializer = new XmlSerializer(typeof (Fruit));

            foreach (var fruit in fruits) {
                serializer.Serialize(writer, fruit);
            }

            writer.WriteEndElement();
            writer.WriteEndDocument();
        }
    }

    [Test]
    public void TestIt () {
        var blender = new Blender();
        blender.XmlBlend(
            new Apple() {
                FellFarFromTree = false,
                IsRotten = true
            },
            new Orange() {
                IsRotten = false,
                NumberOfSegments = 6
            });
    }
}

但是XmlRoot属性似乎完全被忽略了。实际输出看起来像：

<Fruits>
  <Fruit xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:type="Apple">
    <IsRotten>true</IsRotten>
    <FellFarFromTree>false</FellFarFromTree>
  </Fruit>
  <Fruit xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:type="Orange">
    <IsRotten>false</IsRotten>
    <NumberOfSegments>6</NumberOfSegments>
  </Fruit>
</Fruits>

我错过了什么？

Answer 1

回答我自己的解决方法，但如果有人有更好的答案，我会接受它。

我为每个类创建了一个不同的序列化程序，并将它们粘贴在字典中：

public Dictionary<Type, XmlSerializer> ShouldntHaveToDoThis = new Dictionary<Type, XmlSerializer>() {
    {typeof(Apple), new XmlSerializer(typeof(Apple))},
    {typeof(Orange), new XmlSerializer(typeof(Orange))}
};

然后为每个项目获取适当的序列化器：

        foreach (var fruit in fruits) {
            var serializer = ShouldntHaveToDoThis[fruit.GetType()];
            serializer.Serialize(writer, fruit);
        }

Answer 2

执行此操作的一种方法是为Fruits创建一个类型，其中包含每种Fruit类型的列表，并使用XmlElement属性命名项目。

[XmlRoot("Fruits")]
public class Fruits
{
    [XmlElement("AppleFruit")]
    public Apple[] Apples { get; set; }
    [XmlElement("OrangeFruit")]
    public Orange[] Oranges { get; set; }
}
[Serializable]
[XmlInclude(typeof(Apple))]
[XmlInclude(typeof(Orange))]
public abstract class Fruit {
    public bool IsRotten { get; set; }
}

[Serializable]
public class Apple : Fruit {
    public bool FellFarFromTree { get; set; }
}

[Serializable]
public class Orange : Fruit {
    public int NumberOfSegments { get; set; }
}


public void XmlBlend(Fruits fruits) {
    using (var writer = new XmlTextWriter(@"c:\test\blended_fruits.xml", Encoding.UTF8)) {
        writer.Formatting = Formatting.Indented;

        var serializer = new XmlSerializer(typeof(Fruits));
        serializer.Serialize(writer, fruits);
    }
}

生成输出，如：

<?xml version="1.0" encoding="utf-8"?>
<Fruits xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <AppleFruit>
    <IsRotten>true</IsRotten>
    <FellFarFromTree>false</FellFarFromTree>
  </AppleFruit>
  <OrangeFruit>
    <IsRotten>false</IsRotten>
    <NumberOfSegments>6</NumberOfSegments>
  </OrangeFruit>
</Fruits>

必须列出Apple[] Apples等等并不完美，但我认为它类似于[XmlInclude(typeof(Apple))]上需要Fruit。

使用每个具体类的自定义节点名序序列化抽象类

2 个答案: