【问题标题】:How to serialize an object array in C#?如何在 C# 中序列化对象数组?
【发布时间】:2013-10-24 04:10:48
【问题描述】:

我使用 xsd.exe 和这个 xml 文件生成了一个 c# 类:

<?xml version="1.0" encoding="UTF-8"?>
<Mary>
    <Frank>
        <Joe>
            <Susan>
                <Stuff>data</Stuff>
            </Susan>
            <Susan>
                <Stuff>data</Stuff>
            </Susan>
        </Joe>
        <Joe>
            <Susan>
                <Stuff>data</Stuff>
            </Susan>
            <Susan>
                <Stuff>data</Stuff>
            </Susan>
        </Joe>
    </Frank>
</Mary>

The C# class that was generated can be viewed here.

我可以用数据初始化对象:

var susan = new MaryFrankJoeSusan(){Stuff = "my data"};
var frank = new MaryFrank(){Joe = new MaryFrankJoeSusan[1][]};
frank.Joe[0] = new MaryFrankJoeSusan[1]{susan};
var mary = new Mary { Items = new MaryFrank[1] { frank } };

我正在使用以下内容将其序列化到磁盘:

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

using (Stream stream = new FileStream(@"C:\out.xml", FileMode.Create))
{
    var settings = new XmlWriterSettings { Indent = true, NewLineOnAttributes = true, OmitXmlDeclaration = true};
    using (XmlWriter writer = new XmlTextWriter(stream, Encoding.Unicode))
    {
        serializer.Serialize(writer, mary);
        writer.Close();
    }
}

但是在初始化序列化程序时出现以下错误:

error CS0030: Cannot convert type 'MaryFrankJoeSusan[]' to 'MaryFrankJoeSusan'

如何将整个 Mary 对象序列化到磁盘?

【问题讨论】:

    标签: c# .net xml serialization


    【解决方案1】:

    那些生成的类有些问题。

    问题的发生是因为MaryFrank.Joe 被声明为MaryFrankJoeSusan 对象的二维数组,但它用XmlArrayItemAttribute 装饰,它告诉序列化程序该二维数组的每个项目都是@ 类型987654324@ 当然是MaryFrankJoeSusan[]

    如果您在生成的类中更改此行:

    [System.Xml.Serialization.XmlArrayItemAttribute("Susan", typeof(MaryFrankJoeSusan),
     Form=System.Xml.Schema.XmlSchemaForm.Unqualified, IsNullable=false)]
    

    到这里:

    [System.Xml.Serialization.XmlArrayItemAttribute("Susan", typeof(MaryFrankJoeSusan[]),
     Form=System.Xml.Schema.XmlSchemaForm.Unqualified, IsNullable=false)]
    

    然后它将无错误地序列化。但是,您不会得到想要的结果。而不是这个:

    <Mary>
        <Frank>
            <Joe>
                <Susan>
                    <Stuff>my data</Stuff>
                </Susan>
            </Joe>
        </Frank>
    </Mary>
    

    你会得到这个(注意额外的MaryFrankJoeSusan标签):

    <Mary>
        <Frank>
            <Joe>
                <Susan>
                    <MaryFrankJoeSusan>
                        <Stuff>my data</Stuff>
                    </MaryFrankJoeSusan>
                </Susan>
            </Joe>
        </Frank>
    </Mary>
    

    真正的问题似乎是xsd.exe 工具一开始就错误地生成了类结构。它不是在层次结构中创建一个类来代表 Joe,而是尝试将 Joe 和 Susan 组合在一起,这在此处实际上不起作用。

    我通过工具从问题中运行了您的原始 XML 以生成 XSD 架构,我得到了这个:

    <?xml version="1.0" encoding="utf-8"?>
    <xs:schema id="Mary" xmlns="" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:msdata="urn:schemas-microsoft-com:xml-msdata">
      <xs:element name="Mary" msdata:IsDataSet="true" msdata:UseCurrentLocale="true">
        <xs:complexType>
          <xs:choice minOccurs="0" maxOccurs="unbounded">
            <xs:element name="Frank">
              <xs:complexType>
                <xs:sequence>
                  <xs:element name="Joe" minOccurs="0" maxOccurs="unbounded">
                    <xs:complexType>
                      <xs:sequence>
                        <xs:element name="Susan" minOccurs="0" maxOccurs="unbounded">
                          <xs:complexType>
                            <xs:sequence>
                              <xs:element name="Stuff" type="xs:string" minOccurs="0" />
                            </xs:sequence>
                          </xs:complexType>
                        </xs:element>
                      </xs:sequence>
                    </xs:complexType>
                  </xs:element>
                </xs:sequence>
              </xs:complexType>
            </xs:element>
          </xs:choice>
        </xs:complexType>
      </xs:element>
    </xs:schema>
    

    这对我来说看起来不错。然后我采用相同的模式并再次通过该工具运行它以生成 C# 类。我本来希望得到类似的东西:

    [Serializable]
    [XmlRoot(Namespace = "", ElementName = "Mary")]
    public class Mary
    {
        [XmlElement("Frank")]
        public Frank[] Frank { get; set; }
    }
    [Serializable]
    public class Frank
    {
        [XmlElement("Joe")]
        public Joe[] Joe { get; set; }
    }
    [Serializable]
    public class Joe
    {
        [XmlElement("Susan")]
        public Susan[] Susan { get; set; }
    }
    [Serializable]
    public class Susan
    {
        [XmlElement("Stuff")]
        public string Stuff { get; set; }
    }
    

    但是我得到了与您在问题中链接的相同的损坏类。所以对我来说,它看起来像是 xsd 工具中的一个错误。

    要让它工作,您可以使用我上面制作的手动编码类,将初始化代码更改为:

    var susan = new Susan { Stuff = "my data" };
    var joe = new Joe { Susan = new Susan[] { susan } };
    var frank = new Frank { Joe = new Joe[] { joe } };
    var mary = new Mary { Frank = new Frank[] { frank } };
    

    --或--

    另一种选择是更改 xsd。将 FrankJoe 元素的 xs:sequence 指示符替换为 xs:choice,如下所示:

    <?xml version="1.0" encoding="utf-8"?>
    <xs:schema id="Mary" xmlns="" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:msdata="urn:schemas-microsoft-com:xml-msdata">
      <xs:element name="Mary" msdata:IsDataSet="true" msdata:UseCurrentLocale="true">
        <xs:complexType>
          <xs:choice minOccurs="0" maxOccurs="unbounded">
            <xs:element name="Frank">
              <xs:complexType>
                <xs:choice> <!-- was xs:sequence -->
                  <xs:element name="Joe" minOccurs="0" maxOccurs="unbounded">
                    <xs:complexType>
                      <xs:choice> <!-- was xs:sequence -->
                        <xs:element name="Susan" minOccurs="0" maxOccurs="unbounded">
                          <xs:complexType>
                            <xs:sequence>
                              <xs:element name="Stuff" type="xs:string" minOccurs="0" />
                            </xs:sequence>
                          </xs:complexType>
                        </xs:element>
                      </xs:choice> <!-- was /xs:sequence -->
                    </xs:complexType>
                  </xs:element>
                </xs:choice> <!-- was /xs:sequence -->
              </xs:complexType>
            </xs:element>
          </xs:choice>
        </xs:complexType>
      </xs:element>
    </xs:schema>
    

    使用此模式,生成的类会更好:现在有一个类可以代表 Joe。 (为了简洁起见,我在这里简化了生成的代码并删除了一些属性):

    [System.SerializableAttribute()]
    [System.Xml.Serialization.XmlRootAttribute(Namespace="", IsNullable=false)]
    public partial class Mary {
        [System.Xml.Serialization.XmlElementAttribute("Frank", Form=System.Xml.Schema.XmlSchemaForm.Unqualified)]
        public MaryFrank[] Items { get; set; }
    }
    
    [System.SerializableAttribute()]
    public partial class MaryFrank {
        [System.Xml.Serialization.XmlElementAttribute("Joe", Form=System.Xml.Schema.XmlSchemaForm.Unqualified)]
        public MaryFrankJoe[] Items { get; set; }
    }
    
    [System.SerializableAttribute()]
    public partial class MaryFrankJoe {
        [System.Xml.Serialization.XmlElementAttribute("Susan", Form=System.Xml.Schema.XmlSchemaForm.Unqualified)]
        public MaryFrankJoeSusan[] Items { get; set; }
    }
    
    [System.SerializableAttribute()]
    public partial class MaryFrankJoeSusan {
        [System.Xml.Serialization.XmlElementAttribute(Form=System.Xml.Schema.XmlSchemaForm.Unqualified)]
        public string Stuff { get; set; }
    }
    

    然后设置代码变成:

    var susan = new MaryFrankJoeSusan() { Stuff = "my data" };
    var joe = new MaryFrankJoe() { Items = new MaryFrankJoeSusan[] { susan } };
    var frank = new MaryFrank() { Items = new MaryFrankJoe[] { joe } };
    var mary = new Mary { Items = new MaryFrank[] { frank } };
    

    我们得到了预期的输出:

    <?xml version="1.0" encoding="utf-16"?>
    <Mary xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
        <Frank>
            <Joe>
                <Susan>
                    <Stuff>my data</Stuff>
                </Susan>
            </Joe>
        </Frank>
    </Mary>
    

    【讨论】:

    • 很棒的答案布赖恩。谢谢。看起来我会回到手工制作课程,因为它更干净。我在上面测试了你的解决方案,它可以工作。
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