Question

我创建了一个C＃类：

public class books {
    public int bookNum { get; set; }
    public class book {
        public string name { get; set; }
        public class record {
            public string borrowDate { get; set; }
            public string returnDate { get; set; }
        }
        public record[] records { get; set; }
    }
    public book[] books { get; set; }
}

但是当我使用XmlSerializer转换为XML字符串时。结果与下面的xml不同。

我的C＃课有什么问题？我想使用XmlSerializer输出结果而不是使用 XmlDocument的。

有什么想法吗？提前谢谢！

<books>
    <bookNum>2</bookNum>
    <book>
        <name>Book 1</name>
        <record>
            <borrowDate>2013-7-1</borrowDate>
            <returnDate>2013-7-12</returnDate>
        </record>
        <record>            
            <borrowDate>2013-8-1</borrowDate>
            <returnDate>2013-8-5</returnDate>
        </record>
    </book>
    <book>
        <name>Book 2</name>
        <record>
            <borrowDate>2013-6-1</borrowDate>
            <returnDate>2013-6-12</returnDate>
        </record>
        <record>            
            <borrowDate>2013-7-1</borrowDate>
            <returnDate>2013-7-5</returnDate>
        </record>
    </book>
</books>

修改

以下是我的C＃代码和输出结果：

books books = new books {
        bookNum = 2,
        Books = new books.book[] { 
            new books.book {  
                name = "Book1", 
                records = new books.book.record[] {
                    new books.book.record {
                        borrowDate = "2013-1-3",
                        returnDate = "2013-1-5"
                    },
                     new books.book.record {
                        borrowDate = "2013-2-3",
                        returnDate = "2013-4-5"
                    }
                }
            },
             new books.book {  
                name = "Book1", 
                records = new books.book.record[] {
                    new books.book.record {
                        borrowDate = "2013-1-3",
                        returnDate = "2013-1-5"
                    },
                     new books.book.record {
                        borrowDate = "2013-2-3",
                        returnDate = "2013-4-5"
                    }
                }
            }
        }
    };


    XmlSerializer xsSubmit = new XmlSerializer(typeof(books));

    XmlDocument doc = new XmlDocument();

    System.IO.StringWriter sww = new System.IO.StringWriter();
    XmlWriter writer = XmlWriter.Create(sww);
    xsSubmit.Serialize(writer, books);
    var xml = sww.ToString(); // Your xml
    context.Response.Write(xml);

XML：

<books>
    <bookNum>2</bookNum>
    <Books>
        <book>
            <name>Book1</name>
            <records>
                <record>
                    <borrowDate>2013-1-3</borrowDate>
                    <returnDate>2013-1-5</returnDate>
                </record>
                <record>
                    <borrowDate>2013-2-3</borrowDate>
                    <returnDate>2013-4-5</returnDate>
                </record>
            </records>
        </book>
        <book>
            <name>Book1</name>
            <records>
                <record>
                    <borrowDate>2013-1-3</borrowDate>
                    <returnDate>2013-1-5</returnDate>
                </record>
                <record>
                    <borrowDate>2013-2-3</borrowDate>
                    <returnDate>2013-4-5</returnDate>
                </record>
            </records>
         </book>
    </Books>
</books>

Answer 1

您无法使用标准序列化工具从您的问题中序列化类，以便它与<book>节点在同一级别上具有<bookNum>个条目。

当使用标准序列化工具保存类时，<book>节点的列表将始终嵌套到与<bookNum>节点位于同一级别的单独阵列节点中。同样关注records类上的book数组字段。

要生成您想要的 XML 输出 - <book>节点位于与<bookNum>节点相同的级别上 - 您必须在您的网站中实现IXmlSerializable接口用于自定义序列化的books类。要查看IXmlSerializable实施的示例，请访问以下链接：StackOverflow answer，CodeProject article。

另一个解决方案是 - 如我所说user Alexandr对我的回答发表评论 - 从books类型继承您的List<book>类，并拥有book类字段{从records类型继承的类类型的{1}}。

从您的问题序列化类时，假设您已按照以下方式分配了正确的XmlRoot, XmlElement, XmlArray and XmlArrayItem attributes：

List<record>

您将获得 XML 输出，如下所示：

[XmlRoot("books")]
public class books
{
    [XmlElement("bookNum")]
    public int bookNum { get; set; }

    [XmlRoot("book")]
    public class book
    {
        [XmlElement("name")]
        public string name { get; set; }

        [XmlRoot("record")]
        public class record
        {
            [XmlElement("borrowDate")]
            public string borrowDate { get; set; }

            [XmlElement("returnDate")]
            public string returnDate { get; set; }
        }

        [XmlArray("borrowRecords")]
        [XmlArrayItem("record")]
        public record[] records { get; set; }
    }

    [XmlArray("booksList")]
    [XmlArrayItem("book")]
    public book[] books { get; set; }
}

Answer 2

我对您的班级代码进行了以下更改。我无法使用默认的序列化程序复制XML序列化，因为它不会复制＆＃39;记录＆＃39;元素没有给它一个容器元素。

[System.Xml.Serialization.XmlRoot("books")]
public class books 
{
    public int bookNum { get; set; }
    public class book {
        public string name { get; set; }
        public class record {
            public string borrowDate { get; set; }
            public string returnDate { get; set; }
        }
        public record[] records { get; set; }
    }
    public book[] books { get; set; }
}

序列化这给我以下输出

<books xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <bookNum>2</bookNum>
  <books>
    <book>
      <name>first</name>
      <records>
        <record>
          <borrowDate>19/07/2013 4:41:29 PM</borrowDate>
          <returnDate>19/07/2013 4:41:29 PM</returnDate>
        </record>
      </records>
    </book>
  </books>
</books>

使用此测试代码

books bks = new books();
bks.bookNum = 2;
bks.books = new books.book[]{ new books.book{name="first", records = new books.book.record[] {new books.book.record{borrowDate = DateTime.Now.ToString(), returnDate = DateTime.Now.ToString()}}}};

System.Xml.Serialization.XmlSerializer serializer = new System.Xml.Serialization.XmlSerializer(typeof(books));

XmlWriterSettings settings = new XmlWriterSettings();
settings.Encoding = new UnicodeEncoding(false, false); // no BOM in a .NET string
settings.Indent = true;
settings.OmitXmlDeclaration = true;

using(StringWriter textWriter = new StringWriter()) {
    using(XmlWriter xmlWriter = XmlWriter.Create(textWriter, settings)) {
        serializer.Serialize(xmlWriter, bks);
    }
    return textWriter.ToString(); //This is the output as a string
}

Answer 3

我意识到这已经晚了几年，但我已经能够通过使用XmlElementAttribute来实现您想要的结构。

我通过使用XSD.exe从xml生成模式定义并从xsd文件生成.Net代码来发现这一点。据我所知，这适用于.Net 3.5到4.6。

这是我使用的类定义：

public class books
{
    public int bookNum { get; set; }
    public class book {
        public string name { get; set; }
        public class record {
            public string borrowDate { get; set; }
            public string returnDate { get; set; }
        }
        [XmlElement("record")]
        public record[] records { get; set; }
    }
    [XmlElement("book")]
    public book[] allBooks { get; set; }
}

这是一个LinqPad代码段，用于说明序列化/反序列化（基于David Colwell的代码片段，感谢BTW提供有关如何排除BOM的提示，这正是我所寻找的）：

books bks = new books();
books bks2 = null;
bks.bookNum = 2;
bks.allBooks = new books.book[] 
        { 
            new books.book {
                name="book 1", 
                records = new books.book.record[] {
                        new books.book.record{borrowDate = DateTime.Now.ToString(), returnDate = DateTime.Now.ToString()}
                    }
                },
            new books.book { 
                name="book 2", 
                records = new books.book.record[] { 
                        new books.book.record{borrowDate = DateTime.Now.ToString(), returnDate = DateTime.Now.ToString()}, 
                        new books.book.record{borrowDate = DateTime.Now.ToString(), returnDate = DateTime.Now.ToString()}}
                    },
        };
string xmlString;

System.Xml.Serialization.XmlSerializer serializer = new System.Xml.Serialization.XmlSerializer(typeof(books));

XmlWriterSettings settings = new XmlWriterSettings();
settings.Encoding = new UnicodeEncoding(false, false); // no BOM in a .NET string
settings.Indent = true;
settings.OmitXmlDeclaration = true;

XmlSerializerNamespaces ns = new XmlSerializerNamespaces();
// exclude xsi and xsd namespaces by adding the following:
ns.Add(string.Empty, string.Empty);

using(StringWriter textWriter = new StringWriter()) {
    using(XmlWriter xmlWriter = XmlWriter.Create(textWriter, settings)) {
        serializer.Serialize(xmlWriter, bks, ns);
    }
    xmlString = textWriter.ToString(); //This is the output as a string
}

xmlString.Dump();

// Deserialize the xml string now       
using ( TextReader reader = new StringReader(xmlString) ) {
    bks2 = ( books )serializer.Deserialize(reader);
}

bks2.Dump();

这样生成的XML可以在不实现IXmlSerializable的情况下进行序列化和反序列化，例如：

<books>
  <bookNum>2</bookNum>
  <book>
    <name>book 1</name>
    <record>
      <borrowDate>2/2/2016 5:57:25 PM</borrowDate>
      <returnDate>2/2/2016 5:57:25 PM</returnDate>
    </record>
  </book>
  <book>
    <name>book 2</name>
    <record>
      <borrowDate>2/2/2016 5:57:25 PM</borrowDate>
      <returnDate>2/2/2016 5:57:25 PM</returnDate>
    </record>
    <record>
      <borrowDate>2/2/2016 5:57:25 PM</borrowDate>
      <returnDate>2/2/2016 5:57:25 PM</returnDate>
    </record>
  </book>
</books>

Answer 4

如果你需要其他类，比如book类中的book2，你有一些特殊的指令来实现它。示例

public class books
{     
   public int bookNum {get; set; }
   public class book {
         public string name {get; set; }
         public class record {
             public string borrowDate {get; set; }
             public string returnDate {get; set; }
         }
         [XmlElement ("record")]
         public record [] records {get; set; }
     }
     [XmlElement ("book")]
     public book [] allBooks {get; set; }

     public int book2Num {get; set; }
     public class book2 {
         public string name {get; set; }
         public class record {
             public string borrowDate {get; set; }
             public string returnDate {get; set; }
         }
         [XmlElement ("record")]
         public record [] records {get; set; }
     }
     [XmlElement ("book2")]
     public book2 [] allBook2 {get; set; }
}`

当我尝试运行程序时，出现以下错误：

＆＃34;附加信息：反映类型＆＃34;

时出错

XmlSerializer将C＃对象转换为xml字符串

4 个答案: