8

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不一样。

我的 C# 类有什么问题？我想使用 XmlSerializer 来输出结果，而不是使用 XmlDocument。

有任何想法吗？提前致谢！

<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>

编辑

下面是我的 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>
4

4 回答 4

9

您不能使用标准序列化工具从您的问题中序列化类，以便它具有与节点<book>相同级别的条目。<bookNum>

当使用标准序列化工具保存的类时，您的<book>节点列表将始终嵌套到与节点处于同一级别的单独数组节点<bookNum>中。同样关注类records上的数组字段book。

要生成您想要的XML<book>输出 - 节点与节点处于同一级别<bookNum>- 您必须在类中实现IXmlSerializable接口以books进行自定义序列化。要查看实现示例，IXmlSerializable请访问以下链接：StackOverflow 答案、CodeProject 文章。

另一种解决方案是——正如用户 Alexandr在我的回答中所说的那样——books从 type 继承你的类，List<book>并在你的book类字段上拥有从 typerecords继承的类List<record>类型。

从您的问题中序列化类时，假设您分配了正确的XmlRoot、XmlElement、XmlArray 和 XmlArrayItem 属性，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; }
}

您将获得如下XML输出：

<books>
    <bookNum>2</bookNum>
    <booksList>
        <book>
            <name>Book 1</name>
            <borrowRecords>
                <record>
                    <borrowDate>2013-1-3</borrowDate>
                    <returnDate>2013-1-5</returnDate>
                </record>
                <record>            
                    <borrowDate>2013-2-3</borrowDate>
                    <returnDate>2013-4-5</returnDate>
                </record>
            </borrowRecords>
        </book>
        <book>
            <name>Book 2</name>
            <borrowRecords>
                <record>
                    <borrowDate>2013-1-3</borrowDate>
                    <returnDate>2013-1-5</returnDate>
                </record>
                <record>            
                    <borrowDate>2013-2-3</borrowDate>
                    <returnDate>2013-4-5</returnDate>
                </record>
            </borrowRecords>
        </book>
    </booksList>
</books>
于 2013-07-19T06:30:18.443 回答
8

我对您的课程代码进行了以下更改。我无法使用默认序列化程序复制 XML 序列化，因为它不会复制“记录”元素而不给它一个容器元素。 

[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; }
}

序列化这给了我以下输出

<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>

使用这个测试代码

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
}
于 2013-07-19T06:46:11.590 回答
4

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 生成架构定义并从 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; }
}

这是一个说明序列化/反序列化的LinqPad片段（基于 David Colwell 的代码片段，感谢顺便说一句关于如何排除 BOM 的提示，这正是我想要的）：

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>
于 2016-02-03T00:59:49.170 回答
0

如果您需要其他类，例如 books 类中的 book2，则需要一些特殊说明来实现它。例子

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; }
}`

当我尝试运行该程序时，出现以下错误：

“附加信息：反映类型时出错”

于 2018-03-01T18:11:51.933 回答