XmlSerializer将C#对象转换为xml字符串

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; }
}

而是当我使用XmlSerializer转换为XML时。结果与下面的xml不一样。

我的C#课有什么问题？我希望使用XmlSerializer来输出结果，而不是使用XmlDocument。

有什么想法吗？提前感谢！

<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>

Answer 1

您不能使用标准序列化工具从您的问题中序列化类，以便它将具有与<bookNum>节点相同的级别上的<bookNum>条目。

当用标准序列化工具保存类时，<book>节点的列表总是嵌套在与<bookNum>节点相同级别的单独数组节点中。records数组字段在book类上也是如此。

要生成您想要的XML输出--将<book>节点放在与<bookNum>节点相同的级别--您必须在您的books类中实现IXmlSerializable接口，以便进行自定义序列化。要查看IXmlSerializable实现的示例，请访问以下链接：StackOverflow答案，CodeProject文章。

另一种解决方案是--正如我对我的答案的注释中所述的那样--从用户Alexandr类型继承books类，并在book类字段中拥有从List<record>类型继承的类类型的records。

当从您的问题序列化类时，假设您分配的适当XmlRoot、XmlElement、XmlArray和XmlArrayItem属性如下：

[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; }
}

您将得到如下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>

Answer 2

我对您的类代码做了以下更改。我无法使用默认序列化程序复制XML序列化，因为如果不给它一个容器元素，它就不会复制“Record”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; }
}

序列化这将给我以下输出

<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，但是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生成模式定义和从xsd文件生成.Net代码来发现这一点。据我所知，这在.Net 3.5到4.6中是可行的。

下面是我使用的类定义：

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; }
}

下面是一个LinqPad片段，它演示了序列化/反序列化(基于David的代码片段，感谢BTW关于如何排除BOM的技巧，这正是我正在寻找的)：

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();

这个生成的XML可以在不实现IXmlSerializable的情况下被序列化和反序列化，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>