c# - 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; }
}

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.

有什么想法吗？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>

Answer 1

您不能使用标准序列化工具将问题中的类序列化为 <book>与 <bookNum> 处于同一级别的条目节点。

当使用标准序列化工具列表保存类时 <book>节点将始终嵌套到与 <bookNum> 处于同一级别的单独数组节点 中节点。同样的担忧records book 上的数组字段类。

要生成您想要的 XML 输出 - 使用 <book>与 <bookNum> 处于同一级别的节点节点 - 你必须实现 IXmlSerializable books 中的界面用于自定义序列化的类。查看 IXmlSerializable 的示例实现访问这些链接:StackOverflow answer , CodeProject article .

另一种解决方案将是 - 如上所述 user Alexandr评论我的回答 - 继承你的 books来自 List<book> 的类(class)输入并在您的 book 上显示类字段 records从 List<record> 继承的类类型类型。

从您的问题序列化类时，假设您分配的是正确的 XmlRoot, XmlElement, XmlArray and XmlArrayItem attributes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>