如何在ASP.NET Core中用C#生成带METADATA的XML文件？

问题背景 #

我需要从数据库获取数据并转换为对象列表，最终生成带有METADATA标签的XML文件（期望格式如下）。希望DATAPACKET、METADATA这类外层标签无需手动编写，想知道ASP.NET Core中是否有C#类库可在序列化数据时自动生成它们。

尝试的代码 #

public class Book
{
    public string title;
    public string author;
    public string publisher;
    public double price;
}

public void GenerateXML()
{
   var overview = new List<Book>{
        new Book()
        {
            title = "This is a book",
            author = "Somebody wrote it",
            publisher = "Someone published it",
            price = 999.99
        },
        new Book()
        {
            title = "This is a book2",
            author = "Somebody wrote it2",
            publisher = "Someone published it2",
            price = 101010
        }
    };

    System.Xml.Serialization.XmlSerializer writer = new System.Xml.Serialization.XmlSerializer(typeof(Book));

    var path = Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments) + "//myXmFile.xml";
    System.IO.FileStream file = System.IO.File.Create(path);

    foreach (var item in overview)
    {
        writer.Serialize(file, item);
    }

    file.Close();
}

当前输出XML #

<?xml version="1.0" encoding="utf-8"?>
<Book xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <title>
        This is a book
    </title>
    <author>
        Somebody wrote it
    </author>
    <publisher>
        Someone published it
    </publisher>
    <price>
        999.99
    </price>
</Book>
<?xml version="1.0" encoding="utf-8"?>
<Book xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <title>
        This is a book2
    </title>
    <author>
        Somebody wrote it2
    </author>
    <publisher>
        Someone published it2
    </publisher>
    <price>
        101010
    </price>
</Book>

期望的XML格式 #

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>  
<DATAPACKET Version="2.0">
    <METADATA>
            <FIELDS>
                <FIELD attrname="title" fieldtype="string"/>
                <FIELD attrname="author" fieldtype="string"/>
                <FIELD attrname="publisher" fieldtype="string"/>
                <FIELD attrname="price" fieldtype="r8"/>
            </FIELDS>
        <PARAMS/>
    </METADATA>
    
    <ROWDATA>
        <ROW title="This is a book" author="Somebody wrote it" publisher="Someone published it" price="999.99" />
        <ROW title="This is a book2" author="Somebody wrote it2" publisher="Someone published it2" price="101010" />
    </ROWDATA>
</DATAPACKET>

解决方案 #

.NET原生没有专门针对该格式的序列化类库，但你要的格式正是DataSet默认的XML序列化结构，可以利用DataSet和DataTable自动生成包含METADATA的XML，无需手动编写外层标签。

实现步骤 #

1. 将对象列表转换为DataTable
2. 将DataTable添加到DataSet
3. 使用DataSet.WriteXml方法生成XML（指定XmlWriteMode.WriteSchema模式，自动包含元数据）

修改后的代码 #

public class Book
{
    public string title;
    public string author;
    public string publisher;
    public double price;
}

// 将Book列表转换为DataTable
private DataTable ConvertBooksToDataTable(List<Book> books)
{
    DataTable table = new DataTable();
    // 定义列和对应类型
    table.Columns.Add("title", typeof(string));
    table.Columns.Add("author", typeof(string));
    table.Columns.Add("publisher", typeof(string));
    table.Columns.Add("price", typeof(double));

    // 填充数据
    foreach (var book in books)
    {
        table.Rows.Add(book.title, book.author, book.publisher, book.price);
    }
    return table;
}

public void GenerateXML()
{
    var overview = new List<Book>
    {
        new Book()
        {
            title = "This is a book",
            author = "Somebody wrote it",
            publisher = "Someone published it",
            price = 999.99
        },
        new Book()
        {
            title = "This is a book2",
            author = "Somebody wrote it2",
            publisher = "Someone published it2",
            price = 101010
        }
    };

    DataTable bookTable = ConvertBooksToDataTable(overview);
    DataSet dataSet = new DataSet();
    dataSet.Tables.Add(bookTable);

    // 生成XML文件
    var path = Path.Combine(Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments), "myXmFile.xml");
    // WriteSchema模式会自动包含METADATA等结构
    dataSet.WriteXml(path, XmlWriteMode.WriteSchema);
}

替代方案：自定义序列化类 #

如果不想依赖DataSet，可以手动定义匹配目标XML结构的类，并通过System.Xml.Serialization特性标记，再序列化整个对象结构。这种方式需要手动编写所有层级的类，但灵活性更高。示例核心类结构如下：

[XmlRoot("DATAPACKET")]
public class DataPacket
{
    [XmlAttribute("Version")]
    public string Version { get; set; } = "2.0";

    [XmlElement("METADATA")]
    public Metadata Metadata { get; set; }

    [XmlElement("ROWDATA")]
    public RowData RowData { get; set; }
}

public class Metadata
{
    [XmlElement("FIELDS")]
    public Fields Fields { get; set; }

    [XmlElement("PARAMS")]
    public Params Params { get; set; } = new Params();
}

// 其他类（FIELDS、FIELD、ROWDATA、ROW）同理，用Xml特性标记对应节点和属性

总结 #

推荐使用DataSet方案，无需手动编写DATAPACKET、METADATA等外层标签，自动生成符合要求的XML结构，实现成本最低。

内容的提问来源于stack exchange，提问作者Jessie Loke