3

我想通过数据创建 XML，并且我从数据库中接收到数据DataSetor IList<Booking>。现在我正在使用DataSet此代码创建 XML。

string result = String.Empty;

using (StringWriter sw = new StringWriter())
 {
   ds.WriteXml(sw);
   result = sw.ToString();
 }

我的 XML 就是这种形式。

<Booking> 
    <ID>32</ID> 
    <BookingNumber>12120001</BLNumber> 
    <ReferenceNo>ABCED11212280007</ReferenceNo> 
    <Name>Customer Name1</Name> 
    <Address>Customer Address</Address>
</Booking>

<Booking> 
    <ID>33</ID> 
    <BookingNumber>12120002</BLNumber> 
    <ReferenceNo>ABCED11212280008</ReferenceNo> 
    <Name>Customer Name2</Name> 
    <Address>Customer Address2</Address>
</Booking>



<BookingDetail> 
   <ID>206</ID> 
   <BookingID>32</BookingID> 
   <OrderItem>Item1</OrderItem> 
</BookingDetail>

<BookingDetail> 
   <ID>207</ID> 
   <BookingID>32</BookingID> 
   <OrderItem>Item2</OrderItem> 
</BookingDetail>

<BookingDetail> 
   <ID>208</ID> 
   <BookingID>33</BookingID> 
   <OrderItem>Item1</OrderItem> 
</BookingDetail>

<BookingDetail> 
   <ID>209</ID> 
   <BookingID>33</BookingID> 
   <OrderItem>Item2</OrderItem> 
</BookingDetail>

<BookingDetail> 
   <ID>210</ID> 
   <BookingID>33</BookingID> 
   <OrderItem>Item3</OrderItem> 
</BookingDetail>

但我想要这种形式的 XML。

<CompleteBooking>
 <Booking> 
    <ID>32</ID> 
    <BookingNumber>12120001</BLNumber> 
    <ReferenceNo>ABCED11212280007</ReferenceNo> 
    <Name>Customer Name1</Name> 
    <Address>Customer Address</Address>
 </Booking>

 <BookingDetail> 
   <ID>206</ID> 
   <BookingID>32</BookingID> 
   <OrderItem>Item1</OrderItem> 
 </BookingDetail>

 <BookingDetail> 
   <ID>207</ID> 
   <BookingID>32</BookingID> 
   <OrderItem>Item2</OrderItem> 
 </BookingDetail>

</CompleteBooking>

<CompleteBooking>
 <Booking> 
    <ID>33</ID> 
    <BookingNumber>12120002</BLNumber> 
    <ReferenceNo>ABCED11212280008</ReferenceNo> 
    <Name>Customer Name2</Name> 
    <Address>Customer Address2</Address>
 </Booking>

 <BookingDetail> 
   <ID>208</ID> 
   <BookingID>33</BookingID> 
   <OrderItem>Item1</OrderItem> 
 </BookingDetail>

 <BookingDetail> 
   <ID>209</ID> 
   <BookingID>33</BookingID> 
   <OrderItem>Item2</OrderItem> 
 </BookingDetail>

 <BookingDetail> 
   <ID>210</ID> 
   <BookingID>33</BookingID> 
   <OrderItem>Item3</OrderItem> 
 </BookingDetail>

</CompleteBooking>

谁能帮我创建这种类型的 XML？

4

6 回答 6

5

使用 System.Xml.Serialization，创建您的预订对象：

public class XMLEntities
{   
  [XmlRoot(ElementName = "CompleteBooking")]
  public class CompleteBooking
  {
    [XmlElement(ElementName = "Booking")]
    public Booking Bookings { get; set; }

    [XmlElement(ElementName = "BookingDetail")]
    public List<BookingDetail> BookingDetail { get; set; }
  }

  public class Booking
  {
    [XmlElement("ID")]
    public int ID { get; set; }

    [XmlElement("BookingNumber")]
    public int BookingNumber { get; set; }

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

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

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

  public class BookingDetail
  {
    [XmlElement("ID")]
    public int ID { get; set; }

    [XmlElement("BookingID")]
    public int BookingID { get; set; }

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

现在，对于序列化器对象（用于实际将对象序列化为字符串）：

 public class XMLEntitiesSerializer
 {
   public string Serialize(XMLEntities.CompleteBooking completeBooking)
   {
     var serializedXml = string.Empty;

     var serializer = new XmlSerializer(typeof (XMLEntities.CompleteBooking));
     var stringWriter = new System.IO.StringWriter();

     try
     {
       serializer.Serialize(stringWriter, completeBooking);
       serializedXml = stringWriter.ToString();
     }
     catch(Exception ex)
     {
       //Log the stuff
     }
     finally
     {
       stringWriter.Close();
     }

     return serializedXml;
   }
 }

现在，您只需创建正确定义的对象并在某种函数中序列化。例如，在控制台应用程序的 main 方法中：

public static void Main(string[] args)
{
  //Create new booking objects
  var booking1 = new XMLEntities.Booking()
                  {
                    ID = 32,
                    BookingNumber = 1212001,
                    ReferenceNumber = "ABCED11212280007",
                    Name = "Customer Name1",
                    Address = "Customer Address"
                  };
  var booking2 = new XMLEntities.Booking()
                   {
                     ID = 33,
                     BookingNumber = 12120002,
                     ReferenceNumber = "ABCED11212280008",
                     Name = "Customer Name2",
                     Address = "Customer Address2"
                   };

  //Create the booking details objects

  var booking1Detail1 = new XMLEntities.BookingDetail()
                         {
                           ID = 206,
                           BookingID = 32,
                           OrderItem = "Item1"
                         };
  var booking1Detail2 = new XMLEntities.BookingDetail()
                          {
                            ID = 207,
                            BookingID = 32,
                            OrderItem = "Item2"
                          };

  var booking2Detail1 = new XMLEntities.BookingDetail()
                          {
                            ID = 208,
                            BookingID = 33,
                            OrderItem = "Item1"
                          };
  var booking2Detail2 = new XMLEntities.BookingDetail()
                          {
                            ID = 209,
                            BookingID = 32,
                            OrderItem = "Item2"
                          };
  var booking2Detail3 = new XMLEntities.BookingDetail()
                          {
                            ID = 210,
                            BookingID = 32,
                            OrderItem = "Item3"
                          };

  //Smash them together so we can serialize as one

  var completeBooking1 = new XMLEntities.CompleteBooking()
                          {
                            Bookings = booking1,
                            BookingDetail = new List<XMLEntities.BookingDetail>()
                                              {
                                                booking1Detail1,
                                                booking1Detail2
                                              }
                          };
  var completeBooking2 = new XMLEntities.CompleteBooking()
                           {
                             Bookings = booking2,
                             BookingDetail = new List<XMLEntities.BookingDetail>()
                                               {
                                                 booking2Detail1,
                                                 booking2Detail2,
                                                 booking2Detail3
                                               }
                           };

  //Serialize the data for each of the booking objects

  var serializedXML = new XMLEntitiesSerializer();
  var xml = string.Empty;

  var booking1XmlString = serializedXML.Serialize(completeBooking1);
  var booking2XmlString = serializedXML.Serialize(completeBooking2);

  Console.ReadLine();
}

显然，您可以在重构函数中使用它（这将使生活更轻松），但这会为您提供您正在寻找的一般输出。

于 2013-02-22T16:28:33.253 回答
3

这是一种可能的方法：

        DataTable Booking = new DataTable();
        Booking.Columns.AddRange(new DataColumn[]{ new DataColumn("ID"), new DataColumn("BookingNumber"), new DataColumn("ReferenceNo"), new DataColumn("Name"), new DataColumn("Address") });

        DataTable BookingDetail = new DataTable();
        BookingDetail.Columns.AddRange(new DataColumn[] { new DataColumn("ID"), new DataColumn("BookingID"), new DataColumn("OrderItem") });

        Booking.Rows.Add(32, 12120001, "ABCED11212280007", "Customer Name1", "Customer Address");
        BookingDetail.Rows.Add(206, 32, "Item1");
        BookingDetail.Rows.Add(207, 32, "Item2");

        Booking.Rows.Add(33, 12120002, "ABCED11212280008", "Customer Name2", "Customer Address2");
        BookingDetail.Rows.Add(208, 33, "Item1");
        BookingDetail.Rows.Add(209, 33, "Item2");
        BookingDetail.Rows.Add(210, 33, "Item3");

        XElement root = new XElement("Root");

        // For each row from Booking add one CompleteBooking element
        foreach(DataRow BookingRow in Booking.Rows.Cast<DataRow>())
        {
            XElement xeCompleteBooking = new XElement("CompleteBooking");

            XElement xeBooking = new XElement("Booking");
            int BookingID = Convert.ToInt32(BookingRow["ID"]);

            IEnumerable<string> columnNames_Booking = Booking.Columns.Cast<DataColumn>().Select(col => col.ColumnName);
            // Add element under Booking element for every column of table
            foreach (string colName in columnNames_Booking)
                xeBooking.Add(new XElement(colName, BookingRow[colName]));

            xeCompleteBooking.Add(xeBooking);

            IEnumerable<string> columnNames_BookingDetail = BookingDetail.Columns.Cast<DataColumn>().Select(col => col.ColumnName);

            // For Booking.ID find all BookingDetail rows according to BookingDetail.BookingID
            IEnumerable<DataRow> details = BookingDetail.Rows.Cast<DataRow>().Where(BookingDetailRow => Convert.ToInt32(BookingDetailRow["BookingID"]) == BookingID);
            foreach (DataRow BookingDetailRow in details)
            {
                XElement xeBookingDetail = new XElement("BookingDetail");

                // Add element under BookingDetail element for every column of table
                foreach (string colName in columnNames_BookingDetail)
                    xeBookingDetail.Add(new XElement(colName, BookingDetailRow[colName]));

                xeCompleteBooking.Add(xeBookingDetail);
            }

            root.Add(xeCompleteBooking);
        }

        string xml = root.ToString();

它使用 LINQ to XML。它读取列名以创建适当命名的 XML 元素，因此如果您从表中添加或删除某些列，这不应该分解，唯一应该具有固定列名的列是ID( Booking) 和BookingID( BookingDetail)，因为它们用于链接两张桌子。

于 2013-02-21T08:06:04.803 回答
2

你不能在你的数据对象上使用 XmlSerializer 类吗？还是我误解了这个问题？

在没有看到一些对象结构的情况下，我无能为力，只能提供一些“如果我在你的鞋子里……”

您想要的是“非标准”XML，因为当一个是列表（BookingDetail）时，您在同一标签下有两种类型的对象“Booking”和“BookingDetail”。

如果必须采用这种形式，我会采用手动序列化的方式：

public String Serialize(CompleteBooking [] cbs) {
    String FinalXML = "<CompleteBookings>";
    foreach(CompleteBooking cb in cbs) {
        FinalXML += cb.ToXML();
    }
    FinalXML += "</CompleteBookings>";
}

和数据对象：

public class CompleteBooking {
    public Booking Booking;
    public BookingDetail [] BookingDetails

    public String ToXML() {
        String RVal = "<CompleteBooking>" + this.Booking.ToXML();
        foreach(BookingDetail bd in BookingDetails) {
            RVal += bd.ToXML();
        }
        RVal += "</CompleteBooking>"


    }
}

public class Booking {
    // Fields Here
    public String ToXML() {
        return "<Booking>" + [Fields] + "</Booking>";
    }
}

public class BookingDetail {
    // Fields Here
    public String ToXML() {
        return "<BookingDetail>" + [Fields] + "</BookingDetail>";
    }
}
于 2013-02-15T06:52:33.600 回答
2

假设我们有以下数据类：

class Booking
{
    public int ID { get; set;}
    public int BookingNumber { get; set;}
    public string ReferenceNo { get; set;}
    public string Name { get; set;}
    public string Address { get; set;}
}

class BookingDetails
{
    public int ID { get; set;}
    public int BookingId { get; set;}
    public string OrderItem { get; set;}
}

以及以下测试数据：

    static private IList<Booking> _bookings = new List<Booking>() {
        new Booking() { ID = 32, BookingNumber = 12120001, ReferenceNo = "ABCED11212280007", Name = "Customer Name1", Address = "Customer Address" },
        new Booking() { ID = 33, BookingNumber = 12120002, ReferenceNo = "ABCED11212280008", Name = "Customer Name2", Address = "Customer Address2" }
    };

    static private IList<BookingDetails> _details = new List<BookingDetails>() {
        new BookingDetails() { ID = 206, BookingId = 32, OrderItem = "Item1" },
        new BookingDetails() { ID = 207, BookingId = 32, OrderItem = "Item2" },
        new BookingDetails() { ID = 208, BookingId = 33, OrderItem = "Item1" },
        new BookingDetails() { ID = 209, BookingId = 33, OrderItem = "Item2" },
        new BookingDetails() { ID = 210, BookingId = 33, OrderItem = "Item3" }
    };

我们可以使用以下 Linq to XML 查询轻松获得给定格式的输出 XML：

var bookings = _bookings.Join(_details, b => b.ID, d => d.BookingId, (b, d) => new { b, d })
                        .GroupBy(g => g.b, g => g.d)
                        .Select(g => new XElement("CompleteBooking",
                                        new XElement("Booking", 
                                            new XElement("ID", g.Key.ID),
                                            new XElement("BookingNumber", g.Key.BookingNumber),
                                            new XElement("ReferenceNo", g.Key.ReferenceNo),
                                            new XElement("Name", g.Key.Name),
                                            new XElement("Address", g.Key.Address)),
                                        g.Select(d => new XElement("BookingDetail",
                                                            new XElement("ID", d.ID),
                                                            new XElement("BookingID", d.BookingId),
                                                            new XElement("OrderItem", d.OrderItem))).ToArray())).ToArray();

它会给我们一个 XElement 对象数组。获取xml字符串使用String.Join<XElement>方法：

var xmlString = String.Join<XElement>(Environment.NewLine, bookings);

但是，我会建议一些不同的 XML 模式：

<Bookings>
  <Booking>
    <ID>32</ID>
    <BookingNumber>12120001</BookingNumber>
    <ReferenceNo>ABCED11212280007</ReferenceNo>
    <Name>Customer Name1</Name>
    <Address>Customer Address</Address>
    <Details>
      <Detail>
        <ID>206</ID>
        <OrderItem>Item1</OrderItem>
      </Detail>
      <Detail>
        <ID>207</ID>
        <OrderItem>Item2</OrderItem>
      </Detail>
    </Details>
  </Booking>
  <Booking>
    <ID>33</ID>
    <BookingNumber>12120002</BookingNumber>
    <ReferenceNo>ABCED11212280008</ReferenceNo>
    <Name>Customer Name2</Name>
    <Address>Customer Address2</Address>
    <Details>
      <Detail>
        <ID>208</ID>
        <OrderItem>Item1</OrderItem>
      </Detail>
      <Detail>
        <ID>209</ID>
        <OrderItem>Item2</OrderItem>
      </Detail>
      <Detail>
        <ID>210</ID>
        <OrderItem>Item3</OrderItem>
      </Detail>
    </Details>
  </Booking>
</Bookings>

这种格式的数据没有冗余。要获得它，请使用以下查询：

var bookings = _bookings.Join(_details, b => b.ID, d => d.BookingId, (b, d) => new { b, d })
                        .GroupBy(g => g.b, g => g.d)
                        .Select(g => new XElement("Booking", 
                                        new XElement("ID", g.Key.ID),
                                        new XElement("BookingNumber", g.Key.BookingNumber),
                                        new XElement("ReferenceNo", g.Key.ReferenceNo),
                                        new XElement("Name", g.Key.Name),
                                        new XElement("Address", g.Key.Address),
                                        new XElement("Details",
                                            g.Select(d => new XElement("Detail",
                                                              new XElement("ID", d.ID),
                                                              new XElement("OrderItem", d.OrderItem))).ToArray()))).ToArray();

var data = new XDocument(new XElement("Bookings", bookings));
于 2013-02-23T20:53:21.953 回答
1

很抱歉，我对您的问题@Umair Noor 的评论迟到了。

由于您使用DataSet的是 .NET，因此这是我所知道的从其中获取所需 XML 的最简单方法 - 以及相关的在线参考：

  1. 创建一个表示您希望看到的 XML 输出的 XML 文档。 您已经在您的问题中完成了此操作，但您需要 1）用标签替换<\BLNumber>结束标签<\BookingNumber>和 2）解决 @ByteBlast 关于样本中多个根元素的观点。

    通过对您想要的 XML 示例进行最低限度的调整，我很快就得出了以下结论：

    <Bookings> <!-- fix for multiple root elements -->
  <CompleteBooking>
    <Booking>
      <ID>32</ID>
      <BookingNumber>12120001</BookingNumber> <!-- fixed BLNumber closing tag -->
      <ReferenceNo>ABCED11212280007</ReferenceNo>
      <Name>Customer Name1</Name>
      <Address>Customer Address</Address>
    </Booking>

    <BookingDetail>
      <ID>206</ID>
      <BookingID>32</BookingID>
      <OrderItem>Item1</OrderItem>
    </BookingDetail>

    <BookingDetail>
      <ID>207</ID>
      <BookingID>32</BookingID>
      <OrderItem>Item2</OrderItem>
    </BookingDetail>

  </CompleteBooking>

  <CompleteBooking>
    <Booking>
      <ID>33</ID>
      <BookingNumber>12120002</BookingNumber> <!-- fixed BLNumber closing tag -->
      <ReferenceNo>ABCED11212280008</ReferenceNo>
      <Name>Customer Name2</Name>
      <Address>Customer Address2</Address>
    </Booking>

    <BookingDetail>
      <ID>208</ID>
      <BookingID>33</BookingID>
      <OrderItem>Item1</OrderItem>
    </BookingDetail>

    <BookingDetail>
      <ID>209</ID>
      <BookingID>33</BookingID>
      <OrderItem>Item2</OrderItem>
    </BookingDetail>

    <BookingDetail>
      <ID>210</ID>
      <BookingID>33</BookingID>
      <OrderItem>Item3</OrderItem>
    </BookingDetail>

  </CompleteBooking>
</Bookings> <!-- fix for multiple root elements -->

  2. 使用xsd.exe命令行工具从代表性 XML 文件生成 XML 架构定义 (XSD) 文件。

    或者，编写几行代码来读取 XML 文件，用于XmlReadMode.InferSchema为其生成模式；并将生成的架构写入 XSD 文件。例如，这是我在 WPF 暂存应用程序中快速放入按钮事件处理程序的内容：

    private void Button_Click_1(object sender, RoutedEventArgs e)
{
    // BEGIN: the bottom-line logic
    var ds = new DataSet();

    var sr = new StreamReader("Bookings.xml"); // file into which I put your sample, desired XML
    ds.ReadXml(sr, XmlReadMode.InferSchema); // XmlReadMode.InferSchema - key
    sr.Close();

    ds.WriteXmlSchema("Bookings.xsd"); // file into which I got the resulting schema
    // END: the bottom-line logic
}

    这两种方法都足够快。

  3. 使用 XSD 文件...

    <?xml version="1.0" standalone="yes"?>
<xs:schema id="Bookings" xmlns="" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:msdata="urn:schemas-microsoft-com:xml-msdata">
  <xs:element name="Bookings" msdata:IsDataSet="true" msdata:UseCurrentLocale="true">
    <xs:complexType>
      <xs:choice minOccurs="0" maxOccurs="unbounded">
        <xs:element name="CompleteBooking">
          <xs:complexType>
            <xs:sequence>
              <xs:element name="Booking" minOccurs="0" maxOccurs="unbounded">
                <xs:complexType>
                  <xs:sequence>
                    <xs:element name="ID" type="xs:string" minOccurs="0" />
                    <xs:element name="BookingNumber" type="xs:string" minOccurs="0" />
                    <xs:element name="ReferenceNo" type="xs:string" minOccurs="0" />
                    <xs:element name="Name" type="xs:string" minOccurs="0" />
                    <xs:element name="Address" type="xs:string" minOccurs="0" />
                  </xs:sequence>
                </xs:complexType>
              </xs:element>
              <xs:element name="BookingDetail" minOccurs="0" maxOccurs="unbounded">
                <xs:complexType>
                  <xs:sequence>
                    <xs:element name="ID" type="xs:string" minOccurs="0" />
                    <xs:element name="BookingID" type="xs:string" minOccurs="0" />
                    <xs:element name="OrderItem" type="xs:string" minOccurs="0" />
                  </xs:sequence>
                </xs:complexType>
              </xs:element>
            </xs:sequence>
          </xs:complexType>
        </xs:element>
      </xs:choice>
    </xs:complexType>
  </xs:element>
</xs:schema>

    ...定义类型化的DataSet（例如Bookingsor ） - 而不是您可能使用BookingsDataSet的通用目的。DataSet

    您可以将xsd.exe'/d[ataset]选项与步骤 3 中的 XSD 文件一起使用，或者您可以DataSet在 Visual Studio 中将新项目添加到您的项目，然后将步骤 3 派生的架构粘贴到其.xsd文件中。

    相同 - 无论哪种方式都足够快。

  4. 根据需要重复，直到获得所需的确切结果。

作为权威参考，MSDN很好地概述了 ADO.NET中的 XML，它解释了我所描述的大部分内容。

于 2013-02-25T11:23:47.460 回答
1

您可以使用 LINQ to XML 创建自定义结构。使用嵌套的 XElement 构造函数。例如，请参阅如何创建这种结构的 XML并使用 c# 动态构建 XML

于 2013-02-15T06:45:38.850 回答