面对从数据创建的XML结构中的问题

I want to create XML through data, and i have data received from database in DataSet or IList<Booking>. Right now i am using DataSet to create XML by this code.

我想通过数据创建XML,并且我从DataSet或IList 中的数据库接收数据。现在我正在使用DataSet通过此代码创建XML。

string result = String.Empty;

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

And my XML is in this form.

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>

But i want XML in this form.

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>

Can any one help me to create this type of XML?

任何人都可以帮我创建这种类型的XML吗?

6 个解决方案

#1

Using System.Xml.Serialization, create objects of your bookings:

使用System.Xml.Serialization,创建预订对象:

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; }
  }
}

Now, for the serializer object (used to actually serialize the objects into a string):

现在,对于序列化程序对象(用于将对象实际序列化为字符串):

 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;
   }
 }

Now, you simply create the correctly defined objects and serialize in a function of some sort. For instance, in the main method of a console application:

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

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();
}

Obviously, you can use this in a refactored function (which would make life easier) but this gives you the general output you are seeking.

显然,你可以在重构函数中使用它(这会让生活变得更轻松),但这会为你提供你正在寻求的一般输出。

#2

Here is one possible way of doing it:

这是一种可行的方法:

        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();

It uses LINQ to XML. It reads column names to create appropriately named XML elements, so if you add or remove some columns from the table this shouldn't break down, the only columns that should have fixed column names are ID (Booking) and BookingID (BookingDetail) since they are used to link two tables.

它使用LINQ to XML。它读取列名称以创建适当命名的XML元素,因此如果您添加或删除表中的某些列,这不应该分解,应该具有固定列名称的唯一列是ID(预订)和BookingID(BookingDetail),因为它们用于链接两个表。

#3

Can you not use the XmlSerializer class on your data objects? Or am I misunderstanding the question?

你不能在数据对象上使用XmlSerializer类吗?或者我误解了这个问题?

Without seeing some object structure there's little I can do except provide some "If I were in your shoes..."

没有看到一些物体结构,除了提供一些“如果我在你的鞋子里......”之外我几乎无能为力。

What you want is "nonstandard" XML in that you have two types of objects "Booking" and "BookingDetail" under the same tag when one is a list (BookingDetail).

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

If it has to be in this form, the way I'd go about this is manual serialization:

如果必须采用这种形式,我的方法是手动序列化:

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

And the data objects:

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>";
    }
}

#4

Let say we have following classes for data:

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;}
}

And following test data:

并遵循测试数据:

    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" }
    };

We can easily get output XML in given format with following Linq to XML query:

我们可以使用以下Linq to XML查询轻松获取给定格式的输出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();

It would give us an array of XElement objects. To get xml string use String.Join<XElement> method:

它会给我们一个XElement对象数组。要获取xml字符串,请使用String.Join 方法:

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

However, I would suggest a bit different XML schema:

但是,我建议使用一些不同的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>

There is no redundancy in the data with that kind of format. To get it use following query:

这种格式的数据没有冗余。要使用以下查询:

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));

#5

You can create a custom structure using LINQ to XML. Use nested XElement constructors. See for instance How to create an XML of this structure and Build XML Dynamically using c#

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。请参阅参考例如如何使用c#创建此结构的XML和动态构建XML

#6

Sorry for a belated answer to follow my comment on your question, @Umair Noor.

对于迟到的回答,请关注我对您的问题的评论,@ Uirir Noor。

Since you are working with .NET DataSets, here is the easiest way I know to get the XML you want from one - with related online references:

由于您正在使用.NET DataSet,这是我所知道的最简单的方法 - 从相关的在线参考中获取您想要的XML:

1. Create an XML document representing the XML output that you would like to see. You have already done this in your question, but you will need to 1) replace <\BLNumber> closing tags with <\BookingNumber> tags and 2) address @ByteBlast's point regarding multiple root elements in your sample.

   创建一个表示您希望看到的XML输出的XML文档。您已在问题中完成此操作,但您需要1)用<\ BookingNumber>标记替换<\ BLNumber>结束标记,并且2)解决有关样本中多个根元素的@ ByteBlast点。

   Here is what I quickly ended up with by minimally adjusting your example of the XML you would like:

   通过最低限度地调整您想要的XML示例,我最终得到了以下内容:

   <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. Use the xsd.exe command-line tool to generate an XML Schema Definition (XSD) file from the representative XML file.

   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。

   Alternatively, write a few lines of code to read the XML file, using XmlReadMode.InferSchema to generate a schema for it; and write the generated schema to an XSD file. For example, here's what I quickly threw into a button event handler in a WPF scratch app:

   或者,编写几行代码来读取XML文件,使用XmlReadMode.InferSchema为其生成模式;并将生成的模式写入XSD文件。例如,这是我快速投入WPF临时应用程序中的按钮事件处理程序:

   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
   }
   

   Both methods are quick enough.

   这两种方法都足够快。

3. Use the XSD file...

   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>
   

   ...to define a typed DataSet (e.g. Bookings or BookingsDataSet) - rather than the general-purpose DataSet you are likely using.

   ...定义类型化的DataSet(例如Bookings或BookingsDataSet) - 而不是您可能使用的通用DataSet。

   You could use xsd.exe's /d[ataset] option with the XSD file from step 3, or you could add a new DataSet item to your project in Visual Studio, then paste the step-3-derived schema into its .xsd file.

   您可以将xsd.exe的/ d [ataset]选项与步骤3中的XSD文件一起使用,或者可以在Visual Studio中将新的DataSet项添加到项目中,然后将步骤3派生的模式粘贴到其.xsd文件中。

   Same - quick enough either way.

   相同 - 两种方式都足够快。

4. Repeat as necessary until you get the exact results you want.

   根据需要重复,直到获得所需的确切结果。

For authoritative reference, MSDN has a good overview of XML in ADO.NET that explains much of what I have described.

对于权威参考,MSDN对ADO.NET中的XML进行了很好的概述,这解释了我所描述的大部分内容。

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