序列化XML时更改元素的顺序 [英] Change the order of elements when serializing XML

问题描述

我需要将一个对象序列化为XML并返回. XML已修复，我无法更改. 在bookingList之后，我无法生成此结构.

I need to serialize an Object to XML and back. The XML is fix and I can't change it. I fail to generate this structure after bookingList.

如何将这些<booking>元素分组"以显示为LIST并保持<error>& <counter>在此<booking>元素列表之前.

How can I "group" these <booking> elements to appear as a LIST and keep <error> & <counter> before this List of <booking> elements.

在这里查看我的示例:

我需要的结构....

<nicexml>
<key_id>1234567</key_id>
<surname>Jil</surname>
<name>Sander</name>
<station_id>1</station_id>
<ownBookings>
    <bookingList>
        <error></error>
        <counter>20</counter>
        <booking>
             <bookingID>1234567890</bookingID>
        </booking>
        <booking>
             <bookingID>2345678901</bookingID>
        </booking>
    </bookingList>
</ownBookings>
</nicexml>

我使用下面的C#代码获得的结构....

Structure i get with C# code below....

<nicexml>
<key_id>1234567</key_id>
<surname>Jil</surname>
<name>Sander</name>
<station_id>1</station_id>
<ownBookings>
    <bookingList>
           <booking>
        <booking>
             <bookingID>1234567890</bookingID>
        </booking>
        <booking>
             <bookingID>2345678901</bookingID>
        </booking>
             <booking>
        <error></error>
        <counter>20</counter>
    </bookingList>
</ownBookings>
</nicexml>

C#代码:

using System;
using System.Xml.Serialization;
using System.Collections.Generic;

namespace xml_objects_serials
{
    public class bookings
    {
        public class nicexml
        {
            public string key_id
            { get; set; }

            public string surname
            { get; set; }

            public string name
            { get; set; }

            public int station_id
            { get; set; }

            public ownBookings ownBookings
            { get; set; }

        }

        public class ownBookings
        {
            public bookingList bookingList
            { get; set; }

        }
        public class bookingList {
            public string error 
            { get; set; }
            public int counter
            { get; set; }
            public List<booking> booking= new List<booking>();
        }

        public class booking
        {
            public int bookingID
            { get; set; }
        }
    }

推荐答案

尝试使用这是一个例子:

public class bookingList
{
    [XmlElement(Order = 1)]
    public string error { get; set; }
    [XmlElement(Order = 2)]
    public int counter { get; set; }
    [XmlElement(ElementName = "booking", Order = 3)]
    public List<booking> bookings = new List<booking>();
}

public class booking
{
    public int id { get; set; }
}

在我的测试中，我获得了以下输出:

In my test I obtained this output:

<?xml version="1.0" ?> 
<bookingList>
    <error>sample</error>
    <counter>0</counter>
    <booking>
        <id>1</id> 
    </booking>
    <booking>
        <id>2</id> 
    </booking>
    <booking>
        <id>3</id> 
    </booking> 
</bookingList>

相关资源:

• 使用属性控制XML序列化

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