基于.Net MVC模型的自定义XML feed

Question

我们正在使用XMLMVC5重做一个遗留应用程序。该应用程序的一部分会生成.Net提要，供其他应用程序使用。

我们构建了一个名为Call的模型

[XmlRoot("record")]
public class Call
{
    [XmlIgnore]
    public int ID { get; set; }
    [XmlIgnore]
    public string CustomerInitials { get; set; }
    [XmlIgnore]
    public string Prefix { get; set; }
    [XmlIgnore]
    public string Code { get; set; }
    [XmlIgnore]
    public string CustomerNumber { get; set; }
    [XmlElement("starttime")]
    public DateTime Entry { get; set; }
    [XmlElement("endtime")]
    public DateTime Exit { get; set; }
    [XmlElement("cnumber")][NotMapped]
    public string Number
    {
        get { return Prefix + Code + CustomerNumber; } 
    }
}

使用控制器：

public class CallsController : ApiController
{
    private DbContext db = new DbContext();

    // GET: api/Calls
    public IQueryable<Call> GetCalls()
    {
        return db.Calls;
    }
}

它会生成如下XML提要：

  <ArrayOfCall xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <Call>
      <starttime>2016-01-15T14:45:24.447</starttime>
      <endtime>2016-01-15T15:45:24.447</endtime>
    </Call>
    <Call>
      <starttime>2016-01-15T15:46:35.637</starttime>
      <endtime>2016-01-15T16:46:35.637</endtime>
    </Call>
  </ArrayOfCall>

但是，我需要定制XML提要以提供以下输出：

<?xml version="1.0" encoding="UTF-8"?>
<DATA>
    <RECORD>
        <CNUMBER>593042401</CNUMBER>
        <STARTTIME>2016-01-15T14:45:24.447</STARTTIME>
        <ENDTIME>2016-01-15T15:45:24.447</ENDTIME>
    </RECORD>
    <RECORD>
        <CNUMBER>593042401</CNUMBER>
        <STARTTIME>2016-01-15T15:46:35.637</STARTTIME>
        <ENDTIME>2016-01-15T16:46:35.637</ENDTIME>
    </RECORD>
</DATA>

没有遵循XmlRoot数据注释，CNumber也没有包含在我的XML提要中。调用应该在根为<data>的XML提要中显示为<record>。如何解决这些问题？我是不是走错路了？我应该改用视图模型吗？

Answer 1

1.-数据类：

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

namespace StackOverflow
{
    public class Call
    {
        [XmlIgnore]
        public int ID { get; set; }
        [XmlIgnore]
        public string CustomerInitials { get; set; }
        [XmlIgnore]
        public string Prefix { get; set; }
        [XmlIgnore]
        public string Code { get; set; }
        [XmlIgnore]
        public string CustomerNumber { get; set; }

        [XmlElement("CNUMBER")]
        public string Number
        {
            get { return Prefix + Code + CustomerNumber; }
        }
        [XmlElement("STARTTIME")]
        public DateTime Entry { get; set; }
        [XmlElement("ENDTIME")]
        public DateTime Exit { get; set; }
    }

    [XmlRoot(ElementName = "DATA")]
    public class Data
    {
        [XmlElement("RECORD")]
        public List<Call> Calls;
    }
}

2.-使用您的示例数据：

    //  1.- Data
    var data = new Data();
    var calls = new List<Call>
    {
        new Call
        {
            Entry = new DateTime(2016, 1, 15, 14, 45, 24, 447),
            Exit = new DateTime(2016, 1, 15, 15, 45, 24, 447)
        },
        new Call
        {
            Entry = new DateTime(2016, 1, 15, 15, 46, 35, 637),
            Exit = new DateTime(2016, 1, 15, 16, 46, 35, 637)
        }
    };
    data.Calls = new List<Call>(calls);

    //  2.- Serialize the objet to byte[]
    var dataByteArray = new XmlSerializerHelper<Data>().ObjectToByteArray(data, Encoding.GetEncoding("UTF-8"), true);

    //  3.- Save the byte[] to disk
    File.WriteAllBytes("D:/xml.xml", dataByteArray);

3.-序列化助手：

using System;
using System.IO;
using System.Text;
using System.Xml;
using System.Xml.Serialization;

namespace StackOverflow
{
    public class XmlSerializerHelper<T> where T : class
    {
        private readonly XmlSerializer _serializer;

        public XmlSerializerHelper()
        {
            _serializer = new XmlSerializer(typeof(T));
        }

        public byte[] ObjectToByteArray(T obj, Encoding encoding = null, bool ignoreNAmespaces = false)
        {
            var settings = GetSettings(encoding);
            using (var memoryStream = new MemoryStream())
            {
                using (var writer = XmlWriter.Create(memoryStream, settings))
                {
                    if (ignoreNAmespaces)
                    {
                        var serializerNamespaces = new XmlSerializerNamespaces();
                        serializerNamespaces.Add("", "");
                        _serializer.Serialize(writer, obj, serializerNamespaces);
                    }
                    else
                    {
                        _serializer.Serialize(writer, obj);
                    }
                }
                return memoryStream.ToArray();
            }
        }

        private XmlWriterSettings GetSettings(Encoding encoding)
        {
            return new XmlWriterSettings
            {
                Encoding = encoding ?? Encoding.GetEncoding("ISO-8859-1"),
                Indent = true,
                IndentChars = "\t",
                NewLineChars = Environment.NewLine,
                ConformanceLevel = ConformanceLevel.Document
            };
        }
    }
}

4.-输出：

<?xml version="1.0" encoding="utf-8"?>
<DATA>
    <RECORD>
        <STARTTIME>2016-01-15T14:45:24.447</STARTTIME>
        <ENDTIME>2016-01-15T15:45:24.447</ENDTIME>
    </RECORD>
    <RECORD>
        <STARTTIME>2016-01-15T15:46:35.637</STARTTIME>
        <ENDTIME>2016-01-15T16:46:35.637</ENDTIME>
    </RECORD>
</DATA>

干杯

Answer 2

要创建与您想要的内容相匹配的XML，可以创建一个名为Data的对象，该对象可以提供一个包含记录的列表。

public class Data{
   public List<Record> Records{get;set;}
}

目前您正在返回一个调用数组。