使用什么 XML 属性?
What XML attribute to use?
我正在使用 XmlSerializer,我想实现 XML 树:
<request>
<ix>ID</ix>
<content>
<name>NAMEVALUE</name>
<visits>INT</visits>
<dateRequested>yyyy-MM-dd</dateRequested>
</content>
</request>
型号:
[XmlRoot(ElementName = "request")]
public class RequestModel
{
[XmlElement("ix")]
[JsonProperty("ix")]
public int ID { get; set; }
[XmlElement("name")]
[JsonProperty("name")]
public string Name { get; set; }
[XmlElement("visits")]
[JsonProperty("visits")]
public int? Visits { get; set; }
[XmlElement("date")]
[JsonProperty("date")]
public DateTime Date { get; set; }
}
我应该使用什么属性来接收 XMl 树中的 <content>
组?
我的序列化器:
IEnumerable<RequestJSONModel> getModels = _context.Requests.ToList();
foreach (var item in getModels)
{
RequestModel requestModel = new RequestModel();
Content contentModel = new Content();
//serialize
XmlSerializer xmlSerializer = new XmlSerializer(typeof(RequestModel));
var serializedItem = "";
XmlSerializerNamespaces ns = new XmlSerializerNamespaces();
ns.Add("", "");
using (StringWriter writer = new Utf8StringWriter())
{
xmlSerializer.Serialize(writer, xmlModel, ns);
serializedItem = writer.ToString(); // Your XML
}
serializedItem = serializedItem.Replace("\r\n", string.Empty);
}
如果使用jdweng的解决方案,如何解析public class RequestModel
和public class Content
?
试试这个
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Serialization;
namespace ConsoleApplication1
{
class Program
{
const string FILENAME = @"c:\temp\test.xml";
static void Main(string[] args)
{
RequestModel model = new RequestModel() {
ix = 123,
content = new Content() {
Name = "NAMEVALUE",
Visits = 456,
dateRequested = "2018-12-16"
}
};
XmlWriterSettings settings = new XmlWriterSettings();
settings.Indent = true;
XmlWriter writer = XmlWriter.Create(FILENAME, settings);
XmlSerializer serializer = new XmlSerializer(typeof(RequestModel));
serializer.Serialize(writer, model);
}
}
[XmlRoot(ElementName = "request")]
public class RequestModel
{
[XmlElement("ix")]
public int ix { get; set; }
[XmlElement("content")]
public Content content { get; set; }
}
[XmlRoot(ElementName = "content")]
public class Content
{
[XmlElement("name")]
public string Name { get; set; }
[XmlElement("visits")]
public int? Visits { get; set; }
private DateTime Date { get; set; }
[XmlElement("dateRequested")]
public string dateRequested
{
get { return Date.ToString("yyyy-MM-dd"); }
set { Date = DateTime.ParseExact(value, "yyyy-MM-dd", System.Globalization.CultureInfo.InvariantCulture); }
}
}
}
我正在使用 XmlSerializer,我想实现 XML 树:
<request>
<ix>ID</ix>
<content>
<name>NAMEVALUE</name>
<visits>INT</visits>
<dateRequested>yyyy-MM-dd</dateRequested>
</content>
</request>
型号:
[XmlRoot(ElementName = "request")]
public class RequestModel
{
[XmlElement("ix")]
[JsonProperty("ix")]
public int ID { get; set; }
[XmlElement("name")]
[JsonProperty("name")]
public string Name { get; set; }
[XmlElement("visits")]
[JsonProperty("visits")]
public int? Visits { get; set; }
[XmlElement("date")]
[JsonProperty("date")]
public DateTime Date { get; set; }
}
我应该使用什么属性来接收 XMl 树中的 <content>
组?
我的序列化器:
IEnumerable<RequestJSONModel> getModels = _context.Requests.ToList();
foreach (var item in getModels)
{
RequestModel requestModel = new RequestModel();
Content contentModel = new Content();
//serialize
XmlSerializer xmlSerializer = new XmlSerializer(typeof(RequestModel));
var serializedItem = "";
XmlSerializerNamespaces ns = new XmlSerializerNamespaces();
ns.Add("", "");
using (StringWriter writer = new Utf8StringWriter())
{
xmlSerializer.Serialize(writer, xmlModel, ns);
serializedItem = writer.ToString(); // Your XML
}
serializedItem = serializedItem.Replace("\r\n", string.Empty);
}
如果使用jdweng的解决方案,如何解析public class RequestModel
和public class Content
?
试试这个
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Serialization;
namespace ConsoleApplication1
{
class Program
{
const string FILENAME = @"c:\temp\test.xml";
static void Main(string[] args)
{
RequestModel model = new RequestModel() {
ix = 123,
content = new Content() {
Name = "NAMEVALUE",
Visits = 456,
dateRequested = "2018-12-16"
}
};
XmlWriterSettings settings = new XmlWriterSettings();
settings.Indent = true;
XmlWriter writer = XmlWriter.Create(FILENAME, settings);
XmlSerializer serializer = new XmlSerializer(typeof(RequestModel));
serializer.Serialize(writer, model);
}
}
[XmlRoot(ElementName = "request")]
public class RequestModel
{
[XmlElement("ix")]
public int ix { get; set; }
[XmlElement("content")]
public Content content { get; set; }
}
[XmlRoot(ElementName = "content")]
public class Content
{
[XmlElement("name")]
public string Name { get; set; }
[XmlElement("visits")]
public int? Visits { get; set; }
private DateTime Date { get; set; }
[XmlElement("dateRequested")]
public string dateRequested
{
get { return Date.ToString("yyyy-MM-dd"); }
set { Date = DateTime.ParseExact(value, "yyyy-MM-dd", System.Globalization.CultureInfo.InvariantCulture); }
}
}
}