<countries>
<country code="AF" iso="4">Afghanistan</country>
<country code="AL" iso="8">Albania</country>
<country code="DZ" iso="12">Algeria</country>
<country code="AS" iso="16">American Samoa</country>
<country code="AD" iso="20">Andorra</country>
<country code="AO" iso="24">Angola</country>
<country code="AI" iso="660">Anguilla</country>
<country code="AQ" iso="10">Antarctica</country>
<country code="AG" iso="28">Antigua And Barbuda</country>
<country code="AR" iso="32">Argentina</country>
<country code="AM" iso="51">Armenia</country>
<country code="AW" iso="533">Aruba</country>
<country code="AU" iso="36">Australia</country>
<country code="AT" iso="40">Austria</country>
<country code="AZ" iso="31">Azerbaijan</country>
<country code="BS" iso="44">Bahamas</country>
<country code="BH" iso="48">Bahrain</country>
<country code="BD" iso="50">Bangladesh</country>
<country code="BB" iso="52">Barbados</country>
<country code="BY" iso="112">Belarus</country>
<country code="BE" iso="56">Belgium</country>
<country code="BZ" iso="84">Belize</country>
<country code="BJ" iso="204">Benin</country>
<country code="BM" iso="60">Bermuda</country>
<country code="BT" iso="64">Bhutan</country>
<country code="BO" iso="68">Bolivia</country>
<country code="BA" iso="70">Bosnia And Herzegovina</country>
<country code="BW" iso="72">Botswana</country>
<country code="BV" iso="74">Bouvet Island</country>
<country code="BR" iso="76">Brazil</country>
<country code="IO" iso="86">British Indian Ocean Territory</country>
<country code="BN" iso="96">Brunei Darussalam</country>
<country code="BG" iso="100">Bulgaria</country>
<country code="BF" iso="854">Burkina Faso</country>
<country code="BI" iso="108">Burundi</country>
</countries>有人请指导我如何设计我的类来反序列化这个。
这是我目前的课程设计
public class Country
{
public string country { get; set; }
public string code { get; set; }
public int iso { get; set; }
}但这似乎行不通。请有人指点我。
发布于 2015-05-18 15:36:49
与目前为止的其他答案相反,您在使用Xml序列化时不需要使用Serializable属性。但是,您确实需要用代码属性来装饰属性,这些属性描述了Xml文件中的值将从哪个部分获取。
由于您的问题中没有包含Xml文档声明,所以我不确定countries集合是否是文档的根节点。但是,假设您的整个Xml文档实际上如下所示:
<?xml version="1.0" encoding="utf-8" ?>
<countries>
<country code="AF" iso="4">Afghanistan</country>
<country code="AL" iso="8">Albania</country>
<country code="DZ" iso="12">Algeria</country>
</countries>您需要将代码属性应用到类中,这些类描述上述Xml如何映射到属性和对象。这些属性是在System.Xml中定义的。在您的情况下,这些属性可能是这样的:
using System;
using System.IO;
using System.Xml;
using System.Xml.Serialization;
...
[XmlRoot("countries", Namespace="")]
public class countriesDocument
{
[XmlElement("country")]
public country[] countries { get; set; }
}
public class country
{
[XmlText]
public string name { get; set; }
[XmlAttribute]
public string code { get; set; }
[XmlAttribute]
public int iso { get; set; }
}然后,可以使用如下代码对文档进行反序列化:
var serializer = new XmlSerializer(typeof(countriesDocument));
countriesDocument document;
using (var reader = File.OpenText("countries.xml"))
{
document = (countriesDocument)serializer.Deserialize(reader);
}发布于 2015-05-18 15:40:32
Windows 8应用程序没有可序列化的属性,您必须使用DataContractAttribute和DataMemberAttribute来装饰模型类,如下所示:
[DataContractAttribute]
public class Country
{
[DataMemberAttribute]
public string country { get; set; }
[DataMemberAttribute]
public string code { get; set; }
[DataMemberAttribute]
public int iso { get; set; }
}然后您可以序列化这个类,下面是Json的示例
public static string SerializeToJson(object instance)
{
using (MemoryStream _Stream = new MemoryStream())
{
var _Serializer = new DataContractJsonSerializer(instance.GetType());
_Serializer.WriteObject(_Stream, instance);
_Stream.Position = 0;
using (StreamReader _Reader = new StreamReader(_Stream))
{ return _Reader.ReadToEnd(); }
}
}有关XML的示例,请参见此答案:
发布于 2015-05-18 15:02:12
序列化反序列化类的第一步是用Serializable属性标记它。
[Serializable]
public class Country
{
public string country { get; set; }
public string code { get; set; }
public int iso { get; set; }
}https://stackoverflow.com/questions/30306578
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