我希望能够“处理” XML文件中的“车辆”。理想情况下,我想遍历所有车辆并计算它们的价格,并更改它们是否为OnSale。然后,这些值将显示在UI中。我的代码反序列化了XML文件,但是我无法访问Vehicle的任何属性。我不需要将对象序列化回XML。
我试图用Console.WriteLine价格,但是当我运行代码时,它返回为0。是否应该创建一个ResponseGeographyVendorRegionVehicle数组?然后以某种方式将这种类型的对象添加到数组中?
这是XML文件:
<?xml version="1.0" encoding="utf-8" ?>
<Response>
<Geography>
<Vendor id="JOHN">
<Region id="1"></Region>
<Region id="2">
<Vehicle Make="HONDA" Fuel="Gas" Price="12000" OnSale="Y" Account="JOHNH" />
<Vehicle Make="ACURA" Fuel="Gas" Price="14100" OnSale="Y" Account="JOHNH" />
<Vehicle Make="TOYOTA" Fuel="Gas" Price="8000" OnSale="N" Account="JOHNH" />
<Vehicle Make="HYUNDAI" Fuel="Gas" Price="13000" OnSale="Y" Account="JOHNH" />
<Vehicle Make="INFINITY" Fuel="Gas" Price="16000" OnSale="N" Account="JOHNH" />
</Region>
<Region id="3"></Region>
<Region id="4"></Region>
</Vendor>
</Geography>
</Response>
这是我的Program.cs:
namespace XMLDeserializeExample
{
class Program
{
static void Main(string[] args)
{
string path = @"c:\XMLFile1.xml";
XmlRootAttribute xRoot = new XmlRootAttribute();
xRoot.ElementName = "Response";
XmlSerializer ser = new XmlSerializer(typeof(ResponseGeographyVendorRegionVehicle), xRoot);
ResponseGeographyVendorRegionVehicle i;
using (Stream reader = new FileStream(path,FileMode.Open))
{
i = (ResponseGeographyVendorRegionVehicle)ser.Deserialize(reader);
Console.WriteLine(i.Price);
Console.ReadLine();
}
}
}
}
这是创建的“粘贴特殊响应.CS”文件:
namespace XMLDeserializeExample
{
}
// NOTE: Generated code may require at least .NET Framework 4.5 or .NET Core/Standard 2.0.
/// <remarks/>
[System.SerializableAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
[System.Xml.Serialization.XmlRootAttribute(Namespace = "", IsNullable = false)]
public partial class Response
{
private ResponseGeography geographyField;
/// <remarks/>
public ResponseGeography Geography
{
get
{
return this.geographyField;
}
set
{
this.geographyField = value;
}
}
}
/// <remarks/>
[System.SerializableAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
public partial class ResponseGeography
{
private ResponseGeographyVendor vendorField;
/// <remarks/>
public ResponseGeographyVendor Vendor
{
get
{
return this.vendorField;
}
set
{
this.vendorField = value;
}
}
}
/// <remarks/>
[System.SerializableAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
public partial class ResponseGeographyVendor
{
private ResponseGeographyVendorRegion[] regionField;
private string idField;
/// <remarks/>
[System.Xml.Serialization.XmlElementAttribute("Region")]
public ResponseGeographyVendorRegion[] Region
{
get
{
return this.regionField;
}
set
{
this.regionField = value;
}
}
/// <remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string id
{
get
{
return this.idField;
}
set
{
this.idField = value;
}
}
}
/// <remarks/>
[System.SerializableAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
public partial class ResponseGeographyVendorRegion
{
private ResponseGeographyVendorRegionVehicle[] vehicleField;
private byte idField;
/// <remarks/>
[System.Xml.Serialization.XmlElementAttribute("Vehicle")]
public ResponseGeographyVendorRegionVehicle[] Vehicle
{
get
{
return this.vehicleField;
}
set
{
this.vehicleField = value;
}
}
/// <remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public byte id
{
get
{
return this.idField;
}
set
{
this.idField = value;
}
}
}
/// <remarks/>
[System.SerializableAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
public partial class ResponseGeographyVendorRegionVehicle
{
private string makeField;
private string fuelField;
private ushort priceField;
private string onSaleField;
private string accountField;
/// <remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string Make
{
get
{
return this.makeField;
}
set
{
this.makeField = value;
}
}
/// <remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string Fuel
{
get
{
return this.fuelField;
}
set
{
this.fuelField = value;
}
}
/// <remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public ushort Price
{
get
{
return this.priceField;
}
set
{
this.priceField = value;
}
}
/// <remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string OnSale
{
get
{
return this.onSaleField;
}
set
{
this.onSaleField = value;
}
}
/// <remarks/>
[System.Xml.Serialization.XmlAttributeAttribute()]
public string Account
{
get
{
return this.accountField;
}
set
{
this.accountField = value;
}
}
}
请让我知道如何更好地自我解释。道歉,如果这一切都没有道理-欢迎来到我的周末大声笑。
谢谢。
答案 0 :(得分:2)
由于您的xml根目录显然是Response
而不是ResponseGeographyVendor
,因此您必须反序列化为该类型:
string path = @"c:\XMLFile1.xml";
XmlSerializer ser = new XmlSerializer(typeof(response);
ResponseGeographyVendor i;
using (Stream reader = new FileStream(path,FileMode.Open))
{
i = ((Response)ser.Deserialize(reader)).Geography.Vendor;
Console.WriteLine(i.Price);
Console.ReadLine();
}
序列化程序仅适用于整个xml文档。您不能只是编写或阅读其中的一部分。因此,只需使用xml,将其序列化为Response
的实例并获取其Geography
成员即可。
现在,您可以轻松地在第二个Vehicle
中获得第三个Region
:
var vehicle = i.Region[1].Vehicle[2];
请注意,您不需要自己提供xml-root。
答案 1 :(得分:1)
您将要使用.Net内置的XML序列化器。
首先创建一个代表XML文档数据的类:
public class Response
{
public Geography Geography {get; set;}
}
public class Geography
{
public Vendor Vendor{get;set;}
}
public class Vendor
{
public List<Region> Regions {get;set;}
}
public class Region
{
}
,依此类推。 然后将xml作为字符串读取并反序列化:
string myXml = File.ReadAsStringAsync(filepath).Result;
XmlSerializer ser = new XmlSerializer(typeof(Response));
using (TextReader reader = new StringReader(myXml)
{
Response myResponse = ser.Deserialize(reader);
}
然后,您可以遍历地理对象上的所有属性和内容。