如何使用多个属性和选项列表解析 XML
How to parse XML with several Attributes and an options list
我有一个 XML 文件,看起来与此类似:
<root>
<data label="product data" min="0" max="10">
<option>
<id>1</id>
<name>Name1</name>
</option>
<option>
<id>2</id>
<name>Name2</name>
</option>
<option>
<id>3</id>
<name>Name3</name>
</option>
</data>
</root>
我需要检索 data
属性和选项列表。
我试过这个:
[XmlRoot(ElementName = "root")]
public class Data
{
// Retreive data attributes
[XmlElement(ElementName = "data")]
public Options Attributes { get; set; }
// Retrieve option list
[XmlArray("data")]
[XmlArrayItem("option", Type = typeof(GeneralOptions))]
public GeneralOptions[] Options { get; set; }
}
可选类:
选项
public class Options
{
[XmlAttribute("label")]
public string Label{ get; set; }
[XmlAttribute("min")]
public string Min{ get; set; }
[XmlAttribute("max")]
public string Max{ get; set; }
}
常规选项
public class GeneralOptions
{
[XmlElement(ElementName = "id")]
public string Id { get; set; }
[XmlElement(ElementName = "name")]
public string Name{ get; set; }
}
但是当我尝试反序列化该对象时,它会引发以下异常:
The XML element 'data' from namespace '' is already present in the current scope. Use XML attributes to specify another XML name or namespace for the element.
我想问题是我正在尝试检索相同的元素 "twice"。但我需要收回这两件事。我不能使用 [Attribute]
东西,因为有几个属性要检索,我需要用几个具有相同格式的 XML 元素来做到这一点,我想重用它。
那么,我怎样才能取回它们?
您需要稍微重组它:
[XmlRoot("root")]
public class Data
{
[XmlElement("data")]
public OptionsData Options { get; set; }
}
public class OptionsData
{
[XmlAttribute("label")]
public string Label { get; set; }
[XmlAttribute("min")]
public string Min { get; set; }
[XmlAttribute("max")]
public string Max { get; set; }
[XmlElement("option")]
public List<GeneralOptions> Items { get; } = new List<GeneralOptions>();
}
public class GeneralOptions
{
[XmlElement("id")]
public string Id { get; set; }
[XmlElement("name")]
public string Name { get; set; }
}
我建议使用 xmltocsharp 或任何其他工具在几秒钟内将 XML 转换为 C# 模型...(消除所有手动错误)
As mentioned by @canton7, another easy method is using Visual Studio: Edit -> Paste Special -> Paste XML As Classes
[XmlRoot(ElementName="option")]
public class Option {
[XmlElement(ElementName="id")]
public string Id { get; set; }
[XmlElement(ElementName="name")]
public string Name { get; set; }
}
[XmlRoot(ElementName="data")]
public class Data {
[XmlElement(ElementName="option")]
public List<Option> Option { get; set; }
[XmlAttribute(AttributeName="label")]
public string Label { get; set; }
[XmlAttribute(AttributeName="min")]
public string Min { get; set; }
[XmlAttribute(AttributeName="max")]
public string Max { get; set; }
}
[XmlRoot(ElementName="root")]
public class Root {
[XmlElement(ElementName="data")]
public Data Data { get; set; }
}
我有一个 XML 文件,看起来与此类似:
<root>
<data label="product data" min="0" max="10">
<option>
<id>1</id>
<name>Name1</name>
</option>
<option>
<id>2</id>
<name>Name2</name>
</option>
<option>
<id>3</id>
<name>Name3</name>
</option>
</data>
</root>
我需要检索 data
属性和选项列表。
我试过这个:
[XmlRoot(ElementName = "root")]
public class Data
{
// Retreive data attributes
[XmlElement(ElementName = "data")]
public Options Attributes { get; set; }
// Retrieve option list
[XmlArray("data")]
[XmlArrayItem("option", Type = typeof(GeneralOptions))]
public GeneralOptions[] Options { get; set; }
}
可选类:
选项
public class Options
{
[XmlAttribute("label")]
public string Label{ get; set; }
[XmlAttribute("min")]
public string Min{ get; set; }
[XmlAttribute("max")]
public string Max{ get; set; }
}
常规选项
public class GeneralOptions
{
[XmlElement(ElementName = "id")]
public string Id { get; set; }
[XmlElement(ElementName = "name")]
public string Name{ get; set; }
}
但是当我尝试反序列化该对象时,它会引发以下异常:
The XML element 'data' from namespace '' is already present in the current scope. Use XML attributes to specify another XML name or namespace for the element.
我想问题是我正在尝试检索相同的元素 "twice"。但我需要收回这两件事。我不能使用 [Attribute]
东西,因为有几个属性要检索,我需要用几个具有相同格式的 XML 元素来做到这一点,我想重用它。
那么,我怎样才能取回它们?
您需要稍微重组它:
[XmlRoot("root")]
public class Data
{
[XmlElement("data")]
public OptionsData Options { get; set; }
}
public class OptionsData
{
[XmlAttribute("label")]
public string Label { get; set; }
[XmlAttribute("min")]
public string Min { get; set; }
[XmlAttribute("max")]
public string Max { get; set; }
[XmlElement("option")]
public List<GeneralOptions> Items { get; } = new List<GeneralOptions>();
}
public class GeneralOptions
{
[XmlElement("id")]
public string Id { get; set; }
[XmlElement("name")]
public string Name { get; set; }
}
我建议使用 xmltocsharp 或任何其他工具在几秒钟内将 XML 转换为 C# 模型...(消除所有手动错误)
As mentioned by @canton7, another easy method is using
Visual Studio: Edit -> Paste Special -> Paste XML As Classes
[XmlRoot(ElementName="option")]
public class Option {
[XmlElement(ElementName="id")]
public string Id { get; set; }
[XmlElement(ElementName="name")]
public string Name { get; set; }
}
[XmlRoot(ElementName="data")]
public class Data {
[XmlElement(ElementName="option")]
public List<Option> Option { get; set; }
[XmlAttribute(AttributeName="label")]
public string Label { get; set; }
[XmlAttribute(AttributeName="min")]
public string Min { get; set; }
[XmlAttribute(AttributeName="max")]
public string Max { get; set; }
}
[XmlRoot(ElementName="root")]
public class Root {
[XmlElement(ElementName="data")]
public Data Data { get; set; }
}