在不同的元素名称下序列化 class
Serializing a class under a different element name
我有这个class:
[XmlRoot(ElementName ="Lesson")]
public class LessonOld
{
public LessonOld()
{
Students = new List<string>();
}
public string Name { get; set; }
public DateTime FirstLessonDate { get; set; }
public int DurationInMinutes { get; set; }
public List<string> Students { get; set; }
}
我正在使用这段代码对其进行序列化:
TextWriter writer = new StreamWriter(Path.Combine(UserSettings, "Lessons-temp.xml"));
XmlSerializer xmlSerializer = new XmlSerializer(typeof(List<LessonOld>));
xmlSerializer.Serialize(writer, tempList);
writer.Close();
(注意是List<LessonOld>
)
这是我的结果 XML:
<?xml version="1.0" encoding="utf-8"?>
<ArrayOfLessonOld xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<LessonOld>
<FirstLessonDate>0001-01-01T00:00:00</FirstLessonDate>
<DurationInMinutes>0</DurationInMinutes>
<Students />
</LessonOld>
</ArrayOfLessonOld>
我想将其更改为 XML 元素的序列化为 <ArrayOfLesson
和 <Lesson>
。这可能吗? (如您所见,我已经尝试使用 [XmlRoot(ElementName ="Lesson")]
)
你快到了。使用:
[XmlType(TypeName = "Lesson")]
而不是
[XmlRoot(ElementName = "Lesson")]
当然可以轻松测试;您的代码进行了上述更改
[XmlType(TypeName = "Lesson")]
public class LessonOld
{
public LessonOld()
{
Students = new List<string>();
}
public string Name { get; set; }
public DateTime FirstLessonDate { get; set; }
public int DurationInMinutes { get; set; }
public List<string> Students { get; set; }
}
class Program
{
static void Main(string[] args)
{
TextWriter writer = new StreamWriter(Path.Combine(@"C:\Users\Francesco\Desktop\nanovg", "Lessons-temp.xml"));
XmlSerializer xmlSerializer = new XmlSerializer(typeof(List<LessonOld>));
xmlSerializer.Serialize(writer, new List<LessonOld> { new LessonOld() { Name = "name", DurationInMinutes = 0 } });
writer.Close();
}
}
产生这个
<?xml version="1.0" encoding="utf-8"?>
<ArrayOfLesson xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Lesson>
<Name>name</Name>
<FirstLessonDate>0001-01-01T00:00:00</FirstLessonDate>
<DurationInMinutes>0</DurationInMinutes>
<Students />
</Lesson>
</ArrayOfLesson>
正如我所见,XmlRoot
在您想要序列化单个对象时工作正常。
考虑从您的代码派生的这段代码:
[XmlRoot(ElementName = "Lesson")]
public class LessonOld
{
public LessonOld()
{
Students = new List<string>();
}
public string Name { get; set; }
public DateTime FirstLessonDate { get; set; }
public int DurationInMinutes { get; set; }
public List<string> Students { get; set; }
}
class Program
{
static void Main(string[] args)
{
TextWriter writer = new StreamWriter(Path.Combine(@"C:\Users\Francesco\Desktop\nanovg", "Lessons-temp.xml"));
XmlSerializer xmlSerializer = new XmlSerializer(typeof(LessonOld));
xmlSerializer.Serialize(writer, new LessonOld() { Name = "name", DurationInMinutes = 0 });
writer.Close();
}
}
会输出
<?xml version="1.0" encoding="utf-8"?>
<Lesson xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Name>name</Name>
<FirstLessonDate>0001-01-01T00:00:00</FirstLessonDate>
<DurationInMinutes>0</DurationInMinutes>
<Students />
</Lesson>
我有这个class:
[XmlRoot(ElementName ="Lesson")]
public class LessonOld
{
public LessonOld()
{
Students = new List<string>();
}
public string Name { get; set; }
public DateTime FirstLessonDate { get; set; }
public int DurationInMinutes { get; set; }
public List<string> Students { get; set; }
}
我正在使用这段代码对其进行序列化:
TextWriter writer = new StreamWriter(Path.Combine(UserSettings, "Lessons-temp.xml"));
XmlSerializer xmlSerializer = new XmlSerializer(typeof(List<LessonOld>));
xmlSerializer.Serialize(writer, tempList);
writer.Close();
(注意是List<LessonOld>
)
这是我的结果 XML:
<?xml version="1.0" encoding="utf-8"?>
<ArrayOfLessonOld xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<LessonOld>
<FirstLessonDate>0001-01-01T00:00:00</FirstLessonDate>
<DurationInMinutes>0</DurationInMinutes>
<Students />
</LessonOld>
</ArrayOfLessonOld>
我想将其更改为 XML 元素的序列化为 <ArrayOfLesson
和 <Lesson>
。这可能吗? (如您所见,我已经尝试使用 [XmlRoot(ElementName ="Lesson")]
)
你快到了。使用:
[XmlType(TypeName = "Lesson")]
而不是
[XmlRoot(ElementName = "Lesson")]
当然可以轻松测试;您的代码进行了上述更改
[XmlType(TypeName = "Lesson")]
public class LessonOld
{
public LessonOld()
{
Students = new List<string>();
}
public string Name { get; set; }
public DateTime FirstLessonDate { get; set; }
public int DurationInMinutes { get; set; }
public List<string> Students { get; set; }
}
class Program
{
static void Main(string[] args)
{
TextWriter writer = new StreamWriter(Path.Combine(@"C:\Users\Francesco\Desktop\nanovg", "Lessons-temp.xml"));
XmlSerializer xmlSerializer = new XmlSerializer(typeof(List<LessonOld>));
xmlSerializer.Serialize(writer, new List<LessonOld> { new LessonOld() { Name = "name", DurationInMinutes = 0 } });
writer.Close();
}
}
产生这个
<?xml version="1.0" encoding="utf-8"?>
<ArrayOfLesson xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Lesson>
<Name>name</Name>
<FirstLessonDate>0001-01-01T00:00:00</FirstLessonDate>
<DurationInMinutes>0</DurationInMinutes>
<Students />
</Lesson>
</ArrayOfLesson>
正如我所见,XmlRoot
在您想要序列化单个对象时工作正常。
考虑从您的代码派生的这段代码:
[XmlRoot(ElementName = "Lesson")]
public class LessonOld
{
public LessonOld()
{
Students = new List<string>();
}
public string Name { get; set; }
public DateTime FirstLessonDate { get; set; }
public int DurationInMinutes { get; set; }
public List<string> Students { get; set; }
}
class Program
{
static void Main(string[] args)
{
TextWriter writer = new StreamWriter(Path.Combine(@"C:\Users\Francesco\Desktop\nanovg", "Lessons-temp.xml"));
XmlSerializer xmlSerializer = new XmlSerializer(typeof(LessonOld));
xmlSerializer.Serialize(writer, new LessonOld() { Name = "name", DurationInMinutes = 0 });
writer.Close();
}
}
会输出
<?xml version="1.0" encoding="utf-8"?>
<Lesson xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Name>name</Name>
<FirstLessonDate>0001-01-01T00:00:00</FirstLessonDate>
<DurationInMinutes>0</DurationInMinutes>
<Students />
</Lesson>