[英]Deserialize XML to object in C# XmlRoot is not working
我正在使用asp.net(C#)网站,我想使用XmlSerializer
类Deserialize
XML
字符串。
我的模型(.cs文件)
[XmlRoot("MedicalClearanceFormRoot")]
public class MedicalClearanceViewModel
{
[XmlAttribute("PassengerName")]
public string PassengerName { get; set; }
[XmlAttribute("Gender")]
public string Gender { get; set; }
[XmlAttribute("Age")]
public string Age { get; set; }
[XmlAttribute("PhoneNo")]
public string PhoneNo { get; set; }
[XmlAttribute("Email")]
public string Email { get; set; }
[XmlAttribute("BookingRefNo")]
public string BookingRefNo { get; set; }
}
XML字串
<MedicalClearanceFormRoot>
<MedicalClearanceForm PassengerName="AAAAAAAAAAAAA" Age="11" PhoneNo="TTTTTTTTTTT" Email="ZZZZZZZZZZZZZZZZZZZ" BookingRefNo="11111111111111111111" />
</MedicalClearanceFormRoot>
将XML反序列化为对象的代码
string myXMLStringFromDB = GetXMLStringFromDb(); // this method will return XML from db.
XmlSerializer serializer = new XmlSerializer(typeof(MedicalClearanceViewModel));
using (TextReader reader = new StringReader(myXMLStringFromDB))
{
MedicalClearanceViewModel objModel = (MedicalClearanceViewModel)serializer.Deserialize(reader);
}
但是,问题是,当我使用上面的代码反序列化的XML对象......像属性PassengerName
, Age
, PhoneNo
等方面都还处于空白objModel
有人可以帮助我在我的课程上设置适当的XML表示法吗?有人可以帮助我解决此问题。
任何帮助将不胜感激! 谢谢
定义XML的方式,您需要定义两个对象:-一个用于MedicalClearanceFormRoot
xml节点-一个用于MedicalClearanceForm
xml节点
因此,您可以采取两种方法:添加包装器类或更改xml。
要添加包装器类,您将需要一个类来表示MedicalClearanceFormRoot
,该类具有MedicalClearanceForm
对象的属性。 然后将您的序列化器类更改为包装器类。 这是一个例子:
[XmlRoot("MedicalClearanceFormRoot")]
public class Wrapper
{
public MedicalClearanceViewModel MedicalClearanceForm { get; set;}
}
public class MedicalClearanceViewModel
{
[XmlAttribute("PassengerName")]
public string PassengerName { get; set; }
[XmlAttribute("Gender")]
public string Gender { get; set; }
[XmlAttribute("Age")]
public string Age { get; set; }
[XmlAttribute("PhoneNo")]
public string PhoneNo { get; set; }
[XmlAttribute("Email")]
public string Email { get; set; }
[XmlAttribute("BookingRefNo")]
public string BookingRefNo { get; set; }
}
XmlSerializer serializer = new XmlSerializer(typeof(Wrapper));
using (TextReader reader = new StringReader(myXMLStringFromDB))
{
Wrapper objModel = (Wrapper)serializer.Deserialize(reader);
}
选项2:将XML更改为如下所示:
<MedicalClearanceFormRoot PassengerName="AAAAAAAAAAAAA" Age="11" PhoneNo="TTTTTTTTTTT" Email="ZZZZZZZZZZZZZZZZZZZ" BookingRefNo="11111111111111111111" >
</MedicalClearanceFormRoot>
我已经创建了一个示例,其代码将与下面的完全一样。 您的模型不正确。
public class MedicalClearanceForm
{
[XmlAttribute("PassengerName")]
public string PassengerName { get; set; }
[XmlAttribute("Gender")]
public string Gender { get; set; }
[XmlAttribute("Age")]
public string Age { get; set; }
[XmlAttribute("PhoneNo")]
public string PhoneNo { get; set; }
[XmlAttribute("Email")]
public string Email { get; set; }
[XmlAttribute("BookingRefNo")]
public string BookingRefNo { get; set; }
}
[XmlRoot("MedicalClearanceFormRoot")]
public class MedicalClearanceFormRoot
{
[XmlElement("MedicalClearanceForm")]
public MedicalClearanceForm MedicalClearanceForm { get; set; }
}
class Program
{
static void Main(string[] args)
{
string myXMLStringFromDB = @"<MedicalClearanceFormRoot><MedicalClearanceForm PassengerName = 'AAAAAAAAAAAAA' Age = '11' PhoneNo = 'TTTTTTTTTTT' Email = 'ZZZZZZZZZZZZZZZZZZZ' BookingRefNo = '11111111111111111111' /></MedicalClearanceFormRoot >";
XmlSerializer serializer = new XmlSerializer(typeof(MedicalClearanceFormRoot));
using (TextReader reader = new StringReader(myXMLStringFromDB))
{
MedicalClearanceFormRoot objModel = (MedicalClearanceFormRoot)serializer.Deserialize(reader);
}
}
}
我相信,要满足指定的XML,您将需要此类结构。
[XmlRoot("MedicalClearanceFormRoot")]
public class MedicalClearanceViewModel
{
public MedicalClearanceFormElement MedicalClearanceForm { get; set; }
}
[XmlElement]
public class MedicalClearanceFormElement
{
[XmlAttribute("PassengerName")]
public string PassengerName { get; set; }
[XmlAttribute("Gender")]
public string Gender { get; set; }
[XmlAttribute("Age")]
public string Age { get; set; }
[XmlAttribute("PhoneNo")]
public string PhoneNo { get; set; }
[XmlAttribute("Email")]
public string Email { get; set; }
[XmlAttribute("BookingRefNo")]
public string BookingRefNo { get; set; }
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.