XML转换为对象格式

XML转换为对象格式

我想将此XML转换为对象格式

- <information>
- <item>
  <key>Name</key>
  <value>NameValue</value>
  </item>
- <item>
  <key>Age</key>
  <value>17</value>
  </item>
- <item>
  <key>Gender</key>
  <value>MALE</value>
  </item>
- </information>


反对类似的东西,

Person.Name = "Name Value"
Person.Age = 17
Person.Gender = "Male"

最佳答案

您可以通过反射XDocument来实现以下方式:

XDocument XDocument = XDocument.Parse(MyXml);

var nodes = XDocument.Descendants("item");

// Get the type contained in the name string
Type type = typeof(Person);

// create an instance of that type
object instance = Activator.CreateInstance(type);

// iterate on all properties and set each value one by one

foreach (var property in type.GetProperties())
{

    // Set the value of the given property on the given instance
    if (nodes.Descendants("key").Any(x => x.Value == property.Name)) // check if Property is in the xml
    {
        // exists so pick the node
        var node = nodes.First(x => x.Descendants("key").First().Value == property.Name);
        // set property value by converting to that type
        property.SetValue(instance,  Convert.ChangeType(node.Element("value").Value,property.PropertyType), null);
    }
}


var tempPerson = (Person) instance;


我做了一个Example Fiddle

也可以通过使用泛型对其进行重构来使其通用。

关于c# - 如何在C#中将XML转换为对象格式,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/30230045/

10-09 06:42