Question

我面临问题,我想到答案。

我有一个从普通类继承的阶层,而Im试图从这一类中取回T类,但我要说,我可以 t!

例如:

class Products : List<Product>
{}

问题是,从现在起,我不知道T类。因此,我试图照此办理:

Type itemsType = destObject.GetType().GetGenericArguments()[0]

It didn t work out.

我的方法是:

public static object Deserialize(Type destType, XmlNode xmlNode)
    {         
        object destObject = Activator.CreateInstance(destType);

        foreach (PropertyInfo property in destType.GetProperties())
            foreach (object att in property.GetCustomAttributes(false))
                if (att is XmlAttributeAttribute)
                    property.SetValue(destObject, xmlNode.Attributes[property.Name].Value, null);
                else if (att is XmlNodeAttribute)
                {
                    object retObject = Deserialize(property.PropertyType, xmlNode.Nodes[property.Name]);
                    property.SetValue(destObject, retObject, null);
                }

        if (destObject is IList)
        {
            Type itemsType = destObject.GetType().GetGenericArguments()[0];
            foreach (XmlNode xmlChildNode in xmlNode.Nodes)
            {
                object retObject = Deserialize(itemsType, xmlNode);
                ((IList)destObject).Add(retObject);
            }
        }

        return destObject;
    }

想法是阅读Xml文档,将其改为:

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<SETTINGS>
  <PRODUCTS>
    <PRODUCT NAME="ANY" VERSION="ANY" ISCURRENT="TRUE" />
    <PRODUCT NAME="TEST1" VERSION="ANY" ISCURRENT="FALSE" />
    <PRODUCT NAME="TEST2" VERSION="ANY" ISCURRENT="FALSE" />
  </PRODUCTS>
  <DISTRIBUTIONS>
    <DISTRIBUTION NAME="5.32.22" />
  </DISTRIBUTIONS>
</SETTINGS>

在此情况下,我从名单上收集的物品就属于我。

关于如何这样做的想法?

tks guys

Answer 1

你们需要获得基础类型的一般论点,例如:

Type itemType = destObject.GetType().BaseType.GetGenericArguments()[0];

However, this is not resilient; if an intermediary non-generic base type is introduced, it will fail.
Instead, you should find the type parameter of the IList<T> implementation.

例如:

Type listImplementation = destObject.GetType().GetInterface(typeof(IList<>).Name);
if (listImplementation != null) {
    Type itemType = listImplementation.GetGenericArguments()[0];
    ...
}

Answer 2

如果你只是试图说明<代码”的类型。 IList 是:

<编码> 类型项目类型= destType.GetInterface(类型(IList<>);Name.GetGeneric Arguments(0);

在此,你将如何在法典中使用:

var interface = destType.GetInterface(typeof(IList<>).Name);
var destList = destObject as IList;
// make sure that the destination is both IList and IList<T>
if (interface != null && destList != null)
{
    Type itemsType = interface.GetGenericArguments()[0];
    foreach (XmlNode xmlChildNode in xmlNode.Nodes) 
    { 
        object retObject = Deserialize(itemsType, xmlNode); 
        destList.Add(retObject); 
    } 
}

友情链接