Question

我正在尝试传入一个类型对象以传递给序列化程序。

internal void SerializeXML(Object ObjType, String XMLRoot, Object Output, String Filename)
    {
        XmlSerializer serializer = new XmlSerializer(typeof(ObjType), new XmlRootAttribute(XMLRoot));

        StreamReader reader = new StreamReader(System.Reflection.Assembly.GetExecutingAssembly().GetManifestResourceStream(Filename));
        Output = (Type)serializer.Deserialize(reader);
        reader.Close();
    }

我想通过它来调用它（Main.LanguageList.Language是一个类）：

SerializeXML(Main.LanguageList.Language, "Language", LanguageListFile, InternalLangListXML);

我得到的Object是一个变量，但是用作一个类型。

Answer 1

编写泛型方法会更优雅：

internal void DeserializeXML<T>(String XMLRoot, T Output, String Filename)
{
    XmlSerializer serializer = new XmlSerializer(typeof(T), new XmlRootAttribute(XMLRoot));

    StreamReader reader = new StreamReader(System.Reflection.Assembly.GetExecutingAssembly().GetManifestResourceStream(Filename));
    Output = (T)serializer.Deserialize(reader);
    reader.Close();
}

并称之为：

DeserializeXML<Main.LanguageList.Language>("Language", LanguageListFile, InternalLangListXML);

此外，它建议更改方法以返回结果，而不是依赖于输出参数：

internal T DeserializeXML<T>(String XMLRoot, String Filename)
{
    XmlSerializer serializer = new XmlSerializer(typeof(T), new XmlRootAttribute(XMLRoot));

    using (StreamReader reader = new StreamReader(System.Reflection.Assembly.GetExecutingAssembly().GetManifestResourceStream(Filename)))
    {
        return (T)serializer.Deserialize(reader);
    }
}

然后可以这样调用：

var result = DeserializeXML<Main.LanguageList.Language>("Language", InternalLangListXML);

将类传递为类型

1 个答案: