c# - 将任何类型的列表序列化为 XML？

Question

我有 4 个实体类。A、B、C、D。它具有不同的属性，例如 ID、地址甚至复杂类型。我想编写一个通用方法，我可以在其中传递任何列表并将其转换为 XML。让我们说

public string GetAnyListtoXML(Any type of list)
        {
            string myXML=string.Empty;

return myXML;
        }

Answer 1

这个方法允许你序列化你想要的任何东西。你的实体类应该有无参数的构造函数。

此链接可能有助于控制序列化：http: //msdn.microsoft.com/en-us/library/2baksw0z%28v=vs.100%29.aspx

    public string ObjectToXml<T>(T obj)
    {
        var stream = new StringWriter();

        string xmlDoc = string.Empty;
        try
        {
            var xmlSerializer = new XmlSerializer(typeof (T));
            xmlSerializer.Serialize(stream, obj);
            xmlDoc = stream.GetStringBuilder().ToString();
        }
        catch (Exception ex)
        {
            Console.WriteLine("Błąd pliku xml: " + ex);
        }
        finally
        {
            stream.Close();
        }
        return xmlDoc;
    }



    public static T XmlToObject<T>(string xmlDoc)
    {
        var stream = new MemoryStream();
        byte[] xmlObject = Encoding.Unicode.GetBytes(xmlDoc);

        stream.Write(xmlObject, 0, xmlObject.Length);
        stream.Position = 0;
        T message;

        var ss = new XmlSerializer(typeof (T));
        try
        {
            message = (T) ss.Deserialize(stream);
        }
        catch (Exception)
        {
            message = default(T);
        }
        finally
        {
            stream.Close();
        }
        return message;
    }

如果你想这个方法只取列表，你可以使用它：

    public string ObjectToXml<T>(List<T> obj)
    {
        var stream = new StringWriter();

        string xmlDoc = string.Empty;
        try
        {
            var xmlSerializer = new XmlSerializer(typeof (List<T>));
            xmlSerializer.Serialize(stream, obj);
            xmlDoc = stream.GetStringBuilder().ToString();
        }
        catch (Exception ex)
        {
            Console.WriteLine("Błąd pliku xml: " + ex);
        }
        finally
        {
            stream.Close();
        }
        return xmlDoc;
    }