如何对具有多个XmlElementAttributes的对象使用automapper？

Question

我在从.xsd为.cs文件中生成的对象设置automapper时遇到问题。

当对象具有多个属性时，不确定如何解决问题，如下所示：

一直在看TypeConverters等，但不确定如何正确设置它。一直在使用automapper并且没有任何问题，只要没有多个属性连接到一个成员。

public partial class customerInfo {

    private object itemField;

    /// <remarks/>
    [System.Xml.Serialization.XmlElementAttribute("customerInfoBasic", typeof(customerInfoBasic))]
    [System.Xml.Serialization.XmlElementAttribute("customerInfoSimple", typeof(customerInfoSimple))]
    [System.Xml.Serialization.XmlElementAttribute("customerInfoEnhanced", typeof(customerInfoEnhanced))]
    public object Item {
        get {
            return this.itemField;
        }
        set {
            this.itemField = value;
        }
    }
}

public partial class customerInfoBasic{

    private string nameField;

    /// <remarks/>
    public string name {
        get {
            return this.nameField;
        }
        set {
            this.nameField= value;
        }
    }
}

public partial class customerInfoSimple{

    private string nameField;
    private string idField;


    /// <remarks/>
    public string name {
        get {
            return this.nameField;
        }
        set {
            this.nameField= value;
        }
    }

    public string id {
        get {
            return this.idField;
        }
        set {
            this.idField= value;
        }
    }
}

public partial class customerInfoEnhanced{

    private string nameField;
    private string idField;
    private string ageField;

    /// <remarks/>
    public string name {
        get {
            return this.nameField;
        }
        set {
            this.nameField= value;
        }
    }

    public string id {
        get {
            return this.idField;
        }
        set {
            this.idField= value;
        }
    }

    public string age {
        get {
            return this.ageField;
        }
        set {
            this.ageField= value;
        }
    }
}

我遇到的问题是我不知道如何设置它，以便根据“信息”中的某些值正确映射customerInfo。

例如，如果“信息”包含“年龄”和“如果”，则应将其映射到客户信息增强等。

public static void AddSessionTransformationMappings(IMapperConfiguration cfg)
{
    cfg.AllowNullCollections = true;

    cfg.CreateMap<IEnumerable<Info>, customerInfoList>()
        .ForMember(x => x.customerInfo, x => x.MapFrom(y => y));

    cfg.CreateMap<Info, customerInfo>()
        .ForMember(x => x.Item, x => x.MapFrom(y => y));

    cfg.CreateMap<Info, customerInfoBasic>()
        .ForMember(x => x.Name, x => x.MapFrom(y => y.name));

    cfg.CreateMap<Info, customerInfoSimple>()
        .ForMember(x => x.Name, x => x.MapFrom(y => y.name))
        .ForMember(x => x.Id, x => x.MapFrom(y => y.id));

    cfg.CreateMap<Info, customerInfoEnhanced>()
        .ForMember(x => x.Name, x => x.MapFrom(y => y))
        .ForMember(x => x.Id, x => x.MapFrom(y => y.id))
        .ForMember(x => x.Age, x => x.MapFrom(y => y.age));
}

这是序列化器的代码：

var output = provider.Transform(new List<Info> { input });

customerInfoList actual = null;
XmlSerializer serializer = new XmlSerializer(typeof(customerInfoList));
using (MemoryStream ms = new MemoryStream())
{
    serializer.Serialize(ms, output);
    ms.Position = 0;
    actual = (customerInfoList)serializer.Deserialize(ms);
}

如果我设置.ForMember(x => x.customerInfo, x => x.MapFrom(y => (Object)null));代码工作，“实际”给我一个item = null按预期的列表，所以我知道问题是在customerInfo中映射“项目”。

我希望映射器映射到正确的类，现在我得到Missing类型映射或“Info is not expected。使用XmlInclude或SoapInclude属性指定静态未知的类型。

真的很感激如何解决这个问题的一些指示！

Answer 1

解决它是为了我自己的需要，如果其他人遇到同样的问题解决方案。

对我来说，解决方案是使用ResolveUsing而不是MapFrom来获取具有多个标记名称的特定属性，并使用Mapper.Map以及针对不同情况的正确类。

完整代码如下所示：

public static void AddSessionTransformationMappings(IMapperConfiguration cfg)
{
    cfg.AllowNullCollections = true;

    cfg.CreateMap<IEnumerable<Info>, customerInfoList>()
        .ForMember(x => x.customerInfo, x => x.MapFrom(y => y));

    cfg.CreateMap<Info, customerInfo>()
        .ForMember(x => x.Item, x => x.ResolveUsing(y => CustomerInfoLevel(y)));

    cfg.CreateMap<Info, customerInfoBasic>()
        .ForMember(x => x.Name, x => x.MapFrom(y => y.name));

    cfg.CreateMap<Info, customerInfoSimple>()
        .ForMember(x => x.Name, x => x.MapFrom(y => y.name))
        .ForMember(x => x.Id, x => x.MapFrom(y => y.id));

    cfg.CreateMap<Info, customerInfoEnhanced>()
        .ForMember(x => x.Name, x => x.MapFrom(y => y.name))
        .ForMember(x => x.Id, x => x.MapFrom(y => y.id))
        .ForMember(x => x.Age, x => x.MapFrom(y => y.age));
}

private static Object CustomerInfoLevel(Info info)
{
    if (info.age != null)
    {
        return Mapper.Map<customerInfoEnhanced>(info);
    }
    else if (info.id != null)
    {
        return Mapper.Map<customerInfoSimple>(info);
    }
    else
    {
        return Mapper.Map<customerInfoBasic>(info);
    }
}