所以我需要复制对象。我有一个模型在这里“地方”有一个IList HasAndBelongsToMany属性是一个痛苦。我需要获取field属性并复制它,但它只复制引用。这就是我所拥有的
public class place : ActiveRecordBase<place>
{
public place() { }
private int place_id;
[PrimaryKey("place_id")]
virtual public int id
{
get { return place_id; }
set { place_id = value; }
}
private IList<fields> Fields;
[HasAndBelongsToMany(typeof(fields), Lazy = true, Table = "place_to_fields", ColumnKey = "place_id", ColumnRef = "field_id", NotFoundBehaviour = NotFoundBehaviour.Ignore, Cascade = ManyRelationCascadeEnum.AllDeleteOrphan)]
virtual public IList<fields> field
{
get { return Fields; }
set { Fields = value; }
}
}
并使用像这样的automapper
place org = ActiveRecordBase<place>.Find(id);
Mapper.Reset();
Mapper.CreateMap<place, place>().ForMember(dest => dest.id, o => o.Ignore())
.ForMember(dest => dest.field, o => o.Ignore())
;
place copy = new place();
Mapper.Map(org, copy);
copy.SaveAndFlush();
哪个有效,因为我正在滑动这个领域。我所希望的更像是:
Mapper.CreateMap<place, place>().ForMember(dest => dest.id, o => o.Ignore())
.ForMember(dest => dest.field.id, o => o.Ignore())
;
请参阅.ForMember(dest =&gt; dest.id,o =&gt; o.Ignore())的第一行,以便我不复制place对象的引用ID。我需要为place属性字段做同样的事情。我需要忽略id并在其余属性上创建具有相同值的新条目
答案 0 :(得分:0)
您需要为字段类型创建映射并添加&#34;忽略&#34;字段ID的选项,就像您已经为地点类型做了一样。
Mapper.CreateMap<fields, fields>().ForMember(dest => dest.id, o => o.Ignore());
Mapper.CreateMap<place, place>().ForMember(dest => dest.id, o => o.Ignore());