Question

public class SourceExamModel
{
    public int ExamId { get; set; }      

    public List<SectionModel> Sections { get; set; }
}

public class  DesiationExamModel
{
public in ExamId {get;set;}
public System.Collections.Generic.IEnumerable<SectionModel> DestSections
{
    get
    {
    }
    set
    {
    }
}

我尝试了什么：

var config = new MapperConfiguration(cfg => {
    cfg.CreateMap<CrmMapper.SourceExamModel, CrmMapper.DestiationExamModel>()
    .ForMember(v => v.Id, opts => opts.MapFrom(src => src.Id))
    .ForMember(v => v.DestSections, opts => opts.MapFrom(src => src.SourceSections));      
});

IMapper mapper = config.CreateMapper();
var source = new ExamModel();
var dest = mapper.Map<SourceExamModel, CrmMapper.DestiationExamModel>(source);

可以帮助我如何映射复杂对象的复制对象列表

Answer 1

假设您的示例代码中有一半是简单的拼写错误，那么您几乎可以使用它。

如果源和目标上的属性名称相同，则不必明确映射它们。
您的示例中的source没有意义，它需要正确的对象和数据。

这是我尝试的一个工作示例，您可以将其复制到控制台应用程序中。

class Program
{
    static void Main(string[] args)
    {
        var config = new MapperConfiguration(cfg =>
            cfg.CreateMap<SourceExamModel, DestinationExamModel>()
                .ForMember(dest => dest.DestSections, c => c.MapFrom(src => src.Sections))
        );

        config.AssertConfigurationIsValid();
        var mapper = config.CreateMapper();

        var source = new SourceExamModel
        {
            ExamId = 1,
            Sections = new List<SectionModel> { new SectionModel { SectionId = 1 }, new SectionModel { SectionId = 2 } }
        };

        var destination = mapper.Map<SourceExamModel, DestinationExamModel>(source);
    }
}

public class SourceExamModel
{
    public int ExamId { get; set; }

    public List<SectionModel> Sections { get; set; }
}

public class DestinationExamModel
{
    public int ExamId { get; set; }

    public List<SectionModel> DestSections { get; set; }
}

public class SectionModel
{
    public int SectionId { get; set; }
}

automap使用automapper将复杂类型列表复杂类型列表

1 个答案: