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试图将内容从一个项目复制到另一个项目

.net-core generics .net c#

Glory Raj · 技术社区 · 6 年前

我有这样的项目实体结构,包括项目名称和项目编号,以及一些列表对象和属性。

public class DesignProject
{
    [Key, GraphQLNonNullType]
    public string ProjectNumber { get; set; }
    public string Name { get; set; }
    [Column(TypeName = "jsonb")]
    public ProjectSectionStatus SectionStatuses { get; set; } = new ProjectSectionStatus();
    [ForeignKey("AshraeClimateZone"), GraphQLIgnore]
    public Guid? AshraeClimateZoneId { get; set; }
    public virtual AshraeClimateZone AshraeClimateZone { get; set; }
    [Column(TypeName = "jsonb")]
    public List<ProjectObject<classA>> classAList{ get; set; } = new List<ProjectObject<classA>>();
    [Column(TypeName = "jsonb")]
    public List<ProjectObject<classB>> classBList{ get; set; } = new List<ProjectObject<classB>>();
    ...... 
    ......
    ......
   Some more json columns
}

以及 project object 这样的班级

public class ProjectObject<T>
{
    public Guid? Id { get; set; }
    public T OriginalObject { get; set; }
    public T ModifiedObject { get; set; }

    [GraphQLIgnore, JsonIgnore]
    public T TargetObject
    {
        get
        {
            return ModifiedObject ?? OriginalObject;
        }
    }
}

和 ClassA 实体结构如下

public class ClassA
{
    public string Edition { get; set; }
    public string City { get; set; }  
}

我也有一些类似的儿童实体( A级 )与上述相同,我想将内容和状态从一个项目实体复制到另一个项目主体。

我有项目实体 ProjectNumber 1212,还有一个项目的项目编号为23323,所以我想从以下位置复制整个项目内容 1212 到 23323 。那么,有什么方法可以用我正在使用的C#来实现这一点吗。具有实体框架核心的Net Core。

这里,我要复制的源设计项目与目标设计项目具有相同的结构,我可以覆盖目标项目值,我不想在这里更新项目编号。

有人能告诉我如何实现这个复制吗?提前感谢!!

如果我需要为这个问题添加任何细节,请告诉我

更新 :深度复制相关代码

   public InsertResponse<string> CopyBookmarkproject(string SourceProjectNumber, string destinationProjectNumber)
   {
        var sourceDesignProject = this._dbContext.DesignProject.Where(a => a.ProjectNumber == SourceProjectNumber).SingleOrDefault();
        var destinationProject = this._dbContext.DesignProject.Where(a => a.ProjectNumber == destinationProjectNumber).SingleOrDefault();

        CopyProject(sourceDesignProject, destinationProject);

       // Need to update the Db context at here after deep copy
    }

    private void CopyProject(DesignProject sourceDesignProject, DesignProject destinationProject)
    {
        destinationProject.classAList= sourceDesignProject.classAList; // Not sure whether this copy will works 
        destinationProject.AshraeClimateZone = sourceDesignProject.AshraeClimateZone; // not sure whether this copy will works also 
    }

更新的解决方案2:

    var sourceDesignProject = this._dbContext.DesignProjects.AsNoTracking()
                                                    .Where(a => a.ProjectNumber == sourceProjectNumber)
                                                    .Include(a => a.PrimaryBuildingType)
                                                    .Include(a => a.AshraeClimateZone).SingleOrDefault();

     var targetDesignProject = this._dbContext.DesignProjects.Where(a => a.ProjectNumber == targetProjectNumber).SingleOrDefault();

     sourceDesignProject.ProjectNumber = targetDesignProject.ProjectNumber;
     sourceDesignProject.SectionStatuses.AirSystemsSectionStatus = Entities.Enums.ProjectSectionStage.INCOMPLETE;
            
     sourceDesignProject.SectionStatuses.CodesAndGuidelinesSectionStatus = Entities.Enums.ProjectSectionStage.INCOMPLETE;
     sourceDesignProject.SectionStatuses.ExecutiveSummarySectionStatus = Entities.Enums.ProjectSectionStage.INCOMPLETE;
     sourceDesignProject.SectionStatuses.ExhaustEquipmentSectionStatus = Entities.Enums.ProjectSectionStage.INCOMPLETE;
     sourceDesignProject.SectionStatuses.InfiltrationSectionStatus = Entities.Enums.ProjectSectionStage.INCOMPLETE;
      

     this._dbContext.Entry(sourceDesignProject).State = EntityState.Modified; // getting the below error at this line
     this._dbContext.SaveChanges(); 
     ok = true;

出现如下错误

The instance of entity type 'DesignProject' cannot be tracked because another instance with the same key value for {'ProjectNumber'} is already being tracked. When attaching existing entities, ensure that only one entity instance with a given key value is attached.

0 回复 | 直到 6 年前

Amirhossein Mehrvarzi GGregson 6 年前

有一个简单而简短的解决方案,它使用 .AsNoTracking() 对于源条目,分配目标 ProjectNumber 到源,然后更改 EntityState 作为 Modified 它将把所有属性复制到目标条目:

var sourceDesignProject = this._dbContext.DesignProject.AsNoTracking().Where(a => a.ProjectNumber == SourceProjectNumber).SingleOrDefault();
var destinationProject = this._dbContext.DesignProject.Where(a => a.ProjectNumber == destinationProjectNumber).SingleOrDefault();

sourceDesignProject.ProjectNumber = destinationProject.ProjectNumber;
this._dbContext.Entry(sourceDesignProject).State = System.Data.Entity.EntityState.Modified;
this._dbContext.SaveChanges();

确保您的关系属性之前已加载。

Kiran Thokal 6 年前

你可以使用反射。由于您同时拥有源对象和目标对象,因此它们属于同一类型。您可以遍历源对象,获取属性并替换值。类似于:

if (inputObjectA != null && inputObjectB != null)
{
    //create variables to store object values
    object value1, value2;
 
    PropertyInfo[] properties = inputObjectA.GetType().GetProperties(BindingFlags.Public | BindingFlags.Instance);
     
    //get all public properties of the object using reflection  
    foreach (PropertyInfo propertyInfo in properties)
    {
    //get the property values of both the objects
      value1 = propertyInfo.GetValue(inputObjectA, null);
      value2 = propertyInfo.GetValue(inputObjectB, null);
    }

忽略不需要复制的属性。该方法将具有如下签名:

CopyObjects(object inputObjectA, object inputObjectB, string[] ignorePropertiesList)

Arman Ebrahimpour jacek 6 年前

对于对象的深度复制,有两种选择:

1-使用 BinaryFormatter (需求 Serializable 你的类的属性, See original answer ):

public static T DeepClone<T>(this T obj)
{
    using (var ms = new MemoryStream())
    {
        var formatter = new BinaryFormatter();
        formatter.Serialize(ms, obj);
        ms.Position = 0;

        return (T) formatter.Deserialize(ms);
    }
}

2-使用 JsonSerializers 比如NewtoneSoft:

public static T CloneJson<T>(this T source)
{            
    // Don't serialize a null object, simply return the default for that object
    if (Object.ReferenceEquals(source, null))
    {
        return default(T);
    }

    // initialize inner objects individually
    // for example in default constructor some list property initialized with some values,
    // but in 'source' these items are cleaned -
    // without ObjectCreationHandling.Replace default constructor values will be added to result
    var deserializeSettings = new JsonSerializerSettings {ObjectCreationHandling = ObjectCreationHandling.Replace};

    return JsonConvert.DeserializeObject<T>(JsonConvert.SerializeObject(source), deserializeSettings);
}

更新

以下是您可以使用的代码 NewtonSoft Json 图书馆:

private void CopyProject(DesignProject sourceDesignProject, ref DesignProject destinationProject)
{
    if (Object.ReferenceEquals(sourceDesignProject, null))
    {
        destinationProject = null;
        return;
    }

    var deserializeSettings = new JsonSerializerSettings {ObjectCreationHandling = ObjectCreationHandling.Replace};

    var destProjNumber = destinationProject.ProjectNumber;

    destinationProject = JsonConvert.DeserializeObject<DesignProject>
        (JsonConvert.SerializeObject(sourceDesignProject), deserializeSettings);

    destinationProject.ProjectNumber = destProjNumber;
}

Noldor 6 年前

我总是更喜欢使用 AutoMapper 对于这类工作。它经过优化,犯错误的风险更低。而且比序列化/反序列化方法更好,因为这会导致性能问题。

在构造函数中(或在startup.cs中并通过依赖注入传递)

var config = new MapperConfiguration(cfg => { 
                 cfg.CreateMap<DesignProject, DesignProject>().ForMember(x => x.Id, opt => opt.Ignore());
                 cfg.CreateMap<ProjectObject<ClassA>, ProjectObject<ClassA>>().ForMember(x => x.Id, opt => opt.Ignore()));
                 cfg.CreateMap<ProjectObject<ClassA>, ProjectObject<ClassB>>().ForMember(x => x.Id, opt => opt.Ignore()));
                 cfg.CreateMap<ProjectObject<ClassB>, ProjectObject<ClassB>>().ForMember(x => x.Id, opt => opt.Ignore()));
                 cfg.CreateMap<ProjectObject<ClassB>, ProjectObject<ClassA>>().ForMember(x => x.Id, opt => opt.Ignore()));

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

在您的方法中:

 var newProject = mapper.Map(sourceProject, new DesignProject());

或

 mapper.Map(sourceProject, targetProject);

-1

Lazar ÄorÄeviÄ 6 年前

这不应该起作用。若我理解你们想将列表中的所有元素复制到新列表中,但要创建新对象。然后,您可以使用扩展方法克隆完整列表,如下所述 link 。要更新导航属性,请查看此帖子 link .