2

このデータモデルを生成する必要があります（例）：

IList<FeatureGroupFeaturesDto> fcf = new List<FeatureGroupFeaturesDto>();

fcf.Add(new FeatureGroupFeaturesDto
{
    FeatureGroup = new FeatureGroupDto { Id = 1, Name = "Interior" },
    Features = new List<FeatureDto> { 
        new FeatureDto { Id = 7, Name = "Bancos Traseiros Rebatíveis" },
        new FeatureDto { Id = 35, Name = "Computador de Bordo" },
        new FeatureDto { Id = 38, Name = "Suporte para Telemóvel" }
    },
});

fcf.Add(new FeatureGroupFeaturesDto
{
    FeatureGroup = new FeatureGroupDto { Id = 2, Name = "Exterior" },
    Features = new List<FeatureDto> { 
        new FeatureDto { Id = 13, Name = "Barras de Tejadilho" },
        new FeatureDto { Id = 15, Name = "Retrovisores Aquecidos" },
        new FeatureDto { Id = 16, Name = "Retrovisores Elétricos" }
    },
});

エンティティに基づく：

public class Category
{
    public int Id { get; set; }
    public string Name { get; set; }
}

public class CategoryFeatureGroupFeature
{
    [Key]
    [Column("Category_Id", Order = 0)]
    public int Category_Id { get; set; }

    [Key]
    [Column("FeatureGroup_Id", Order = 1)]
    public int FeatureGroup_Id { get; set; }

    [Key]
    [Column("Feature_Id", Order = 2)]
    public int Feature_Id { get; set; }
}

public class Feature
{
    public int Id { get; set; }
    public string Name { get; set; }
}

public class FeatureGroup
{
    public int Id { get; set; }
    public string Name { get; set; }
}

基本的には、FeatureGroupによってグループ化されたカテゴリのすべての機能を取得するという考え方です。

編集：

私はこのモデルに情報を入れようとしています：

public class FeatureCategoryFeatures
{
    public FeatureGroup FeatureGroup { get; set; }
    public IList<Feature> Features { get; set; }
}

LINQとEntityFrameworkを使用してこれを実現するにはどうすればよいですか？

ありがとう。

4

2 に答える 2

0

カテゴリに複数のグループを含めることができると仮定して、回答を少し修正しました（私の仮定が正しいことを願っています）。また、返される結果として匿名オブジェクトを使用しましたが、意図は明らかです。

/*** DATA ***/
IList<Feature> featuresList = new List<Feature> { 
        new Feature { Id = 13, Name = "Barras de Tejadilho" },
        new Feature { Id = 15, Name = "Retrovisores Aquecidos" },
        new Feature { Id = 16, Name = "Retrovisores Elétricos" },
        new Feature { Id = 7, Name = "Bancos Traseiros Rebatíveis" },
        new Feature { Id = 35, Name = "Computador de Bordo" },
        new Feature { Id = 38, Name = "Suporte para Telemóvel" },
        new Feature { Id = 1, Name = "2nd Exterior Feature" }
};
IList<FeatureGroup> featureGroupList = new List<FeatureGroup>{
    new FeatureGroup { Id = 1, Name = "Interior" },
    new FeatureGroup { Id = 2, Name = "Exterior" },
    new FeatureGroup { Id = 3, Name = "2nd Exterior" }
};
IList<Category> categoryList = new List<Category>{
    new Category{ Id=1, Name="All interior" },
    new Category { Id=2, Name="All exterior" }
};
IList<CategoryFeatureGroupFeature> cfcList = new List<CategoryFeatureGroupFeature>
{
    new CategoryFeatureGroupFeature { Category_Id = 1, FeatureGroup_Id = 1, Feature_Id = 7 },
    new CategoryFeatureGroupFeature { Category_Id = 1, FeatureGroup_Id = 1, Feature_Id = 35 },
    new CategoryFeatureGroupFeature { Category_Id = 1, FeatureGroup_Id = 1, Feature_Id = 38 },
    new CategoryFeatureGroupFeature { Category_Id = 2, FeatureGroup_Id = 2, Feature_Id = 13 },
    new CategoryFeatureGroupFeature { Category_Id = 2, FeatureGroup_Id = 2, Feature_Id = 15 },
    new CategoryFeatureGroupFeature { Category_Id = 2, FeatureGroup_Id = 2, Feature_Id = 16 },
    new CategoryFeatureGroupFeature { Category_Id = 2, FeatureGroup_Id = 3, Feature_Id = 1 }
};

/*** QUERY ***/
var result = from c in categoryList
                select new {
                    Id = c.Id,
                    Name = c.Name,
                    FeatureGroups = from fg in featureGroupList
                                    where (from cfc in cfcList
                                        where cfc.Category_Id == c.Id
                                        select cfc.FeatureGroup_Id).Distinct()
                                    .Contains(fg.Id)
                                    select new {
                                        Id = fg.Id,
                                        Name = fg.Name,
                                        Features = (from f in featuresList
                                                join cfc2 in cfcList on f.Id equals cfc2.Feature_Id
                                                where cfc2.FeatureGroup_Id == fg.Id
                                                select f).Distinct()
                                    }
                };

joinとを組み合わせても同じ結果が得られますgroup byが、私は を選びましたDistinct()。

于 2013-03-20T19:17:37.043 に答える