包含和選擇的實體框架

Question

我有以下實體（節省空間的偽代碼）

Program [ int Id, 
          string Name, 
          List<ProgramFoodType> ProgramFoodTypes, 
          List<ProgramFood> ProgramFoods]

ProgramFoodType[ int Id, int ProgramId, int Type, bool IsActive]
ProgramFood [ int Id, int ProgramId, Food Food, FoodType FoodType]
Food [int Id, string Name]
FoodType [int Id, string Name]

我的任務是獲取單個Program及其相關ProgramFoodTypes ，條件 ProgramFoodType 應處於活動狀態，並且ProgramFoods與相關實體Food和FoodType

到目前為止，我使用了以下內容

1- 以下查詢將檢索ProgramFoodTypes和ProgramFoods的詳細信息，但它將帶來所有活動和非活動ProgramFoodTypes

var program = mEntities.Programs
                          .Include(p =>p.ProgramFoodTypes)
                          .Include(p =>p.ProgramFoods.Select(f =>f.Food))
                          .InClude(p =>p.ProgramFoods.Select( f =>f.FoodType))
                          .Where(m =>m.Id== Id);

2- 以下查詢將檢索詳細信息但缺少Food和FoodType

var program = (from p in mEntities.Programs
              where p.Id ==Id
              select new {
                 Program = p,
                 ProgramFoodTypes = from pf in p.ProgramFoodTypes
                                    where pf.IsActive
                                    select pf,                  
                 ProgramFoods = p.ProgramFoods // here i can't add include statement
              }).ToArray().Select(m => m.Program);

如何在第二個查詢中包含食物和食物類型？

Answer 1

對於您的第二個解決方案，我認為您可以使用：

var program = (from p in mEntities.Programs
                  .Include(p => p.ProgramFoods.Select(f => f.Food))
                  .Include(p => p.ProgramFoods.Select(f => f.FoodType))
               where p.Id == Id
               select new {
                  Program = p,
                  ProgramFoodTypes = from pf in p.ProgramFoodTypes
                                     where pf.IsActive
                                     select pf,                  
                  p.ProgramFoods 
               }).ToArray().Select(m => m.Program);

更新：由於您在 linq 查詢中使用匿名類型， Include 語句被忽略。
您必須在客戶端加載所有相關的ProgramFoodTypes ，然后進行過濾：

var program = mEntities.Programs
                   .Include(p => p.ProgramFoodTypes)
                   .Include(p => p.ProgramFoods.Select(f => f.Food))
                   .Include(p => p.ProgramFoods.Select(f => f.FoodType))
                   .SingleOrDefault(m => m.Id == Id);

program.ProgramFoodTypes = program.ProgramFoodTypes.Where(pft => pft.IsActive);  

您可以使用AsNoTracking()或在新對象中克隆返回的Program對象，以防您想確保數據在 db 端完好無損。

Answer 2

嘗試這個：

var program = mEntities.Programs
                       .Include(p => p.ProgramFoodTypes)
                       .Include(p => p.ProgramFoods.Select(f => f.Food))
                       .InClude(p => p.ProgramFoods.Select(f => f.FoodType))
                       .SingleOrDefault(m => m.Id == Id && m.ProgramFoodTypes.All(t => t.IsActive));

Answer 3

或許：

var program = (from p in mEntities.Programs
          where p.Id ==Id
          select new {
             Program = p,
             ProgramFoodTypes = from pf in p.ProgramFoodTypes
                                where pf.IsActive
                                select pf,                  
             ProgramFoods = p.ProgramFoods.Select(y => new {
                 Food = y.Food,
                 Type = y.FoodType
             }) 
          }).ToArray().Select(m => m.Program);