[t0].[Id1], [t0].[TimeStamp1], [t0].[FirstName], [t0].[FullName],

       [t0].[LastName], [i].[Id], [m].[Id]

FROM [dbo].[Inventory] AS [i]

INNER JOIN [dbo].[Makes] AS [m] ON [i].[MakeId] = [m].[Id]

INNER JOIN (

  SELECT [o].[Id], [o].[CarId], [o].[CustomerId], [o].[TimeStamp],

         [c].[Id] AS [Id1], [c].[TimeStamp] AS [TimeStamp1], [c].[FirstName],

         [c].[FullName], [c].[LastName]

  FROM [Dbo].[Orders] AS [o]

  INNER JOIN (

    SELECT [i0].[Id], [i0].[IsDrivable]

    FROM [dbo].[Inventory] AS [i0]

    WHERE [i0].[IsDrivable] = CAST(1 AS bit)

  ) AS [t] ON [o].[CarId] = [t].[Id]

    INNER JOIN [Dbo].[Customers] AS [c] ON [o].[CustomerId] = [c].[Id]

    WHERE [t].[IsDrivable] = CAST(1 AS bit)

) AS [t0] ON [i].[Id] = [t0].[CarId]

WHERE ([i].[IsDrivable] = CAST(1 AS bit)) AND EXISTS (

    SELECT 1

    FROM [Dbo].[Orders] AS [o0]

    INNER JOIN (

      SELECT [i1].[Id], [i1].[Color], [i1].[IsDrivable], [i1].[MakeId],

             [i1].[PetName], [i1].[TimeStamp]

      FROM [dbo].[Inventory] AS [i1]

      WHERE [i1].[IsDrivable] = CAST(1 AS bit)

    ) AS [t1] ON [o0].[CarId] = [t1].[Id]

    WHERE ([t1].[IsDrivable] = CAST(1 AS bit)) AND ([i].[Id] = [o0].[CarId]))

ORDER BY [i].[Id], [m].[Id]

Будете вы разделять свои запросы или нет, зависит от существующих бизнес-требований.

Фильтрация связанных данных

В версии EF Core 5 появилась возможность фильтрации при включении навигационных свойств типа коллекций. До выхода EF Core 5 единственным способом получения отфильтрованного списка для навигационного свойства типа коллекций было использование явной загрузки. Добавьте в MakeTests.cs следующий тест, который демонстрирует получение записей производителей, выпускающих автомобили желтого цвета:

[Fact]

public void ShouldGetAllMakesAndCarsThatAreYellow

{

  var query = Context.Makes.IgnoreQueryFilters

      .Include(x => x.Cars.Where(x => x.Color == "Yellow"));

     var qs = query.ToQueryString;

  var makes = query.ToList;

  Assert.NotNull(makes);

  Assert.NotEmpty(makes);

  Assert.NotEmpty(makes.Where(x => x.Cars.Any));

  Assert.Empty(makes.First(m => m.Id == 1).Cars);

  Assert.Empty(makes.First(m => m.Id == 2).Cars);

  Assert.Empty(makes.First(m => m.Id == 3).Cars);

  Assert.Single(makes.First(m => m.Id == 4).Cars);

  Assert.Empty(makes.First(m => m.Id == 5).Cars);

}

Ниже показан сгенерированный код SQL:

SELECT [m].[Id], [m].[Name], [m].[TimeStamp], [t].[Id], [t].[Color],

       [t].[IsDrivable], [t].[MakeId], [t].[PetName], [t].[TimeStamp]

FROM [dbo].[Makes] AS [m]

LEFT JOIN (

  SELECT [i].[Id], [i].[Color], [i].[IsDrivable], [i].[MakeId],

         [i].[PetName], [i].[TimeStamp]

  FROM [dbo].[Inventory] AS [i]

  WHERE [i].[Color] = N'Yellow') AS [t] ON [m].[Id] = [t].[MakeId]

ORDER BY [m].[Id], [t].[Id]

Изменение запроса на разделенный приводит к выдаче такого кода SQL (получен с использованием профилировщика SQL Server):

SELECT [m].[Id], [m].[Name], [m].[TimeStamp]

FROM [dbo].[Makes] AS [m]

ORDER BY [m].[Id]

SELECT [t].[Id], [t].[Color], [t].[IsDrivable], [t].[MakeId],