实体框架 – 实体框架 – 加载相关实体

对不起,标题不是更具体 – 我不知道如何描述这个简洁.
除了位置不需要知道使用它们的行程之外,我已经有多对多关系的旅程和位置是直接的.我创建了这些实体来表示:

public class Trip
{
    public int TripId { get; set; }
    public virtual IList<TripLocation> TripLocations { get; set; }
}

public class TripLocation
{
    public int TripId { get; set; }
    public int LocationId { get; set; }

    public virtual Location Location { get; set; }
}

public class Location
{
    public int LocationId { get; set; }
    // Note: Intentionally no collection of Trips
}

我可以得到旅行渴望加载它的TripLocations,但我不能得到TripLocations渴望加载他们的位置.我已经尝试了一大堆组合流畅的配置和“包括”在查询中

IQueryable<Trip> query = from trip in context
                              .Include(r =>r.TripLocations)
                              .Include(r => r.TripLocations.Select(tl => tl.Location))
                         select ride;

任何建议非常感谢!

最佳答案
我在这里重新创建了您的场景,我可以在单个查询中获取所有结果.

var a = from trip in context.Trips.Include("TripLocations.Location")
        select trip;

就这样.这是对我的数据库的查询:

SELECT 
[Project1].[TripId] AS [TripId], 
[Project1].[Name] AS [Name], 
[Project1].[C1] AS [C1], 
[Project1].[TripId1] AS [TripId1], 
[Project1].[LocationId] AS [LocationId], 
[Project1].[LocationId1] AS [LocationId1], 
[Project1].[Name1] AS [Name1]
FROM ( SELECT 
    [Extent1].[TripId] AS [TripId], 
    [Extent1].[Name] AS [Name], 
    [Join1].[TripId] AS [TripId1], 
    [Join1].[LocationId1] AS [LocationId], 
    [Join1].[LocationId2] AS [LocationId1], 
    [Join1].[Name] AS [Name1], 
    CASE WHEN ([Join1].[TripId] IS NULL) THEN CAST(NULL AS int) ELSE 1 END AS [C1]
    FROM  [dbo].[Trips] AS [Extent1]
    LEFT OUTER JOIN  (SELECT [Extent2].[TripId] AS [TripId], [Extent2].[LocationId] AS [LocationId1], [Extent3].[LocationId] AS [LocationId2], [Extent3].[Name] AS [Name]
        FROM  [dbo].[TripLocations] AS [Extent2]
        INNER JOIN [dbo].[Locations] AS [Extent3] ON [Extent2].[LocationId] = [Extent3].[LocationId] ) AS [Join1] ON [Extent1].[TripId] = [Join1].[TripId]
)  AS [Project1]
ORDER BY [Project1].[TripId] ASC, [Project1].[C1] ASC

更新:

如果你想保持与lambda版本,这将做的工作:

IQueryable<Trip> query = from ride in context.Set<Trip>()
                             .Include(t=>t.TripLocations.Select(l=>l.Location))                                     
                         select ride;

有关MSDN blog的更多信息.

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