在我的数据库中,我有 4 个具有如下关系的表:
检查 --> Correspondence --> CorrespondenceTo, User
现在,当InspectionID
等于某物时,我想得到Subject
, CorrespondenceType
, SentDate
from Correspondence
, DisplayName
from User
, FirstName
, LastName
from CorrespondenceTo
。所以我写这样的代码:
var result = from i in context.Inspections
join correspondence in context.Correspondences
on i.Id equals correspondence.InspectionId
join correspondenceTo in context.CorrespondenceToes
on correspondence.Id equals correspondenceTo.CorrespondenceId
join user in context.Users
on correspondence.SenderUserId equals user.Id
where i.Id == inspectionID
select new
{
Subject = correspondence.Subject,
CorrespondenceType = correspondence.CorrespondenceType,
SentDate = correspondence.SentDate,
Sender = user.DisplayName,
Receiver = new
{
correspondenceTo.FirstName,
correspondenceTo.LastName
}
};
return Json(result);
我的代码有什么问题?如何在 Visual Studio 2012 中查看结果?此外,我想离开加入上面的所有表格。我应该怎么办?如果我想使用Include
而不是Join
,该怎么做?