如何根据二级关系过滤实体

Question

我使用 EntityFramework Core 的数据库架构简化了这一点：

public class Parent
{
    public int ID { get; set; }
}

public class Child
{
    public int ID { get; set; }
    public bool Include { get; set; }
}

public class Link
{
    public int FromID { get; set; }
    public int ToID { get; set; }
}

父项通过第三个表中的条目链接到子项。 父链接是一对多的 与孩子的链接是一对一的

作为这项工作的一部分，我无法更改数据库架构。

我正在尝试返回与至少一个孩子相关联且所有孩子都满足特定条件的 Parent 实例（在此示例中为 include == true）

我尝试了很多不同的查询，包括下面的一个，它在这个测试场景中有效，但似乎不能很好地扩展。

var ParentList = new[]
{
    new Parent {ID = 1}, //3 children, 1 fails criteria
    new Parent {ID = 2}, //2 children, all pass criteria
    new Parent {ID = 3}  //no children
};
var ChildList = new[]
{
    new Child {ID = 11, Include = true},
    new Child {ID = 12, Include = true},
    new Child {ID = 13, Include = false},
    new Child {ID = 21, Include = true},
    new Child {ID = 22, Include = true},
};
var LinkList = new[]
{
    new Link {FromID = 1, ToID = 11},
    new Link {FromID = 1, ToID = 12},
    new Link {FromID = 1, ToID = 13},
    new Link {FromID = 2, ToID = 21},
    new Link {FromID = 2, ToID = 22},
};

var relevant = (from parent in ParentList
    join l in LinkList on parent.ID equals l.FromID into links
    from linkedChildren in (
        from link in links
        join child in ChildList on link.ToID equals child.ID into children
        from child in children
        group child by link.FromID into kids
        select kids.AsEnumerable())
    where linkedChildren.All(x => x.Include)
    select new { parent, linkedChildren }).ToList();

Assert.Single(relevant);
Assert.Equal(2, relevant.First().parent.ID);

Answer 1

其中任何一个都应该为您指明正确的方向

ParentList.Where(s => s.LinkList .Any(e => e.ChildList.Include == true))

from s in dc.ParentList
from e in s.LinkList 
where e.ChildList.Include == true
select s;

看看这是否更符合您的需要：

var result = from l in LinkList
             join p in ParentList on l.FromID equals p.ID
             join c in ChildList on l.ToID equals c.ID and c.Include == true
             select new { p.ID, c.ID, l.FromID, l.ToID};