Question

我正在尝试在LINQ中编写此选择但是我很长时间没有成功修复它。我也试过LINQ - join with Group By and get average，但它在我的代码中不起作用。显然我错了。

SQL：

SELECT name_type, AVG(t.price) as avgPrice FROM type tp
JOIN location l ON l.ID_type = tp.ID 
JOIN event e ON e.ID_location = l.ID
JOIN ticket t ON t.ID_event = e.ID
GROUP BY tp.name_type

LINQ：

var q3 = from l in db.location
join tp in db.type on l.ID_type equals tp.ID
join e in db.event on l.ID equals u.ID_location
join t in db.ticket on e.ID equals t.ID_event 
group tp by new {Type_name = tp.type_name} into grp
select new
{
     Type_name = grp.Key.type_name,
     avgPrice = grp.Average( x => x.ticket.price)
};

Answer 1

有一些问题：

第二次加入时出错 - 我认为u.ID_location需要e.ID_location。
我认为您正在对错误的实体进行分组，请尝试按t而不是tp进行分组。
您不需要group by中的匿名类型。

试试这个：

var results = 
     from l in db.location
     join tp in db.type on l.ID_type equals tp.ID
     join e in db.event on l.ID equals e.ID_location
     join t in db.ticket on e.ID equals t.ID_event 
     group t by new tp.type_name into grp
     select new
     {
          Type_name = grp.Key,
          avgPrice = grp.Average(x => x.price)
     };

如果您碰巧在实体之间设置了导航属性，这将更加容易。很难说实体假设是如何相关的，但我认为这样的事情会起作用：

// average ticket price per location type
var results = 
    from t in db.ticket
    group t by t.event.location.type.type_name into g
    select new
    {
         Type_name = g.Key,
         avgPrice = g.Average(x => x.price)
    };

或者用流利的语法：

var results = db.ticket.GroupBy(t => t.event.location.type.type_name)
                       .Select(g => new 
                               { 
                                   Type_name = g.Key, 
                                   avgPrice = g.Average(x => x.price) 
                               });

LINQ - 与Group by进行多连接并获得平均值

1 个答案: