简体   繁体   English

复杂组按linq查询

[英]Complex Group By linq Query

I have a question... i have 3 tables... 我有一个问题...我有3张桌子...

user(id,name) 用户(ID,名称)

Answers(id,title) 答案(id,标题)

UserAnswers(id,user_id,right_answer_id,user_answer_id) UserAnswers(id,user_id,right_answer_id,user_answer_id)

now i want to get user name and total score which is count(where (right_answer_id==user_answer_id)) * 10 so result can be like this 现在我想获取用户名和总得分为count(其中(right_answer_id == user_answer_id))* 10,因此结果可以像这样

UserName Score 用户名分数

Imran 20 伊姆兰20

john 30 约翰30

I don't have such experience in Linq. 我在Linq中没有这样的经验。 how can i do that? 我怎样才能做到这一点? i know in sql i can do like 我知道在SQL中我可以做

select u.Name, 
       sum (case 
              when q.Right_Anwer_id = q.user_answerr_id then 1 
              else 0 
            end)
 from quiz q
 inner join [User] u 
 on u.id = q.user_id
 group by u.Name,
          q.test_id
 having q.test_id = 1

Classes: 类:

class Answer
{
    public int Id { get; set; }
    public string Title{ get; set; }
}

class User
{
    public int Id { get; set; }
    public string Name{ get; set; }
}

class UserAnswers
{
    public int id { get; set; }
    public int user_id { get; set; }
    public int user_answer_id { get; set; }
    public int right_answerr_id { get; set; }
    public virtual Answer Answer { get; set; }
    public virtual User User { get; set; }
}

If you also add a navigation property User.UserAnswers you can use this query: 如果还添加导航属性User.UserAnswers ,则可以使用以下查询:

from u in context.Users
where u.UserId == userid
select new
{
    u.Name,
    Score = u.UserAnswers
             .Count(ua => ua.user_answerr_id == ua.right_answerr_id) * 10
}

I don't see a TestId property in UserAnswer (please use the singular name for the class), or anywhere, so this test_id = 1 condition in your SQL can't be applied in LINQ (yet). 我在UserAnswer (请为类使用单数名称)或任何地方都看不到TestId属性,因此SQL中的test_id = 1条件无法在LINQ中应用(尚未)。

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM