我有两张桌子。首先是比赛用户和比赛点。 表与ParticipateID之间存在外键关系。 在CompetitionPoints表中,有多个参与点的不同点数。所以我想根据总积分获取总积分和等级。因此,如果一个参与ID有多个相同的总积分,那个参与者的等级应该相同。它相同喜欢该学生的总分和成绩。 这是我的代码。
var competitionusers = (from c in db.CompetitionUsers
group c by new { c.ParicipateId, c.CompetitionPoints.FirstOrDefault().Points }
into g orderby g.Key.Points descending select
new { Points = db.CompetitionPoints.Where
(x => x.ParticiapteId == g.FirstOrDefault().ParicipateId).Sum(x => x.Points),
Rank = (from o in db.CompetitionUsers
group o by o.CompetitionPoints.FirstOrDefault().Points into l
select l).Count(s => s.Key > db.CompetitionPoints.
Where(x => x.ParticiapteId == g.FirstOrDefault().ParicipateId).Sum(x => x.Points)) + 1,
}).Where(x => x.Points != null).OrderByDescending(x => x.Points).AsQueryable();
答案 0 :(得分:0)
如果我正确理解您的数据模型,我认为您可以简化为以下内容:
var result = db.CompetitionUsers
// group by total points
.GroupBy(cu => cu.CompetitionPoints.Sum(cp => cp.Points))
// order by total points descending
.OrderByDescending(g => g.Key)
// calculate rank based on position in grouped results
.SelectMany((g, i) => g.Select(cu => new { Rank = i+1, TotalPoints = g.Key, CompetitionUser = cu }));
答案 1 :(得分:0)
IQueryable<CompetitionLaderMadel> competitionUsers;
competitionUsers = (from c in db.CompetitionUsers
select new CompetitionLaderMadel
{
CompetitionName = c.Competition.CompetitionName,
CompetitionId = c.CompetitionId,
Points = db.CompetitionPoints.Where(x => x.ParticiapteId == c.ParicipateId).Sum(x => x.Points),
IsFollow = db.CrowdMember.Any(x => x.Following == userid && x.UserCrowd.UserID == c.UserId && x.Status != Constants.Deleted),
}).Where(x => x.Points != null && x.UserId != null).OrderByDescending(x => x.Points);
然后写了这个查询
var q = from s in competitionUsers
orderby s.Points descending
select new
{
CompetitionName = s.CompetitionName,
CompetitionId = s.CompetitionId,
HeadLine = s.HeadLine,
UserId = s.UserId,
Points = s.Points,
Image = s.Image,
IsFollow = s.IsFollow,
UserName = s.UserName,
Rank = (from o in competitionUsers
where o.Points > s.Points
select o).Count() + 1
};