这样的表:
id | userid | contentid
-------------------------
41 | 92 | 1187
42 | 92 | 1189
43 | 92 | 1190
44 | 92 | 1193
45 | 92 | 1200
46 | 92 | 1201
47 | 92 | 1202
48 | 104 | 1200
49 | 104 | 1201
50 | 104 | 1202
51 | 103 | 1200
52 | 103 | 1201
53 | 103 | 1202
我试图用1202(例如)获取所有相关的内容ID。
很快我想得到以下列表:
1201 - count: 3
1200 - count: 3
1187 - count: 1
1189 - count: 1
1190 - count: 1
1193 - count: 1
我尝试了类似以下的查询,但我还需要做更多其他事情。
(from x in IRepository<ContentRelation>().Query().ToList()
where x.Content.Id == content.Id
group x by x.GUser.Id into c
select new
{
a = c.Key,
b = (from d in IRepository<ContentRelation>().Query()
where d.GUser.Id == c.Key && d.Content.Id != content.Id
select d)
})
编辑: 我得到了我想要的以下查询,但我不确定这是正确的方式:
var q = DependencyResolver.Current.GetService<IRepository<ContentRelation>>().Query();
List<int> gh = new List<int>();
foreach (var item in q.Where(x => x.Content.Id == content.Id).GroupBy(x => x.GUser.Id).Select(x => x.Key))
{
foreach (var a in q.Where(x => x.GUser.Id == item && x.Content.Id != content.Id).ToList())
{
gh.Add(a.Content.Id);
}
}
foreach (var hhj in gh.GroupBy(x => x).OrderByDescending(x => x.Count()))
{
Response.Write(hhj.Key + "-" + hhj.Count()+ "<br />");
}
答案 0 :(得分:1)
有了你想要的东西(理论上:) :)
IRepository<ContentRelation>().Query().GroupBy(x => x.Content.Id).Select(x => new Tuple<int, int>(x.Key, x.Count())).OrderBy(x => x.First)
答案 1 :(得分:0)
var result = IRepository<ContentRelation>().Query()
.GroupBy(p => p.contentid)
.Select(p => new
{
contentid=p.Key,
counter=p.Count()
});