计算在当前团队之前个人与其他团队成员的先前联系

Question

我想计算在当前团队成立之前每个成员与其他团队成员的先前联系数。

我有下表

TeamId	UserId	Date
1	    100	  2/1/2017 
1	    101	  2/1/2017
1	    102	  2/1/2017 
2	    100	  2//1/2016
2	    105   2//1/2016
3   	    100	  2//1/2015
3	    101	  2//1/2015
3   	    102	  2//1/2015
3   	    105	  2//1/2015
3	    106	  2//1/2015
4	    101   1/3/2015
4	    102	  1/3/2015
4	    105   1/3/2015
4   	    106	  1/3/2015

我想得到下表

TeamId	number of previous collaborations
1       	4
2	        1
3       	6
4	        0

以团队 1 为例： 用户 100 在团队 3 中与用户 101 合作过一次，在团队 3 中与用户 102 合作过一次，因此 = 1+1=2 用户 101 与用户 102 合作了两次（一次在团队 3 中，一次在团队 4 中）= 1+1=2

所以这个团队之前的合作次数是 4。

我不知道如何编写代码。任何帮助表示赞赏

SELECT [TeamId],
COUNT( SELECT [Date] FROM [submissions] s1
       WHERE s.[UserId] = [submissions].[UserId]
       AND [Date]< [submissions].[Date]
       ORDER BY [DateSubmitted] DESC) as Num_Collaborations
     )
FROM [submissions]

Answer 1

这是一个具有挑战性的问题。您必须在查询中包含同一个表 4 次：

select t1.TeamId, t1.UserId, t2.TeamId as OtherTeam, t3.UserId as Collaborator
from @table t1, @table t2, @table t3, @table t4
where t1.UserId = t2.UserId and
      t2.TheDate < t1.TheDate and
      t3.TeamId = t2.TeamId and
      t4.TeamId = t1.TeamId and
      t4.UserId = t3.UserId and
      t4.UserId < t1.UserId

http://rextester.com/LHZS40204

（然后你可以分组并计数得到最终结果）

Answer 2

您可以通过自加入获得所有用户协作：

select s.userid, s2.userid as userid2, s.date,
       row_number() over (order by (select null)) as c_id
from submissions s join
     submissions s2
     on s2.teamid = s.teamid;

让我们将其用作子查询来回答您的问题。这个想法是将协作与提交相匹配——限制两个用户在以后日期在同一个团队中。然后，您可以聚合：

with c as (
      select s.userid, s2.userid as userid2, s.date,
             row_number() over (order by (select null)) as c_id
      from submissions s join
           submissions s2
           on s2.teamid = s.teamid
     )
select s1.teamid, count(distinct c_id) as num_previous_collaborations
from c c join
     submissions s1
     on s1.userid = c.userid and
        s1.date > c.date join  -- they are later on a team
     submissions s2
     on s2.userid = c.userid2 and
        s2.teamid = s1.teamid 
group by s1.teami;