SQL group by select

Question

我正在使用SQL Server,我有一个包含以下列的表:

SessionId | Date | first name | last name 

我想做group by sessionId,然后得到最大日期的行.

例如:

xxx | 21/12/2012 | f1 | l1
xxx | 20/12/2012 | f2 | l2
yyy | 21/12/2012 | f3 | l3
yyy | 20/12/2012 | f4 | l4

我想获得以下行:

xxx | 21/12/2012 | f1 | l1
yyy | 21/12/2012 | f3 | l3

谢谢

Answer 1

试试这个:

WITH MAXSessions
AS
(
  SELECT 
    *,
    ROW_NUMBER() OVER(PARTITION BY SessionID ORDER BY Date DESC) rownum
  FROM Sessions
)
SELECT
  SessionId,
  Date,
  firstname,
  lastname 
FROM MAXSessions
WHERE rownum = 1;

要么:

SELECT 
  s.SessionId,
  s.Date,
  s.firstname,
  s.lastname 
FROM Sessions s
INNER JOIN
(
   SELECT SessionID, MAX(Date) LatestDate
   FROM sessions
   GROUP BY SessionID
) MAxs  ON maxs.SessionID  = s.SessionID
       AND maxs.LatestDate = s.Date;

更新:要获取会话计数,您可以执行以下操作:

SELECT 
  s.SessionId,
  s.Date,
  s.firstname,
  s.lastname,
  maxs.SessionsCount
FROM Sessions s
INNER JOIN
(
   SELECT SessionID, COUNT(SessionID), SessionsCount, MAX(Date) LatestDate
   FROM sessions
   GROUP BY SessionID
) MAxs  ON maxs.SessionID  = s.SessionID
       AND maxs.LatestDate = s.Date;

Answer 2

有几个选项，其中一个是使用按日期对行进行排名的CTE ：

编辑以包括会话计数

WITH Sessions AS (
    SELECT SessionId, [Date], FirstName, LastName,
    ROW_NUMBER() OVER (PARTITION BY SessionId ORDER BY [Date] DESC) AS Ord
    FROM YourTable
)
SELECT S.SessionId, S.Date, S.FirstName, S.LastName, X.SessionCount
FROM Sessions S
INNER JOIN (
    SELECT SessionId, COUNT(*) AS SessionCount
    FROM Sessions
    GROUP BY SessionId 
) X ON X.SessionId = S.SessionId  
WHERE S.Ord = 1

在这种情况下，排名函数是您的朋友，您想要获取满足某些“有序”条件（例如最大日期）的整行。