Question

我正在处理查询。我有以下表格：

CREATE TABLE #House 
( 
     HouseName Varchar(1), 
     Address Varchar(255)
)

CREATE TABLE #Events 
(
    Id Integer, 
    HouseName Varchar(1), 
    EventName Varchar(255), 
    Date Date, 
    Family Varchar(255)
)

INSERT #House VALUES('A', '1 Market St')
INSERT #House VALUES('B', '2 Market St')
INSERT #House VALUES('X', '3 Market St')
INSERT #House VALUES('D', '4 Market St')

INSERT #Events VALUES(1,'A','MoveIn','2005-01-31','Smith')
INSERT #Events VALUES(2,'A','Remodel','2005-03-31','Smith')
INSERT #Events VALUES(3,'A','Remodel','2008-03-31','Smith')
INSERT #Events VALUES(4,'A','CleanUp','2008-05-31','Smith')
INSERT #Events VALUES(5,'B','MoveIn','2005-01-31','Newton')
INSERT #Events VALUES(6,'B','MoveOut','2008-05-01','Newton')
INSERT #Events VALUES(7,'B','MoveIn','2008-12-01','Einstein')
INSERT #Events VALUES(8,'D','MoveIn','2007-08-31','Soo')
INSERT #Events VALUES(9,'D','MoveOut','2010-08-13','Soo')

我期待以下结果：

  HouseName Address       MostRecentResident MoveInDate
  A         1 Market      StSmith            2005-01-31
  B         2 Market St   Einstein           2008-12-01
  D         4 Market St   Soo                2007-08-31

注意：众议院B有两个MoveIn事件。

这是查询。您认为有更简单的方法可以获得相同的结果吗？

Select 
    a.HouseName, a.Address, b.Family as [MostRecentResident], 
    mid.MoveInDate
from
    #House a, #Events b, 
    --MoveInDate of the last family:
    (Select HouseName, Max(MoveInDate) MoveInDate 
     from
         (Select HouseName, Family, MIN(Date) MoveInDate 
          From #Events 
          Group by HouseName, Family) ff 
     group by HouseName) mid,
    --Distinct house and the last move in date:
    (Select HouseName, MAX(Date) LastMoveInDate 
     From #Events 
     Group by HouseName) lmid
where 
    a.HouseName = b.HouseName
    and b.HouseName = lmid.HouseName
    and b.Date = lmid.LastMoveInDate
    and b.HouseName = mid.HouseName

Answer 1

我认为你只想要一个条件聚合的连接：

select h.housename, h.address,
       max(case when seqnum = 1 and eventname = 'Movein' then family end) as MostRecentResident,
       max(case when seqnum = 1 and eventname = 'Movein' then date end) as MoveInDate
from #house h join
     (select e.*,
             row_number() over (partition by e.housename, e.eventname order by date desc) seqnum
      from #event e
     ) e
     on h.housename = e.housename
group by h.housename,h.address;

编辑：

以上内容可以进一步简化为：

select h.housename, h.address,
       max(case when seqnum = 1 then family end) as MostRecentResident,
       max(case when seqnum = 1 then date end) as MoveInDate
from #house h join
     (select e.*,
             row_number() over (partition by e.housename, e.eventname order by date desc) seqnum
      from #event e
      where eventname = 'Movein'
     ) e
     on h.housename = e.housename
group by h.housename,h.address;

或者：

select h.housename, h.address,
       e.family as MostRecentResident,
       e.date as MoveInDate
from #house h outer apply
     (select top 1 e.*
      from #event e
      where e.eventname = 'Movein' and h.housename = e.housename
     ) e;

这似乎是最简单的方法。

Answer 2

我认为输出像

这样的数据集是不可能的

  HouseName Address       MostRecentResident MoveInDate
  A         1 Market      StSmith            2005-01-31
  B         2 Market St   Einstein           2008-12-01
  D         4 Market St   Soo                2007-08-31

因为MostRecentResident包含 StSmith ，如果是拼写错误，则应该史密斯。

那么也许你想要

select a.HouseName,a.Address,b.Family as MostRecentResident,b.Date as MoveInDate from #House a 
CROSS APPLY (SELECT TOP(1) Family,Date FROM #Events WHERE EventName='MoveIn' and HouseName=a.HouseName order by Date desc) as b

应用交叉申请可以简化此查询。

Answer 3

SELECT * FROM (
    SELECT h.*,e.Family AS MostRecentResident,e.Date AS MoveInDate,ROW_NUMBER()OVER(PARTITION BY h.HouseName ORDER BY e.Date desc) AS rn
    FROM #House AS h INNER JOIN #Events AS e ON h.HouseName=e.HouseName
    WHERE e.EventName='MoveIn'
) AS t WHERE t.rn=1


HouseName Address              MostRecentResident   MoveInDate rn
--------- -------------------- -------------------- ---------- ----------
A         1 Market St          Smith                2005-01-31 1
B         2 Market St          Einstein             2008-12-01 1
D         4 Market St          Soo                  2007-08-31 1

复杂的SQL查询。有什么方法可以简化？

3 个答案: