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Find all scores for the 2nd to most recent date

开发者 https://www.devze.com 2023-03-13 03:26 出处:网络
I have a history table that contains a score per group per date (PK is group, date). What is the SQL query that can retrieve the scores for all groups for the 2nd most recent date?

I have a history table that contains a score per group per date (PK is group, date). What is the SQL query that can retrieve the scores for all groups for the 2nd most recent date?

ETA: The dates are the same a开发者_Go百科cross groups (each score is entered into the history table at the same time for each group).


select *
from ScoreHistory sc1
where exists
(
    select GroupId, max(ScoreDate) RecentScoreDate
    from ScoreHistory sc2
    where not exists
    (
        select GroupId, max(ScoreDate) RecentScoreDate
        from ScoreHistory sc3
        group by GroupId
        having GroupId = sc2.GroupId and max(ScoreDate) = sc2.ScoreDate
    )
    group by GroupId
    having GroupId = sc1.GroupId and max(ScoreDate) = sc1.ScoreDate
)

Setup:

create table ScoreHistory(GroupId int, ScoreDate datetime)

insert ScoreHistory
    select 1, '2011-06-14' union all
    select 1, '2011-06-15' union all
    select 1, '2011-06-16' union all
    select 2, '2011-06-15' union all
    select 2, '2011-06-16' union all
    select 2, '2011-06-17' 

The query would looks as simple as below for MS SQL 2005 +

;with cte
as
(
    select *, row_number() over(partition by GroupId order by ScoreDate desc) RowNumber
    from ScoreHistory
)
select *
from cte
where RowNumber = 2


You need two aggregates

  1. get max dates per group
  2. get max dates per group that are less then the dates from step 1
  3. join back to the score from this aggregate

Something like

SELECT
    Group, Date, Score
FROM
    ( ..2nd max date per group
    SELECT
       Group, MAX(Date) AS TakeMe
    FROM
        ( --max date per group
        SELECT
           Group, MAX(Date) AS IgnoreMe
        FROM
           MyTable
        GROUP BY
           Group
        ) ex
        JOIN
        MyTable M ON ex.Group = M.Group AND ex.IgnoreMe > M.Date
    GROUP BY
        M.Group
    ) inc
    JOIN
    MyTable M2 ON inc.Group = M2.Group AND inc.TakeMe = M2.Date

This is so much easier on SQL Server 2005 with ROW_NUMBER()...


SELECT *
FROM tblScore
WHERE EXISTS
(
    SELECT NULL
    FROM tblScore as tblOuter
    WHERE NOT EXISTS
    (
        SELECT NULL
        FROM tblScore As tblInner
        WHERE tblInner.[group] = tblOuter.[group]
        GROUP BY [group]
        HAVING MAX(tblInner.[date]) = tblOuter.[date]
    ) 
    AND tblOuter.[group] = tblScore.[group]
    GROUP BY [group]
    HAVING MAX(tblOuter.[date]) = tblScore.[date]
)


Try this. I am trying to get the TOP 2 DISTINCT Dates Desc first which will work if you are using just dates and not datetimes. Then reversing that table and getting the TOP 1 and using that result as the 2nd most recent date to get the groups scores.

SELECT *
FROM YourTable
INNER JOIN 
(SELECT TOP 1 x.[date]
FROM
    (SELECT TOP 2 DISTINCT [date]
    FROM YourTable
    ORDER BY [date] DESC) AS x
ORDER BY [date] ASC) AS y
ON y.[date] = YourTable.[date]

I think this may need a WHERE y.date = YourTable.date but I am not sure

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