SQL 查询：最后一个、倒数第二个、倒数第三个..值 (Microsoft SQL)答案

【问题标题】：SQL Query: last, second last, third last..value (Microsoft SQL)SQL 查询：最后一个、倒数第二个、倒数第三个..值 (Microsoft SQL)
【发布时间】：2012-10-31 14:23:03
【问题描述】：

我正在尝试为运动队构建报告。结果存储在 SQL 数据库中，如下所示：

对于报告，我需要在 Home2-Away4 列中填充相应的结果。这意味着：

想象一下：ID=10 和 Home(no.) (Team#) = 1 --> Home1 = 结果一场比赛之前 ID=3 和球队客场比赛，结果为 0 (=home1)。这是我已经做过的，但是 Home2 = 结果两场比赛前，因为 Home3=result 三场场比赛以前等等。

我是这样做的：

SELECT top 1 [result_home] from (SELECT Top 2 [result_home] FROM 
(SELECT top 2 [result_home]  from [Table]
where [ID] < 10  and ( [Home(no.)]=1 or [Away(No.)]=1 ) order by [ID] desc  ) top2     )top1

（Microsoft SQL Server Express 2012 顺便说一句。）

我编写了一个小的 Java 代码，它决定是从 result_home 还是 result_away 中进行选择。

基本问题是：如何查询两天前的比赛结果？

提前致谢。

亲切的问候

斯蒂芬

【问题讨论】：

标签： sql sql-server-2012-express

【解决方案1】：

我认为您需要LAG 函数。最简单的情况是这样的：

[Home1] = LAG(result_Home, 1) OVER(PARTITION BY [Home(no.)] ORDER BY ID),

但是，它并不像这样简单，因为您需要引用 home 或 away，因此您需要首先取消透视数据

SELECT  ID,
        Type,
        TeamID,
        Result
FROM    T
        CROSS APPLY 
        (   VALUES 
                ('Home', [Home(no.)], Result_Home), 
                ('Away', [Away(no.)], Result_Away)
        ) Upvt (Type, TeamID, Result);

这意味着您可以在一列中访问每个团队的结果，无论是主场还是客场（对于滞后功能）。

然后使用这些未透视的数据，您应用滞后来获得前 4 个结果（我使用了 ORDER BY ID，但我假设我缺少一个日期字段？）：

SELECT  *,
        [Result1] = LAG(Result, 1) OVER(PARTITION BY TeamID ORDER BY ID),
        [Result2] = LAG(Result, 2) OVER(PARTITION BY TeamID ORDER BY ID),
        [Result3] = LAG(Result, 3) OVER(PARTITION BY TeamID ORDER BY ID),
        [Result4] = LAG(Result, 4) OVER(PARTITION BY TeamID ORDER BY ID)
FROM    (Previous Query)

然后，一旦您获得了每支球队每场比赛的前 4 个结果，您就可以将未透视的数据重新组合在一起，以使主队和客队再次位于同一行，从而提供完整的查询：

WITH Results AS
(   SELECT  ID,
            Type,
            TeamID,
            Result
    FROM    T
            CROSS APPLY 
            (   VALUES 
                    ('Home', [Home(no.)], Result_Home), 
                    ('Away', [Away(no.)], Result_Away)
            ) Upvt (Type, TeamID, Result)
), Results2 AS
(   SELECT  *,
            [Result1] = LAG(Result, 1) OVER(PARTITION BY TeamID ORDER BY ID),
            [Result2] = LAG(Result, 2) OVER(PARTITION BY TeamID ORDER BY ID),
            [Result3] = LAG(Result, 3) OVER(PARTITION BY TeamID ORDER BY ID),
            [Result4] = LAG(Result, 4) OVER(PARTITION BY TeamID ORDER BY ID)
    FROM    Results
)
SELECT  Home.ID,
        [Home(no.)] = Home.TeamID ,
        [Away(no.)] = Away.TeamID,
        [Result_Home] = Home.Result,
        [Result_Away] = Away.Result,
        [Home1] = Home.Result1,
        [Home2] = Home.Result2,
        [Home3] = Home.Result3,
        [Home4] = Home.Result4,
        [Away1] = Away.Result1,
        [Away2] = Away.Result2,
        [Away3] = Away.Result3,
        [Away4] = Away.Result4
FROM    Results2 AS Home
        INNER JOIN results2 AS Away
            ON Away.ID = Home.ID
            AND Away.Type = 'Away'
WHERE   Home.Type = 'Home'
ORDER BY ID;

Example on SQL Fiddle

【讨论】：

【解决方案2】：

您似乎正在寻找 ROW_NUMBER() 函数或其他窗口函数

类似

;with cte as 
(
    select id, home as team, result_home as score from yourtable
    union
    select id, away, result_away from yourtable
),
games as
(
    select *, ROW_NUMBER() over (partition by team order by id) as gameday
    from cte
)   
    select  *
    from (select team, score, gameday from games) src
    pivot
    (max(score) for gameday in ([1],[2],[3],[4],[5],[6]))p

如果您知道每支球队的个人比赛比赛日，那么两场比赛前就是本场比赛的比赛日 - 2，依此类推。

【讨论】：