Question

我有一张名为 #TimeAtHome 的表。它包括一个 address、date 和一个标志 atHome，以表明该人当天是否在家。我需要为每个 min 的人不在家 (max) 的每个分组捕获 date 和 0 address。

这是一些示例代码：

create table #TimeAtHome (
    [address] varchar(100),
    [date] date,
    [atHome] bit
)

insert into #TimeAtHome
values ('123 ABC Street', '2020-01-01', '1'),
       ('123 ABC Street', '2020-01-02', '1'),
       ('123 ABC Street', '2020-01-03', '0'),
       ('123 ABC Street', '2020-01-04', '0'),
       ('123 ABC Street', '2020-01-05', '0'),
       ('123 ABC Street', '2020-01-06', '0'),
       ('123 ABC Street', '2020-01-07', '1'),
       ('123 ABC Street', '2020-01-08', '0'),
       ('123 ABC Street', '2020-01-09', '0'),
       ('123 ABC Street', '2020-01-10', '1'),
       ('777 Hello Ct', '2020-01-01', '1'),
       ('777 Hello Ct', '2020-01-02', '1'),
       ('777 Hello Ct', '2020-01-03', '1'),
       ('777 Hello Ct', '2020-01-04', '0'),
       ('777 Hello Ct', '2020-01-05', '1'),
       ('777 Hello Ct', '2020-01-06', '1')

这是我想要的结果：

Answer 1

我想我使用了一个更简单的解决方案，因为这似乎是一个间隙和岛屿问题。因此，我使用 LAG() 函数根据 AtHome 标志查找岛屿的起点和终点。然后我使用 SUM() 函数创建一个组并从那里聚合日期：

SELECT Address,Min(Date) minDate, Max(date) maxDate
FROM
(
    SELECT *, SUM(CASE WHEN AtHome <> PrevAtHome THEN 1 ELSE 0 END) OVER(PARTITION BY Address order by date) Grp
    FROM(
      SELECT *, LAG(ATHome,1,AtHome) OVER(PARTITION BY address order by date) PrevAtHome
      from #TimeAtHome
      ) T
) Final
WHERE Athome = 0
GROUP BY Address,Grp
ORDER BY Address

Answer 2

我们可以尝试以下方法：

获取与自身连接的所有 minDate 值，并检查当前日期该人是否在家而不是在下一个日期不在家（将是子查询 1）。
以与第 1 点相同的方式获取所有 maxDate，除了检查此人是否在下一个日期回来（子查询 2）。
匹配每个地址的第一个 minDate 与第一个 maxDate，第二个 minDate 与第二个 maxDate，依此类推（加入子查询 1 和 2）。

SELECT q1.address,
    q1.minDate,
    q2.maxDate
FROM (
        SELECT ROW_NUMBER() OVER(
                PARTITION BY t2.address
                ORDER BY t2.date
            ) as row,
            t2.address,
            t2.date as minDate
        FROM #TimeAtHome t1 inner join #TimeAtHome t2 ON t1.address = t2.address and t1.date = DATEADD(DAY, -1, t2.date)
        WHERE t1.atHome = 1
            AND t2.atHome = 0
    ) q1
    INNER JOIN (
        SELECT ROW_NUMBER() OVER(
                PARTITION BY t1.address
                ORDER BY t1.date
            ) as row,
            t1.address,
            t1.date as maxDate
        FROM #TimeAtHome t1 INNER JOIN #TimeAtHome t2 ON t1.address = t2.address and t1.date = DATEADD(DAY, -1, t2.date)
        WHERE t1.atHome = 0
            AND t2.atHome = 1
    ) q2 ON q1.address = q2.address
    AND q1.row = q2.row

请注意此查询的约束

表中的日期应该是连续的，因此为了找到表中的下一条记录，我们只需减去一天 t1.date = DATEADD(DAY, -1, t2.date)。
这个人从家里开始，所以他出去时的第一个 minDate 与他回来时的第一个 maxDate 相匹配。

Answer 3

查询是这样的，Cte1用于获取下一步要使用的数据的完整视图。 Cte2用于查找mindate，Cte3用于获取maxDate，Rank func用于最后加入

;WITH cte1
AS
(
    SELECT *, 
        LEAD(date) OVER (PARTITION BY address ORDER BY date) AS nextDate, 
        LEAD(atHome) OVER (PARTITION BY address ORDER  BY date) AS NextAtHome 
    FROM #TimeAtHome
    --ORDER BY address, date
),
CTE2 AS
(
    SELECT 
        address, 
        cte1.nextDate AS minDate,
        ROW_NUMBER() OVER (ORDER BY cte1.address , cte1.date) AS R1
    FROM cte1 
    WHERE cte1.atHome = 1 AND cte1.NextAtHome = 0
),
CTE3 AS
(
    SELECT 
        address, 
        date AS maxDate,
        ROW_NUMBER() OVER (ORDER BY cte1.address, cte1.date) AS R2
    FROM cte1 
    WHERE cte1.atHome = 0 AND cte1.NextAtHome = 1
)
SELECT CTE2.address,CTE2.minDate,CTE3.maxDate
FROM cte2
INNER JOIN cte3 ON cte2.R1 = Cte3.R2

Answer 4

另一种可能性：

create table #TimeAtHome (
    [address] varchar(100),
    [date] date,
    [atHome] bit
)

insert into #TimeAtHome
values ('123 ABC Street', '2020-01-01', '1'),
       ('123 ABC Street', '2020-01-02', '1'),
       ('123 ABC Street', '2020-01-03', '0'),
       ('123 ABC Street', '2020-01-04', '0'),
       ('123 ABC Street', '2020-01-05', '0'),
       ('123 ABC Street', '2020-01-06', '0'),
       ('123 ABC Street', '2020-01-07', '1'),
       ('123 ABC Street', '2020-01-08', '0'),
       ('123 ABC Street', '2020-01-09', '0'),
       ('123 ABC Street', '2020-01-10', '1'),
       ('777 Hello Ct', '2020-01-01', '1'),
       ('777 Hello Ct', '2020-01-02', '1'),
       ('777 Hello Ct', '2020-01-03', '1'),
       ('777 Hello Ct', '2020-01-04', '0'),
       ('777 Hello Ct', '2020-01-05', '1'),
       ('777 Hello Ct', '2020-01-06', '1')
   




SELECT dt.address,
       MIN(dt.Dt) AS minDate,
       MAX(dt.Dt) AS maxDate
FROM (
        SELECT address, 
               t.Date AS Dt,
               DATEDIFF(D, ROW_NUMBER() OVER(partition by t.address ORDER BY t.Date), 
t.Date) AS DtRange
        FROM #TimeAtHome t
        WHERE t.atHome = 0
    ) AS dt
GROUP BY dt.address, dt.DtRange
ORDER BY address, minDate;

Answer 5

这是另一种方式

    SELECT
    *
FROM 
(
    SELECT
        *
        ,ROW_NUMBER() OVER(PARTITION BY address order by date) PrevAtHome_A
        ,ROW_NUMBER() OVER(PARTITION BY address order by date DESC) PrevAtHome_D
    from #TimeAtHome
    WHERE AtHome = 0
)A
WHERE PrevAtHome_A =1 OR PrevAtHome_D =1
ORDER BY [address], [date]

T-SQL：选择最小值和最大值的每个实例

5 个答案: