了解GroupBY

Question

我在SQL Server中有一个表，其中记录了个人输入和输出。 同一天有多个小时的进入或退出。我需要的是恢复当天的第一个入口和最后一个出口。

Date       hour   Clock
------------------------
01/01/2017 09:00  1
01/01/2017 11:30  2
01/01/2017 17:00  2
02/01/2017  7:59  1
02/01/2017 16:00  1

我有这个SQL查询可以正常工作。

SELECT  
    d.Date,
    MIN(d.hour) as Entry, 
    MAX(dt.hour) as Exit   
FROM 
    #temp1 AS d
LEFT JOIN 
    #temp1 AS dt ON d.Date = dt.Date
GROUP BY 
    d.Date
ORDER BY
    Date DESC

但如果我在查询中再添加2列

SELECT  
    d.Date,
    d.clock as ClockEntry,       -- Aggregated column to display
    MIN(d.hour) as Entry, 
    dt.clock as ClockExit,       -- Aggregated column to display
    MAX(dt.hour) as Exit
FROM 
    #temp1 AS d
LEFT JOIN 
    #temp1 AS dt ON d.Date = dt.Date
GROUP BY   
    d.Date
ORDER BY
    Date DESC

我收到此错误：

  

专栏＆＃39;＃temp1.clock＆＃39;在选择列表中无效，因为它不包含在聚合函数或GROUP BY子句中。

我只需按字段分组＆＃34;日期＆＃34;，我不想为GROUP BY添加更多条件..我怎么能解决它？

我想要这个结果

DATE        ClockEntry     Entry    ClockExit    Exit     
-------------------------------------------------------
01/01/2017     1           09:00        2        17:00
02/01/2017     1            7:59        1        16:00

Answer 1

所以有一种简单的方法可以做到这一点 - 使用2个排名功能：

1. 按日期分区，按小时升序排序
2. 按日期分区，按小时降序排序

此时，可以将行连接在一起，它们的值均为1，并且日期匹配。

我倾向于使用CTE：

with temp2 (MinId, MaxId, Date, Hour, Clock)
AS
(
    select ROW_NUMBER() Over (partition by date order by hour), 
     ROW_NUMBER() Over (partition by date order by hour desc), 
    *
    from temp1
)
select distinct 
       d1.Date,
       d1.Clock,
       d1.Hour,
       d2.Clock,
       d2.Hour
FROM temp2 d1
LEFT JOIN temp2 d2 
    ON d1.Date = d2.Date  -- dates match
    AND d1.MinId=d2.MaxId -- minId=earliest record MaxId=latest record
WHERE d1.MinId=1

Answer 2

GROUP BY方法

如果您知道只有一个值会出现clock列， 只需使用MAX或MIN聚合函数汇总值，例如：

SELECT  
d.Date,
MIN(d.clock) as ClockEntry,
MIN(d.hour) as Entry, 
MAX(dt.clock) as ClockExit,
MAX(dt.hour) as Exit

FROM #temp1 AS d
LEFT JOIN #temp1 AS dt 
ON d.Date= dt.Date
GROUP BY   d.Date
order by Date desc

或者，如果您有多个时钟值并希望全部看到它们， 将它们添加到GROUP BY语句中：

SELECT  
d.Date,
d.clock as ClockEntry,
MIN(d.hour) as Entry, 
dt.clock as ClockExit,
MAX(dt.hour) as Exit

FROM #temp1 AS d
LEFT JOIN #temp1 AS dt 
ON d.Date= dt.Date
GROUP BY   d.Date, d.clock, dt.clock
order by Date desc

ORDER BY方法

每个日期使用cursor或Common Table Expression来获得第一个条目和最后一个条目。

指定日期的首次参赛

SELECT TOP 1 d.date, d.clock as ClockEntry, d.hour as Entry
FROM #temp1 AS d
WHERE d.date = @myDate
ORDER BY d.hour ASC

指定日期的最后一次退出

SELECT TOP 1 d.date, d.clock as ClockExit, d.hour as Exit
FROM #temp1 AS d
WHERE d.date = @myDate
ORDER BY d.hour DESC

参考：

GROUP BY Documentation
AGGREGATE FUNCTIONS