Question

需要一些帮助：）

所以我有一个包含以下列的记录表：

Key (PK, FK, int) DT (smalldatetime) Value (real)

DT是一天中每半小时的日期时间，具有相关值

E.g。

Key       DT                       VALUE
1000      2010-01-01 08:00:00      80
1000      2010-01-01 08:30:00      75
1000      2010-01-01 09:00:00      100

我有一个查询，它每24小时找到一个最大值及其相关时间，但是，在一天最大值出现两次，因此重复导致处理问题的日期。我尝试过使用rownumber（），但我不能在where子句中使用计算列？目前我有：

SELECT       cast(T1.DT as date) as 'Date',Cast(T1.DT as time(0)) as 'HH', ROW_NUMBER() over (PARTITION BY  cast(DT as date) ORDER BY DT) AS 'RowNumber'
FROM        TABLE_1 AS T1
INNER JOIN  (
                SELECT CAST([DT] as date) as 'DATE'
                ,       MAX([VALUE]) as 'MAX_HH'
                FROM    TABLE_1
                WHERE   DT > '6-nov-2016' and [KEY] = '1000'
                GROUP BY CAST([DT] as date)
            ) AS MAX_DT
        ON  MAX_DT.[DATE] = CAST(T1.[DT] as date)
        AND T1.VALUE = MAX_DT.MAX_HH
WHERE       DT > '6-nov-2016' and [KEY] = '1000'
ORDER BY DT

这导致

Key       DT               VALUE       HH
1000      2010-01-01       80          07:00:00
1000      2010-02-01       100         17:30:00
1000      2010-02-01       100         18:00:00

我需要删除重复的日期（我没有偏好HH它）

我想我已经解释得非常糟糕，让我知道它是否毫无意义，我会尝试重写

有什么想法吗？

Answer 1

你可以尝试这个新代码是** **：

 SELECT       cast(T1.DT as date) as 'Date', ** MIN(Cast(T1.DT as time(0))) as 'HH' **
    FROM        TABLE_1 AS T1
    INNER JOIN  (
                    SELECT CAST([DT] as date) as 'DATE'
                    ,       MAX([VALUE]) as 'MAX_HH'
                FROM    TABLE_1
                WHERE   DT > '6-nov-2016' and [KEY] = '1000'
                GROUP BY CAST([DT] as date)
            ) AS MAX_DT
        ON  MAX_DT.[DATE] = CAST(T1.[DT] as date)
        AND T1.VALUE = MAX_DT.MAX_HH
WHERE       DT > '6-nov-2016' and [KEY] = '1000'

这里把小组放在

GROUP BY cast(T1.DT as date)
ORDER BY DT

Answer 2

我会做这样的事情我没试过，但我认为这是正确的。

SELECT  cast(T1.DT as date) as 'Date',Cast(T1.DT as time(0)) as 'HH', VALUE 
FROM TABLE_1 T1      
       WHERE [DT] IN (       
       --select the max date from Table_1 for each day
            SELECT MAX([DT]) max_date FROM TABLE_1
            WHERE (CAST([DT] as date) ,value) IN 
            (
             SELECT CAST([DT] as date) as 'CAST_DATE'
              ,MAX([VALUE]) as 'MAX_HH'
              FROM    TABLE_1
              WHERE   DT > '6-nov-2016' and [KEY] = '1000'
             GROUP BY CAST([DT] as date
            )group by [DT]
           )
 WHERE       DT > '6-nov-2016' and [KEY] = '1000'

Answer 3

将JOIN更改为APPLY。

APPLY操作允许您将连接关系限制为每个源关系只有一个结果。

SELECT v.[Key], cast(v.DT As Date) as "Date", v.[Value], cast(v.DT as Time(0)) as "HH"
FROM
(   -- First a projection to get just the exact dates you want
    SELECT DISTINCT [Key], CAST(DT as DATE) as DT 
    FROM Table_1 
    WHERE [Key] = '1000' AMD DT > '20161106'
) dates
CROSS APPLY (
    -- Then use APPLY rather than JOIN to find just the exact one record you need for each date
    SELECT TOP 1 * 
    FROM Table_1 
    WHERE [Key] = dates.[Key] AND cast(DT as DATE) = dates.DT ORDER BY [Value] DESC
) v

最后一点：此查询和问题中的示例查询都将包含2016年11月6日的值。查询显示> 2016-11-05具有独特的不等式，但原始仍在使用完整的DateTime值进行比较，意味着隐含的0作为时间成分。因此，11月6日的12:01 AM仍然比11月6日的12:00:00.001 AM还要大。如果您想要从查询中排除所有11月6日的日期，您需要更改此项以在结尾处使用时间值进行>比较之前的日期或演员

Answer 4

使用SQL，您可以使用SELECT DISTINCT，

SELECT DISTINCT语句用于仅返回不同的（不同的）值。

在表格中，列通常包含许多重复值;有时您只想列出不同的（不同的）值。

SELECT DISTINCT语句用于仅返回不同的（不同的）值。

忽略重复记录SQL

4 个答案: