Question

这是我的表记录

field_date  field_time  field_id 
20161202    12:56:00    309073 
20161202    12:57:00    208901 
20161202    12:58:00    208901 
20161202    13:15:00    208901

具有相同的ID（208901）我想删除时间间隔低于15分钟的ID，例如那里，这是我想要的结果

field_date  field_time  field_id 
20161202    12:56:00    309073 
20161202    12:57:00    208901 
20161202    13:15:00    208901

id与12:58:00时间被删除，因为它与id第一条记录只有1分钟的间隔，但13:15:00没有被删除，因为间隔距离第一条记录超过15分钟。

这是我的代码，它仍然显示12:58:00记录，我想删除它所以我只会显示间隔少于15分钟的记录，此代码只显示所有记录（不是我想要的重复记录，那里有什么问题？

SELECT *
FROM tbl_name WHERE EXISTS (SELECT 1 FROM tbl_name t2 
        WHERE(t2.field_id = tbl_name.field_id AND 
        DATEDIFF(MINUTE, t2.field_time, tbl_name.field_time) <= 15) 
        )
        ORDER BY field_time

Answer 1

这个（对于SQL Server 2012 +）：

DECLARE @DataSource TABLE
(
    [field_date] INT
   ,[field_time] VARCHAR(8)
   ,[field_id] INT
);

INSERT INTO @DataSource ([field_date], [field_time], [field_id])
VALUES (20161202, '12:56:00', 309073 )
      ,(20161202, '12:57:00', 208901)
      ,(20161202, '12:58:00', 208901)
      ,(20161202, '13:15:00', 208901);

WITH DataSource ([field_date], [field_time], [field_id], [timeDiff]) AS
(
    SELECT *
         ,DATEDIFF(MINUTE, MIN([field_time]) OVER(PARTITION BY [field_date], [field_id] ORDER BY [field_time] ASC), [field_time])
    FROM @DataSource
)
SELECT *
FROM DataSource
WHERE [timeDiff] = 0
    OR [timeDiff] > 15
ORDER BY field_time;

Answer 2

DECLARE @Data TABLE ([field_date] int, [field_time] varchar(8), [field_id] int);

INSERT INTO @Data
VALUES
    (20161202, '12:56:00', 309073),
    (20161202, '12:57:00', 208901),
    (20161202, '12:58:00', 208901),
    (20161202, '13:15:00', 208901)
;


;WITH data AS 
(
    SELECT 
        *, 
        -- A field easier to use for comparison, e.g. cross date
        CONVERT(datetime, CAST(field_date AS char(8)), 112) + CONVERT(time, field_time) AS combined_time 
    FROM @Data
)
DELETE curr 
FROM 
    data curr 
    OUTER APPLY 
    (
        SELECT TOP 1 * 
        FROM data prev 
        WHERE 
            prev.field_id = curr.field_id 
            -- assume the record does not contain duplicate record of field_time
            AND prev.combined_time < curr.combined_time 
        ORDER BY combined_time DESC
    ) AS prev
WHERE DATEDIFF(MINUTE, prev.combined_time, curr.combined_time) <= 15

-- Result
SELECT * FROM @Data

Answer 3

尝试如下：

0xff

在SQL Server中删除具有特定时间的记录

3 个答案: