Mysql 从最接近选定日期的数据库中获取唯一数据

我想得到最接近某个特定日期的数据，无论是在过去还是未来，以更接近该日期的为准。但我也需要它的类型不同。例如：

id |type |data | date                
 1 |0    |1904 |2018-08-19 00:14:32
 2 |0    |1904 |2018-08-19 00:24:47
 3 |0    |1904 |2018-08-19 05:02:30
 4 |0    |1904 |2018-08-22 00:05:58
 5 |0    |1904 |2018-08-22 00:08:34
 6 |4    |1903 |2018-08-19 00:14:31
 7 |6    |1926 |2018-08-19 00:14:32
 8 |6    |1926 |2018-08-19 04:44:10
 9 |6    |1926 |2018-08-19 04:52:58
10 |6    |1926 |2018-08-19 04:59:23
11 |6    |1926 |2018-08-22 00:05:58
12 |6    |1926 |2018-08-22 00:07:52
13 |6    |1926 |2018-08-22 00:08:34
14 |7    |1564 |2018-08-19 00:14:32
15 |8    |1900 |2018-08-19 00:14:32 

如果我指定日期时间：2018-08-19 00:00:00。我应该得到以下结果：

id | type
 1 | 0    
 6 | 4
 7 | 6
14 | 7
15 | 8

或者如果我指定日期时间：2018-09-03 00:00:00。我期望：

id | type
 5 | 0    
 6 | 4
13 | 6
14 | 7
15 | 8

或者最后：2018-08-22 00:05:58

我尝试了一些毫无意义的问题。。。比如：

SELECT *
FROM history 
WHERE (ABS(TIMESTAMPDIFF(DAY, date, "2018-08-19 00:08:34")) < 4) AND user_id = 1299
GROUP BY type
ORDER BY date

不管怎么说，仅使用MySQL查询就可以做到这一点吗？我需要使用一些应用程序逻辑来实现这一点吗？我如何质疑这一点

---

您可以在子查询上使用内部联接来实现最小差异

select 
    h.id, h.type 
from 
    history  
inner join  
    (select
         type, min(TIMESTAMPDIFF(SECOND, date, '2012-06-06 15:20:18')) min_diff 
     from  
         history
     where
         (abs(TIMESTAMPDIFF(DAY, date, "2018-08-19 00:08:34")) < 4) 
         and user_id = 1299
     group by  
         type) t on t.type = h.type 
                 and TIMESTAMPDIFF(SECOND, h.date, '2012-06-06 15:20:18') = t.min_diff

---

这采用了一种标准的查询模式来查找每种类型都具有极值的行。在这种情况下，“极限”表示最接近目标值

首先，需要一个聚合子查询来获取极值：每种类型的行与目标日期之间的最小增量时间

然后将其连接到原始表，使用ON子句匹配类型和增量时间。把它们放在一起

编辑仅获取日期最近的未来或过去项目更容易。最近过去的子查询为

         SELECT type, MAX(date) date
           FROM history
         WHERE date <= @target
         GROUP BY date

然后整个查询是

  SELECT history.*
    FROM history
    JOIN (
             SELECT type, MAX(date) date
               FROM history
             WHERE date <= @target
             GROUP BY date
         ) m ON history.type = m.type AND history.date = m.date 

---

通过abstimestampdiffsecond，date，@dt:

---

这应该能奏效

create table test2 as
select c.id,c.type from(
select a.*,b.min
from test as a
left join
(select id, min(abs(date-"2018-09-01 00:00:00"dt)) as min from test
group by type ) as b
on a.id = b.id
) as c
where min = abs(date-"2018-09-01 00:00:00"dt)
order by id;

---

我们将id上的join返回到原始表min abs datediff，然后选择所需的记录。

它没有给出正确的结果。它似乎只提供最新的数据。如果我需要添加额外的过滤。我可以这样做：选择历史。*从历史中加入选择类型，MINABSTIMESTAMPDIFFMINUTE，date，@target delta FROM history，其中user_id=1299按历史上的delta类型分组。type=delta.type和ABSTIMESTAMPDIFFMINUTE，date，@target=delta其中user_id=1299是否也可以只获取最近的未来或过去？不是两者都有。@HABO我没有用tsql标记它。

         SELECT type, MAX(date) date
           FROM history
         WHERE date <= @target
         GROUP BY date

  SELECT history.*
    FROM history
    JOIN (
             SELECT type, MAX(date) date
               FROM history
             WHERE date <= @target
             GROUP BY date
         ) m ON history.type = m.type AND history.date = m.date 

select @rn := 0, @dt := cast('2018-08-19 04:52:00' as datetime);
select id, type from (
    select id,
           type,
           data,
           date,
           @rn := @rn + 1 rn
    from history
    order by abs(timestampdiff(second, date, @dt))
) a where rn = 1;

create table test2 as
select c.id,c.type from(
select a.*,b.min
from test as a
left join
(select id, min(abs(date-"2018-09-01 00:00:00"dt)) as min from test
group by type ) as b
on a.id = b.id
) as c
where min = abs(date-"2018-09-01 00:00:00"dt)
order by id;