在mysql查询中查找与指定日期时间最接近的日期时间

Question

我试图在mysql数据库中找到一个datetime值,它与我指定的日期时间最接近,我遇到了一些麻烦.

以下伪代码是我想要实现的:

SELECT one FROM table WHERE datetimefield is closest to "2014-12-10 09:45:00" LIMIT 1

提前致谢

编辑

感谢到目前为止的回复,因为事实证明查询比我最初想要达到预期结果要复杂得多,ABS方法在添加一些功能时效果很好.

接下来的问题是,无论如何都要降低对以下查询的性能要求？

SELECT DISTINCT timegenerated, *other values*  
FROM table1 e INNER JOIN table2 dt
ON e.circuit = dt.circuit 
WHERE dt.circuit IN ("2", "3", "4", "5", "6", "7", "8") 
AND e.circuit != 1
AND dt.siteid = 435 
ORDER BY ABS(TIMESTAMPDIFF(MINUTE, timegenerated, "2014-12-09 14:15:00")) LIMIT 7

或者将其合并以匹配多个日期,因为我需要能够匹配多个特定日期时间(可能多达90天)它当前正在从另一个阵列运行每个单独的日期时间但我意识到这不是最佳的.该查询是在PHP中动态构建的.

查询时间目前约为每个查询1.1秒,因此运行30次或更多次会产生问题,该表有数十万行.

非常感谢!

Answer 1

关键的想法是使用order by和limit:

如果你想要最近的一个:

SELECT one
FROM table
WHERE datetimefield <= '2014-12-10 09:45:00'
ORDER BY datetimefield DESC
LIMIT 1;

如果你想要最接近的方向,那么使用TIMESTAMPDIFF():

ORDER BY abs(TIMESTAMPDIFF(second, datetimefield, '2014-12-10 09:45:00'))
LIMIT 1

Answer 2

使用 abs() 可以防止使用日期时间字段索引。我建议为最接近的之前选择一个，为最接近的之后选择一个，两者都使用索引，并在之后选择最接近的：

create table `table` (datetimefield datetime key, one varchar(99));
insert into `table` values
  ('2014-06-01', 'a'), ('2014-12-01', 'b'),
  ('2015-01-01', 'c'), ('2015-02-01', 'd');

set @d = '2014-12-10 09:45:00';

select * from
(
  ( select *, TIMESTAMPDIFF(SECOND, @d, datetimefield) as diff
    from `table` where datetimefield >= @d
    order by datetimefield asc  limit 1
  )
  union
  ( select *, TIMESTAMPDIFF(SECOND, datetimefield, @d) as diff
    from `table` where datetimefield < @d
    order by datetimefield desc limit 1
  )
) x
order by diff
limit 1;

http://sqlfiddle.com/#!2/bddb4/1