是否可以做一个简单的查询来计算我在一个确定的时间段内有多少记录,比如一年,一个月,或者一天,有一个TIMESTAMP字段,比如:
SELECT COUNT(id)
FROM stats
WHERE record_date.YEAR = 2009
GROUP BY record_date.YEAR
甚至:
SELECT COUNT(id)
FROM stats
GROUP BY record_date.YEAR, record_date.MONTH
每月进行统计。
谢谢!
是否可以做一个简单的查询来计算我在一个确定的时间段内有多少记录,比如一年,一个月,或者一天,有一个TIMESTAMP字段,比如:
SELECT COUNT(id)
FROM stats
WHERE record_date.YEAR = 2009
GROUP BY record_date.YEAR
甚至:
SELECT COUNT(id)
FROM stats
GROUP BY record_date.YEAR, record_date.MONTH
每月进行统计。
谢谢!
当前回答
.... group by to_char(date, 'YYYY')——> 1989
.... group by to_char(date,'MM')——>05
.... 3 .用to_char(date,'DD')——>
.... group by to_char(date,'MON')——>
.... 9 . group by to_char(date,'YY')——>
其他回答
GROUP BY DATE_FORMAT(record_date, '%Y%m')
Note (primarily, to potential downvoters). Presently, this may not be as efficient as other suggestions. Still, I leave it as an alternative, and a one, too, that can serve in seeing how faster other solutions are. (For you can't really tell fast from slow until you see the difference.) Also, as time goes on, changes could be made to MySQL's engine with regard to optimisation so as to make this solution, at some (perhaps, not so distant) point in future, to become quite comparable in efficiency with most others.
如果你想在MySQL中按日期分组,那么使用下面的代码:
SELECT COUNT(id)
FROM stats
GROUP BY DAYOFMONTH(record_date)
希望这为那些要找到这个帖子的人节省了一些时间。
我希望每天都能得到类似的数据,经过一些试验,这是我在这个场景中能找到的最快的数据
SELECT COUNT(id)
FROM stats
GROUP BY record_date DIV 1000000;
如果你想每个月有一次,添加额外的零(00) 我不建议从“使代码可读”的角度考虑,它也可能在不同的版本中中断。但在我们的例子中,与我测试的其他更清晰的查询相比,这只花了不到一半的时间。
这是一个MySQL的答案(因为MySQL被标记在问题中),并在手册https://dev.mysql.com/doc/refman/8.0/en/date-and-time-type-conversion.html中有详细的说明
GROUP BY YEAR(record_date), MONTH(record_date)
查看MySQL中的日期和时间函数。
如果你想过滤特定年份(例如2000年)的记录,那么优化WHERE子句,如下所示:
SELECT MONTH(date_column), COUNT(*)
FROM date_table
WHERE date_column >= '2000-01-01' AND date_column < '2001-01-01'
GROUP BY MONTH(date_column)
-- average 0.016 sec.
而不是:
WHERE YEAR(date_column) = 2000
-- average 0.132 sec.
结果是根据一个包含300k行和date列索引的表生成的。
至于GROUP BY子句,我根据上面提到的表测试了三个变体;以下是调查结果:
SELECT YEAR(date_column), MONTH(date_column), COUNT(*)
FROM date_table
GROUP BY YEAR(date_column), MONTH(date_column)
-- codelogic
-- average 0.250 sec.
SELECT YEAR(date_column), MONTH(date_column), COUNT(*)
FROM date_table
GROUP BY DATE_FORMAT(date_column, '%Y%m')
-- Andriy M
-- average 0.468 sec.
SELECT YEAR(date_column), MONTH(date_column), COUNT(*)
FROM date_table
GROUP BY EXTRACT(YEAR_MONTH FROM date_column)
-- fu-chi
-- average 0.203 sec.
最后一个是赢家。