每天选择10行 [英] select 10 rows per day with order

问题描述

我有一个数据库，其中记录了日期(时间戳) 我每天需要选择10条记录(每天还有更多记录) 并按几列对其进行排序...

i have a db with records with date (timestamp) i need to select 10 records for each day (there are many more per day) and order them by few columns...

该查询应如何显示?

推荐答案

您每天必须在子查询中每天获取10条记录，并通过左连接将它们连接到主表中，因此您将获得最大值每天10条记录. SQL看起来像这样:

You have to get your 10 records per day in a subquery for each day and join them to the main table by a left join, so you'll get max 10 records per day. The SQL would look like this:

SELECT t1.columns
FROM mytable t1 
  LEFT JOIN 
     (SELECT pk FROM mytable t2 
     WHERE t2.datecol = t1.datecol 
     ORDER BY t2.orderFor10Rows LIMIT 10) t3
  ON t1.pk = t3.pk
ORDER BY t1.anyOtherColumns

我不习惯正确的MySQL语法，因此不予保证.

No warranty for proper MySQL-syntax as I'm not used to it.

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