mysql 查询 uid 按time 最新的前3条记录

也就是 查询某个字段的前3条记录

select uid,time from 表名 order by time desc limit 0,3

再按 uid 分组

如下：

SELECT uid,time from table_x as tbx_o 
where 3 > (
    SELECT count(*) from table_x as tbx_i 
    where tbx_o.uid = tbx_i.uid and tbx_o.time < tbx_i.time
) ORDER BY uid;

# 更新的语句
select table_out.uid,sum(table_out.score) from (
	SELECT uid,time,score from table_x as tbx_o 
	where 3 > (
	    SELECT count(*) from table_x as tbx_i 
	    where tbx_o.uid = tbx_i.uid and tbx_o.time < tbx_i.time
	) ORDER BY uid
) as table_out
group by table_out.uid
order by 2 desc;

order by 子句后跟一个 limit n，就是限制前 n 条。

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