SQL 查询提取给定时间范围内的最大值？

Question

我在 sql 查询中遇到问题。我需要获取最近 30 天的数据，然后仅删除 "voting" 值大于

的 4 个结果

数据库结构

Id |    time    | voting
1  | unix time  | 3
2  | unix time  | 2
3  | unix time  | 4
4  | unix time  | 1
5  | unix time  | 6

我只想带我的数据：5-3-1-2

我试过

select a.* 
from table a 
inner join 
    ( select votingng, max(time) as latest from table group by voting) v 
        on a.time = v.latest 
        and a.voting = v.voting 
order by time desc limit

Answer 1

我想这就是你想要的：

select v.*
from voting v
where timestamp >= unix_timestamp(date_sub(curdate(), interval 1 month)
order by voting desc
limit 4;

Answer 2

听起来您正在尝试获得过去 30 天内的前 4 名投票结果。这些是否都能满足您的需求？

SELECT a.*
  FROM table a
 WHERE a.time > UNIX_TIMESTAMP(DATE_SUB(CURDATE(), INTERVAL 30 DAY))
 ORDER BY a.voting DESC
 LIMIT 4;

或

SELECT a.*
  FROM table a
 WHERE DATEDIFF( NOW(), FROM_UNIXTIME(a.time) ) <= 30 
 ORDER BY a.voting DESC
 LIMIT 4;

Answer 3

希望这就是您要找的：

Select *
From
    Voting
Where
    time Between CURDATE() And DATE_SUB(CURDATE(),INTERVAL 30 DAY)
Order By voting Desc 
Limit 4

Answer 4

如果您使用的是 teradata，请试试这个

SELECT * from table
qualify row_number () over(order by time desc)=1 ;

或

select * from 
(select table.*, row_number () over(order by time desc) as RANK from table)
where RANK=1

SQL 查询提取给定时间范围内的最大值？

SQL query for extract the highest value at a given time range?

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