如何从MySQL获取最新行

Question

我有下面提到的表：

ID      Date                    Value
AL-1    2017-04-01 22:04:08     154
AL-1    2017-04-05 15:08:45     158
AL-1    2017-04-06 18:09:15     225
AL-1    2017-04-08 20:08:17     254
AL-2    2017-04-01 22:04:08     154

我正在尝试： select ID, Value from table1 where id in ('AL-1','AL-2') and group by max(date(Date)); 但是它给我的输出像：

ID      Date                    Value
AL-1    2017-04-01 22:04:08     154
AL-2    2017-04-01 22:04:08     154

所需输出：

ID      Date                    Value
AL-1    2017-04-08 20:08:17     254
AL-2    2017-04-01 22:04:08     154

Answer 1

如果要获取每个ID最新行，则可以使用相关 subquery ：

select t1.*
from table1 t1
where Date = (select max(t2.Date) from tabel1 t2 where t1.ID = t2.ID);

Answer 2

有一个子查询，返回每个ID及其最大日期。 JOIN该结果：

select t1.ID, t1.Value, t1.date
from table1 t1
join (select id, max(Date) maxdate
      from table1
      where id in ('AL-1','AL-2')
      group by id) t2 on t1.ID = t2.ID and date(t1.Date) = date(t2.maxdate)
where t1.id in ('AL-1','AL-2')

Answer 3

使用order by desc有limit

select ID, Value from table1 order by date desc limit 1

Answer 4

尝试如下：

SELECT * FROM table t
WHERE t.date =(SELECT MAX(date) FROM table )

Answer 5

尝试这个：

select ID, Value 
from table1 
where id='AL-1' 
group by date desc 
limit 1

Answer 6

SELECT ID, SUBSTRING_INDEX(GROUP_CONCAT(DATE ORDER BY DATE DESC SEPARATOR '##'),'##',-1) AS DATE, VALUE
FROM table1 GROUP BY ID