不是mySQL，但对于其他发现此问题并使用SQL的人，另一种解决最大的每组问题的方法是在MS SQL中使用交叉应用

WITH DocIds AS (SELECT DISTINCT id FROM docs)

SELECT d2.id, d2.rev, d2.content
FROM DocIds d1
CROSS APPLY (
  SELECT Top 1 * FROM docs d
  WHERE d.id = d1.id
  ORDER BY rev DESC
) d2

下面是SqlFiddle中的一个示例

这是一个很好的方法

使用以下代码：

with temp as  ( 
select count(field1) as summ , field1
from table_name
group by field1 )
select * from temp where summ = (select max(summ) from temp)

SELECT *
FROM Employee
where Employee.Salary in (select max(salary) from Employee group by Employe_id)
ORDER BY Employee.Salary

如果select语句中有多个字段，并且希望通过优化代码获得所有这些字段的最新值：

select * from
(select * from table_name
order by id,rev desc) temp
group by id 

这是另一个解决方案，希望它能帮助某人

Select a.id , a.rev, a.content from Table1 a
inner join 
(SELECT id, max(rev) rev FROM Table1 GROUP BY id) x on x.id =a.id and x.rev =a.rev

像这样吗？

SELECT yourtable.id, rev, content
FROM yourtable
INNER JOIN (
    SELECT id, max(rev) as maxrev
    FROM yourtable
    GROUP BY id
) AS child ON (yourtable.id = child.id) AND (yourtable.rev = maxrev)