得用子查询了

select id, category, max(value) from t group by category;
max(value) 查出来不用就行了

这样查不出 id 吧

SELECT
category,
id
FROM
test_table
WHERE
_value IN (
SELECT
max( _value )
FROM
test_table
GROUP BY
category)

select * from (select id, category, value from t order by category, value desc) b group by b.category;

SELECT
    a.id,
    a.category 
FROM
    t a
    INNER JOIN ( SELECT category, MAX( `value` ) AS 'value' FROM t GROUP BY category ) AS b ON a.category = b.category 
    AND a.`value` = b.`value`

有什么方法可以不用到 group by 语句吗

可以用 row_number() over(partition by category order by value desc) 结合子查询

嗯，查不出

select id from category
where concat(value,category) in
(select concat(max(value),category) from table a
group by category)

如果不同的 category 的 max（value）是相同的怎么办

SELECT
    P.id,
    P.`value`,
    P.category 
FROM
    table_p AS P
    INNER JOIN ( SELECT MAX(`value`) AS M, category FROM table_p GROUP BY category ) AS P1 
WHERE
    P1.M = P.`value` 
    AND P1.category = P.category 
GROUP BY
    P.category

如果最大值有多个需要同时保留就不要最后的 group by

我在 LeetCode 刷到过，原题及题解帮你找到了。
https://leetcode-cn.com/problems/department-highest-salary/solution/bu-men-gong-zi-zui-gao-de-yuan-gong-by-leetcode/

===> 什么数据库呢？
如果 mysql ， Tester_谜城 的完全正确。
如果其他数据库 (如 oracle,sqlserver) ,要求 value 最大有几个并且只保留第一个的话要复杂些。

是 MySQL，如果是其他数据库会有什么不同

可以妙用 exists，select d.name Department,e.name Employee,e.Salary from Employee e join Department d on e.DepartmentId = d.Id where not exists(select * from Employee c where e.DepartmentId = c.DepartmentId and c.Salary > e.Salary) https://leetcode-cn.com/problems/department-highest-salary/

666

SELECT category,id
FROM table
WHERE value IN (
SELECT max( value )
FROM table
GROUP BY category)

mysql 中相关子查询可以搞定，不需要分组

select * from (select * from categorys order by value desc) c group by category;
或者
select distinct(category), value from (select * from categorys order by value desc) c;