作为初学者：如果您运行的是mysql 8.0，那么您可以通过窗口函数轻松地完成这项工作：

select 
    date,
    ant,
    sum(num_habit) as hab,
    sum(sum(num_habit)) over(order by date) cumulative_hab,
    sum(sum(num_habit)) over(order by date desc) reverse_cumulative_hab
from mytable
where date between current_date - interval 5 day and current_date
group by date, ant
order by date

在早期版本中，它更为复杂。我建议加入两个问题：

select t.*, r.reverse_cumulative_hab
from (
    select t.*, @csum := @csum + hab cumulative_hab
    from (
        select date, ant, sum(num_habit) as hab
        from mytable
        where date between current_date - interval 5 day and current_date
        group by date, ant
        order by date
    ) t
    cross join (select @csum := 0) x
) t
inner join (
    select t.*, @rcsum := @rcsum + hab reverse_cumulative_hab
    from (
        select date, ant, sum(num_habit) as hab
        from mytable
        where date between current_date - interval 5 day and current_date
        group by date, ant
        order by date desc
    ) t
    cross join (select @rcsum := 0) x
) r on r.date = t.date
order by t.date

这假设没有重复 ant 每 date .
也可以简化逻辑，通过计算累积和与总和之间的差值来计算逆和：

select t.*, z.total_hab - t.cumulative_hab reverse_cumulative_hab
from (
    select t.*, @csum := @csum + hab cumulative_hab
    from (
        select date, ant, sum(num_habit) as hab
        from mytable
        where date between current_date - interval 5 day and current_date
        group by date, ant
        order by date
    ) t
    cross join (select @csum := 0) x
) t
cross join (
    select sum(num_habit) as total_hab
    from mytable
    where date between current_date - interval 5 day and current_date
) z
order by date

注意，在行的排序方面，这些查询比原始代码更安全：在计算累计和之前，记录在子查询中排序。