SQL-最大ID的总和

Question

我有这样的桌子，

| id | name | subtask | maintask |
|----|------|---------|----------|
| 1  | t1   | 11      | 20       |
| 1  | t1   | 12      | 20       |
| 1  | t1   | 1       | 30       |
| 2  | t1   | 2       | 20       |
| 2  | t1   | 2       | 20       |

我想准备这样的结果

| id | name | sum_of_subtask | sum_of_maintask | diff |
|----|------|----------------|-----------------|------|
| 2  | t1   | 4              | 40              | 36   |

需要选择最大ID，然后对子任务和主任务进行求和，然后最后一列是sum（子任务）与sum（主任务）的差

我在下面的查询中尝试了此方法，但是它计算所有列的总和。

select max(id), name, sum(subtask),sum(maintask),sum(subtask-maintask) from tbl
group by name

Answer 1

您只想要一行吗？如果是这样，请使用order by和limit：

select id, name, sum(subtask), sum(maintask), sum(subtask-maintask)
from tbl
group by id, name
order by id desc
limit 1;

如果数据很大，则过滤[[之前聚合可能会更有效：

select id, name, sum(subtask), sum(maintask), sum(subtask-maintask)
from tbl
where id = (select max(id) from tbl)
group by id, name;

如果要每个id的最大name，则过滤逻辑为：
select id, name, sum(subtask), sum(maintask), sum(subtask-maintask)
from tbl t
where t.id = (select max(t2.id) from tbl t2 where t2.name = t.name)
group by id, name;

Answer 2

请使用以下查询，

select id, name, sum(subtask), sum (maintask), sum(subtask)-sum (maintask)
where id in
(select max(id) from table)
group by id, name;

Answer 3

select id, name, sum(subtask), sum(maintask), sum(subtask-maintask)
from tbl
where id = (select max(id) from tbl)
group by id, name