减去不同表的两列

Question

我有两个不相关的表：

contribution(id,amount, create_at, user_id)

solicitude(id, amount, create_at, status_id, type_id, user_id)

我需要从用户中减去贡献和求知的总和，但是结果不能为负。

我怎样才能做到这一点？ 功能还是查询？
我试过这个查询：

SELECT  sum(contribution.amount)
- (SELECT sum(solicitude.amount) 
   FROM solicitude 
   WHERE user_id = 1 AND status_id = 1) as total
FROM contribution 
WHERE contribution.user_id = 1

Answer 1

您可以添加外部查询以检查总值：

SELECT CASE WHEN total > 0 THEN total ELSE 0 END AS total
FROM (
    SELECT 
          sum(contribution.amount) - (SELECT sum(solicitude.amount) 
          FROM solicitude 
          WHERE user_id = 1 AND status_id = 1) as total
       FROM contribution 
       WHERE 
       contribution .user_id = 1
    ) alias;

这个解决方案还可以，但是我建议您使用另一种方法。 检查此查询的工作方式：

with contribution as (
    select user_id, sum(amount) as amount from contribution
    group by 1),
solicitude as (
    select user_id, sum(amount) as amount from solicitude
    where status_id = 1
    group by 1)
select 
    c.user_id, c.amount as contribution, s.amount as solitude,
    case when c.amount > s.amount then c.amount - s.amount else 0 end as total
from contribution c
join solicitude s on c.user_id = s.user_id;

---

出于好奇，我对此设置进行了一个简单的测试：

create table public.solicitude (
    id integer,
    amount numeric,
    create_at timestamp without time zone,
    status_id integer,
    type_id integer,
    user_id integer
);

create table public.contribution (
    id integer,
    amount numeric,
    create_at timestamp without time zone,
    user_id integer
);

insert into contribution (user_id, amount)
select (random()* 50)::int, (random()* 100)::int
from generate_series(1, 4000000);

insert into solicitude (user_id, amount, status_id)
select (random()* 50)::int, (random()* 100)::int, 1
from generate_series(1, 4000000);

结果（毫秒）：

Erwin's solution with greatest():  922, 905, 922, 904, 904, 904, 905, 912, 905, 922
My solution with an outer query:   796, 795, 814, 814, 815, 795, 815, 796, 815, 796

Answer 2

我解释您的意见， but that result can't to be negative因为要求返回0而不是负数结果。 简单的解决方案是GREATEST() ：

SELECT GREATEST(sum(amount)
      - (SELECT sum(amount)
         FROM   solicitude 
         WHERE  status_id = 1
         AND    user_id = 1), 0) AS total
FROM   contribution
WHERE  user_id = 1;

否则，我保留您的原始查询，这很好。

对于可能无法返回任何行的其他情况，我将替换为两个子选择。 但是，即使根本找不到给定的user_id ，使用聚合函数也可以保证结果行。 相比：

• 获取n个分组类别并将其他类别合计为一个

如果减法的结果为NULL （因为未找到行或总和为NULL ），则GREATEST()也将返回0 。

Answer 3

你必须参加桌子

SELECT sum(c.amount) - s.total 
FROM contribution c inner join  (SELECT user_id, sum(solicitude.amount) total
FROM solicitude  GROUP BY c.user_id HAVING user_id = 1 AND status_id = 1  )s
on c.user_id=s.user_id GROUP BY c.user_id,s.total HAVING c.user_id = 1 

编辑：我忘记了一个别名