Sql 减去不同表的两列
我有两个不相关的表:Sql 减去不同表的两列,sql,postgresql,sum,aggregate-functions,subtraction,Sql,Postgresql,Sum,Aggregate Functions,Subtraction,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) 我需要减去用户的贡献和关心的总和,但结果不能是负数 我该怎么做?功能还是查询? 我尝试了以下查询: SELECT sum(contribution.amount) - (SELECT sum(solicitude.amount) FROM solic
contribution(id,amount, create_at, user_id)
solicitude(id, amount, create_at, status_id, type_id, user_id)
我需要减去用户的贡献和关心的总和,但结果不能是负数
我该怎么做?功能还是查询?我尝试了以下查询:
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
你必须加入表格
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:我忘记了别名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,但我建议另一种方法。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,我在这个设置上做了一个简单的测试:
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):
我将您的评论解释为
,但该结果不能为负
,因为要求返回0而不是负结果。简单的解决办法是:
否则,我保留了您最初的查询,这很好
对于可能导致无法返回任何行的其他情况,我将替换为两个子选择。但是使用聚合函数可以保证结果行,即使根本找不到给定的用户id
。比较:
如果减法的结果为
NULL
(因为找不到行或总和为NULL
),则magest()
也将返回0
,但该结果不能为负。请详细说明。你的询问有什么问题?语法可以简化,但做得很好。谢谢您的时间;)我认为这比需要的更昂贵、更冗长。@ErwinBrandstetter-你是说第二个问题吗?@klin:是的,特别是第二个问题,CTE更昂贵,在这里没有必要。但也是第一个:比必要的查询级别多出一个查询级别,并且grest()
将进一步简化。@ErwinBrandstetter-第二个查询故意冗长。这是一个关于如何连接聚合表的易于理解和修改的示例。此外,有时我觉得有必要写点好东西。这个查询很漂亮,不是吗?关于
,我完全同意。我只是忘了有这样的东西存在;)@ErwinBrandstetter—实际上,我的外部查询解决方案似乎更快。
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
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;