[英]Subtract two columns of different tables
I have two unrelated tables: 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)
I need to subtract the sum of the amount of the contribution and of the solicitude from a user, but that result can't to be negative. 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。
How can I do this? 我怎样才能做到这一点? Function or query?
功能还是查询?
I tried this query: 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
You can add an outer query to check the total value: 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;
This solution is OK, but I suggest an alternative approach. 这个解决方案还可以,但是我建议您使用另一种方法。 Check how this query works:
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;
I made a simple test, just out of curiosity, on this setup: 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);
Results (msecs): 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
I interpret your remark but that result can't to be negative
as requirement to return 0 instead of negative results. 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。 The simple solution is GREATEST()
: 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;
Otherwise, I kept your original query, which is fine. 否则,我保留您的原始查询,这很好。
For other cases with the possible result that no row could be returned I would replace with two sub-selects. 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。 But the use of the aggregate function guarantees a result row, even if the given
user_id
is not found at all. 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。 Compare: 相比:
If the result of the subtraction would be NULL
(because no row is found or the sum is NULL
), GREATEST()
will also return 0
. 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
。
You have to join the tables 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
EDIT: I forgot an alias 编辑:我忘记了一个别名
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