[英]mysql - How do we get the sum of the values on two tables?
I have two tables 我有两张桌子
table: a --------------------- id amount status --------------------- 4031 3000 1 4032 4000 1 4033 5000 1 4034 6000 1 4035 7000 1 4036 8000 0
table: s -------------- id a_id b_id -------------- 1 4031 20 2 4031 21 3 4032 23 4 4032 24 5 4033 25 6 4033 26 7 4034 21 8 4034 20 9 4035 25 10 4035 29 11 4036 21 12 4036 20
How do we get the sum of the a.amount where have ( b_id = 20 AND b_id = 21) AND a.status = 1? 我们如何获得a.amount的总和,其中有(b_id = 20 AND b_id = 21)和a.status = 1?
The answer should be 9000. 答案应该是9000。
SELECT SUM(amount) FROM (
JOIN s ON a.id = s.id
WHERE STATUS =1
AND (b_id = 20 OR b_id = 21) GROUP BY a.id
) AS amounts
total : 9000 合计:9000
In the case you can add several times the same amount, I guess this should work without join: 如果您可以添加相同数量的数倍,我想这应该在不加入的情况下起作用:
SELECT SUM(amount) AS total
FROM `a`, `s`
WHERE a_id = a.id AND (b_id = 20 OR b_id = 21) AND status = 1
total : 18000 合计:18000
You can get the answer using a subquery: 您可以使用子查询获得答案:
SELECT SUM(a.amount)
FROM a
WHERE a.status=1 AND
EXISTS (SELECT 1
FROM s
WHERE s.a_id=a.id AND s.b_id in (20,21));
There is no need to group the data as we want the global sum of the amounts selected. 不需要对数据进行分组,因为我们希望选择的金额为全球总和。
Try this: 尝试这个:
select sum(a.amount)
from a
join b on a.id = b.a_id
where b.b_id IN ( 20, 21 ) and a.status = 1
SELECT SUM(a.amount)
FROM a
WHERE a.status=1 AND
EXISTS (SELECT 1
FROM s
WHERE s.a_id=a.id AND s.b_id=20) AND
EXISTS (SELECT 1
FROM s
WHERE s.a_id=a.id AND s.b_id=21) ;
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