本文介绍了从两个连接表中选择 Sum的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
有结构:
CREATE TABLE `invoices` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`date` date NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB;
INSERT INTO `invoices` VALUES (1,'2018-09-22');
CREATE TABLE `products` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`invoice_id` int(10) unsigned NOT NULL,
`amount` decimal(10,2) unsigned NOT NULL,
`quantity` smallint(5) unsigned NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB;
INSERT INTO `products` VALUES (1,1,150.00,2),(2,1,60.00,3),(3,1,50.00,1);
CREATE TABLE `payments` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`invoice_id` int(10) unsigned NOT NULL,
`amount` decimal(10,2) unsigned NOT NULL,
`date` date NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB;
INSERT INTO `payments` VALUES (1,1,400.00,'2018-09-23'),(2,1,80.00,'2018-09-23');
我有这个问题:
select i.id, sum(pr.amount * pr.quantity) as productAmount,
sum(pm.amount) as paymentAmount
from invoices as i
left join products as pr on pr.invoice_id=i.id
left join payments as pm on pm.invoice_id=i.id
group by i.id
并得到这个结果:
+----+---------------+---------------+
| id | productAmount | paymentAmount |
+----+---------------+---------------+
| 1 | 1060.00 | 1440.00 |
+----+---------------+---------------+
1 row in set (0,00 sec)
但是,我想得到以下结果:
However, I want to get the following result:
+----+---------------+---------------+
| id | productAmount | paymentAmount |
+----+---------------+---------------+
| 1 | 530.00 | 480.00 |
+----+---------------+---------------+
1 row in set (0,00 sec)
我想要按 invoice.id 分组的产品总金额和付款总金额.
在这种情况下应该查询什么?
I want sum amount of products and sum amount of payments grouped by invoice.id.
What should be the query in this case?
推荐答案
我确实有时会遇到这种查询.由于多个连接,来自特定表的值会重复、三次等.为了解决这个问题,我通常通过将总和(在 特定表上)除以不同 Id 的计数来做一个小技巧(s) 来自其他表.这抵消了多次重复发生的影响.
I do face this kind of queries at times. Due to multiple joins, values from a particular table get duplicated, triplicated etc. To fix this, I normally do a small hack by dividing the sum (on a particular table) by the count of distinct Id(s) from the other table. This negates the effect of multiple duplicates happening.
尝试以下查询:
select i.id,
(sum(pr.amount * pr.quantity) / IF(count(distinct pm.id) > 0, count(distinct pm.id), 1) as productAmount,
(sum(pm.amount) / IF(count(distinct pr.id) > 0, count(distinct pr.id), 1) as paymentAmount
from invoices as i
left join products as pr on pr.invoice_id=i.id
left join payments as pm on pm.invoice_id=i.id
group by i.id
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