從兩個聯接表中選擇總和

Question

有結構：

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');

我有這個查詢：

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

並得到以下結果：

+----+---------------+---------------+
| id | productAmount | paymentAmount |
+----+---------------+---------------+
|  1 |       1060.00 |       1440.00 |
+----+---------------+---------------+
1 row in set (0,00 sec)

但是，我想得到以下結果：

+----+---------------+---------------+
| id | productAmount | paymentAmount |
+----+---------------+---------------+
|  1 |        530.00 |        480.00 |
+----+---------------+---------------+
1 row in set (0,00 sec)

我想要按發票.id分組的產品總金額和付款總金額。
在這種情況下，查詢應該是什么？

Answer 1

我確實有時會遇到這種查詢。 由於存在多個聯接，因此特定表中的值將被重復，三重等。要解決此問題，我通常通過將總和（在特定表上 ）除以與另一個表不同的Id的數量來進行小技巧。 這消除了發生多次重復的影響。

請嘗試以下查詢：

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

Answer 2

使用下面的子查詢來獲得您期望的結果

SELECT id, 
(select sum(pr.amount * pr.quantity) from products as pr where pr.invoice_id=i.id ) as productAmt, 
(select sum(amount) from payments where invoice_id=i.id ) as PaymentAmt 
FROM `invoices` i order by id asc

Answer 3

試試這個： https : //dbfiddle.uk/?rdbms=mysql_8.0&fiddle=3b496214bf0ffb517dcaa9be7f0b6bb7

select a.id,pramount,sum(amount) as paymentamount from 
(select i.id,sum(pr.amount *pr.quantity) as pramount
from invoices as i
join products as pr on pr.invoice_id=i.id
group by i.id)a
join payments pm on a.id=pm.invoice_id
group by a.id

Answer 4

我可以想象這樣一種情況：創建發票但沒有創建付款，類似地，創建發票卻沒有創建付款。因此您可以為產品和付款創建子查詢

drop table if exists i,pr,pm;
CREATE TABLE `i` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`date` date NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB;

INSERT INTO `i` VALUES (1,'2018-09-22'),(2,'2018-09-22'),(3,'2018-09-22'),(4,'2018-09-22');

CREATE TABLE `pr` (
`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 `pr` VALUES (1,1,150.00,2),(2,1,60.00,3),(3,1,50.00,1),(4,3,50.00,1);

CREATE TABLE `pm` (
`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 `pm` VALUES (1,1,400.00,'2018-09-23'),(2,1,80.00,'2018-09-23'),(3,4,80.00,'2018-09-23');


select i.id, (select sum(pr.amount * pr.quantity) from pr where pr.invoice_id = i.id) as productAmount, 
                 (select sum(pm.amount) from pm where pm.invoice_id = i.id) as paymentAmount
from i
having productAmount > 0 or paymentAmount > 0;

您可能（或可能不希望擁有having子句）

+----+---------------+---------------+
| id | productAmount | paymentAmount |
+----+---------------+---------------+
|  1 |        530.00 |        480.00 |
|  3 |         50.00 |          NULL |
|  4 |          NULL |         80.00 |
+----+---------------+---------------+
3 rows in set (0.00 sec)