MySQL - 总和正在成倍增加答案

【问题标题】：MySQL - sums are getting multipliedMySQL - 总和正在成倍增加
【发布时间】：2021-08-02 03:54:54
【问题描述】：

我正在尝试从table_a 获取数据，左侧的sum() 加入table_b，另一个左侧的sum() 加入table_c

所以，加入a 和b：

SELECT a.id, a.name, sum(b.quantity)
FROM table_a a
LEFT JOIN table_b b on b.id_table_a = a.id
GROUP BY a.id;

输出

id	name	sum(b.quantity)
1	test	14

并且，加入a 和c：

SELECT a.id, a.name, sum(c.quantity)
FROM table_a a
LEFT JOIN table_c c on c.id_table_a = a.id
GROUP BY a.id;

输出

id	name	sum(c.quantity)
1	test	8

当我尝试同时加入a 和b 和c 时出现问题

SELECT a.id, a.name, SUM(b.quantity), SUM(c.quantity)
FROM table_a a
LEFT JOIN table_b b ON b.id_table_a = a.id
LEFT JOIN table_c c ON c.id_table_a = a.id
GROUP BY a.id;

输出

id	name	sum(b.quantity)	sum(c.quantity)
1	test	28	24

sum() 乘以连接表中的行数（14x2 和 8x3）

此时的问题是：我怎样才能得到正确的总和？

这是我所期望的：

id	name	sum(b.quantity)	sum(c.quantity)
1	test	14	8

现在我正在执行以下操作，但我想知道使用上一个查询是否可以实现相同的结果，也许我在 group by 中遗漏了一些东西？

SELECT 
    x.*, SUM(c.quantity)
FROM
    (SELECT 
        a.id, a.name, SUM(b.quantity)
    FROM
        table_a a
    LEFT JOIN table_b b ON b.id_table_a = a.id
    GROUP BY a.id) x
        LEFT JOIN
    table_c c ON c.id_table_a = x.id
GROUP BY x.id;

这是一个指向小提琴的链接，位于 SQL 下方

http://sqlfiddle.com/#!9/b40639

CREATE TABLE `table_a` (
  `id` INT UNSIGNED NOT NULL AUTO_INCREMENT,
  `name` VARCHAR(50) NULL,
  PRIMARY KEY (`id`));

CREATE TABLE `table_b` (
  `id` INT UNSIGNED NOT NULL AUTO_INCREMENT,
  `id_table_a` INT UNSIGNED NOT NULL,
  `quantity` TINYINT UNSIGNED NULL,
  PRIMARY KEY (`id`),
  INDEX `fk_table_b_id_table_a1_idx` (`id_table_a` ASC),
  CONSTRAINT `fk_table_b_id_table_a1`
    FOREIGN KEY (`id_table_a`)
    REFERENCES `table_a` (`id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION);

CREATE TABLE `table_c` (
  `id` INT UNSIGNED NOT NULL AUTO_INCREMENT,
  `id_table_a` INT UNSIGNED NOT NULL,
  `quantity` TINYINT UNSIGNED NULL,
  PRIMARY KEY (`id`),
  INDEX `fk_table_c_id_table_a1_idx` (`id_table_a` ASC),
  CONSTRAINT `fk_table_c_id_table_a1`
    FOREIGN KEY (`id_table_a`)
    REFERENCES `table_a` (`id`)
    ON DELETE NO ACTION
    ON UPDATE NO ACTION);

INSERT INTO `table_a` (`name`) VALUES ('test');
INSERT INTO `table_b` (`id_table_a`, `quantity`) VALUES ('1', '2');
INSERT INTO `table_b` (`id_table_a`, `quantity`) VALUES ('1', '4');
INSERT INTO `table_b` (`id_table_a`, `quantity`) VALUES ('1', '8');
INSERT INTO `table_c` (`id_table_a`, `quantity`) VALUES ('1', '3');
INSERT INTO `table_c` (`id_table_a`, `quantity`) VALUES ('1', '5');

【问题讨论】：

标签： mysql sql join sum

【解决方案1】：

你可能会发现最好的表现是：

SELECT a.id, a.name, 
       (SELECT SUM(b.quantity) 
        FROM table_b b
        WHERE b.id_table_a = a.id
       ) as sum_b,
       (SELECT SUM(c.quantity) 
        FROM table_c c
        WHERE c.id_table_a = a.id
       ) as sum_c
FROM table_a a;

为了提高性能，您需要在 table_b(id_table_a, quantity) 和 table_c(id_table_a, quantity) 上建立索引。

【讨论】：

【解决方案2】：

这是常见的“连接乘法”。可能的解决方案 - 在子查询中聚合单个表（对于每个聚合表）然后加入。

SELECT a.id, a.name, sum_b, sum_c
FROM table_a a
LEFT JOIN ( SELECT id_table_a, SUM(quantity) sum_b
            FROM table_b
            GROUP BY id_table_a ) b ON b.id_table_a = a.id
LEFT JOIN ( SELECT id_table_a, SUM(quantity) sum_c
            FROM table_c
            GROUP BY id_table_a ) c ON c.id_table_a = a.id;

https://dbfiddle.uk/?rdbms=mysql_5.6&fiddle=32c4b2898693a21e799a84970a87b3c8

【讨论】：