Question

我有3张桌子：

表a =＆gt; id，name

表b =＆gt; id，id_a，金额

表c =＆gt; id，id_a，金额

如何从表b和表c？

中选择表a和总金额

我已经尝试过了：

SELECT a.`name`, SUM(b.`amount`) AS Total
FROM a
LEFT JOIN b ON a.`id` = b.`id_a`
UNION
SELECT a.`name`, SUM(c.`amount`) AS Total
FROM a
RIGHT JOIN c ON a.`id` = c.`id_a`
GROUP BY a.`id`

但它显示了表a中的多个记录，其中表b和表c中都有id_a值。

仅供参考：如果插入表a，则表b或表c或表b和表c都将插入详细信息。因此，表a或表b中没有可能的表a没有详细信息

这是示例数据

表格a：

id | name
1 | name 1
2 | name 2

表格b：

id | id_a | amount
1 | 1 | 300
2 | 1 | 200

表格c：

id | id_a | amount
1 | 1 | 200
2 | 1 | 100
3 | 2 | 300
4 | 2 | 100

首选结果：

name | amount
name 1 | 800
name 2 | 400

更新答案=＆gt;感谢@Jens

SELECT d.NAME, SUM(d.Total) FROM (
SELECT a.id AS id , a.`name` AS NAME, SUM(b.`amount`) AS Total
FROM a
LEFT JOIN b ON a.`id` = b.`id_a`
GROUP BY a.`id`
UNION
SELECT a.`id` AS id, a.`name`, SUM(c.`amount`) AS Total
FROM a
RIGHT JOIN c ON a.`id` = c.`id_a`
GROUP BY a.`id`
) AS d
GROUP BY d.`id`

来自@Alex的另一个回答是工作和简单感谢@Jens和@Alex。这增加了我的知识

Answer 1

尝试

select d.name, sum(d.total) from (
SELECT a.id as id , a.`name` as name, b.`amount` AS Total
FROM a
LEFT JOIN b ON a.`id` = b.`id_a`
UNION
SELECT a.id as id a.`name`, c.`amount` AS Total
FROM a
RIGHT JOIN c ON a.`id` = c.`id_a`
) as d
GROUP BY d.`id`

<强>更新：

SELECT d.NAME, SUM(d.Total) FROM (
SELECT a.id AS id , a.`name` AS NAME, SUM(b.`amount`) AS Total
FROM a
LEFT JOIN b ON a.`id` = b.`id_a`
GROUP BY a.`id`
UNION
SELECT a.`id` AS id, a.`name`, SUM(c.`amount`) AS Total
FROM a
RIGHT JOIN c ON a.`id` = c.`id_a`
GROUP BY a.`id`
) AS d
GROUP BY d.`id`

Answer 2

你试过这个吗？

SELECT a.name, sum(b.amount), sum(c.amount)
FROM a
JOIN b on a.id = b.id_a
JOIN c on a.id = c.id_a
GROUP BY a.id

不确定关系，但如果你的情况如此，你可以用LEFT JOIN替换JOIN。

<强>更新

这应该有效：

SELECT a.name as "name", sum(d.amount) as "amount"
FROM a
JOIN (SELECT id_a, amount FROM b UNION ALL SELECT id_a, amount FROM c) as d
on d.id_a = a.id
group by a.name

这会产生以下结果：

name1   |    800
name2   |    400

MYSQL - 选择3个连接表的总数量

2 个答案: