如何将 sum() 的结果与 select 查询连接起来

Question

我有一个基本的 transactions table，总共有 4 列，如下所示：

`transaction_id` int(11) NOT NULL AUTO_INCREMENT,
`fruit_id` int(11),
`person_id` int(11),
`quantity_bought` int(11),

示例select *查询：

transaction_id	fruit_id	person_id	quantity_bought
1	banana	alex	65
2	banana	joe	25
3	banana	jenny	70
4	apple	dan	80
4	apple	danny	50

我需要将每个水果的 total_quantity_bought 添加到此 select 查询的每一行：

SELECT transactions.* FROM transactions;

这是所需输出的示例：

transaction_id	fruit_id	person_id	quantity_bought	total_quantity_bought
1	banana	alex	65	160
2	banana	joe	25	160
3	banana	jenny	70	160
4	apple	dan	80	130
4	apple	danny	50	130

这是我迄今为止尝试过的方法，但惨遭失败：

select x.* , y.total_quantity_bought
from
   (SELECT *
    FROM transactions) x
    cross join
   (SELECT fruit_id , SUM(quantity_bought) AS total_quantity_bought
    FROM transactions
    GROUP BY fruit_id) y

我想为这样的总和创建一个 View，但我正在寻找一个不需要 View

的解决方案

fruit_id	total_quantity_bought
banana	160
apple	130

感谢任何帮助。

Answer 1

你基本上就在那里。您的 JOIN 想要指定左侧的所有内容（来自 transactions table）应该连接到右侧的一个关联总数（totals 子查询）...并且该连接应该以匹配 fruit_id.

为条件

随着数据集的增长，此查询的性能可能会（严重）下降。您可能想要查看缓存总计（使用数据库触发器等）或在其他地方的代码中呈现总计。

如果您选择后者（在别处渲染），您可能会对 WITH ROLLUP 函数感兴趣：

https://mariadb.com/kb/en/select-with-rollup/

SELECT
   transactions.*,
   totals.total_quantity_bought
FROM
   transactions   -- These are the individual transactions

   INNER JOIN     -- Join each individual transaction against the calculated totals

   (
    SELECT
       fruit_id,
       SUM(quantity_bought) AS total_quantity_bought
    FROM transactions
    GROUP BY fruit_id
   ) as totals    -- These are the calculated totals (subqueried)

   ON totals.fruit_id = transactions.fruit_id
                  -- This joins the individual records to the totals by `fruit_id`

（演示：http://sqlfiddle.com/#!9/74ae7e/10）

Answer 2

可能我只是被打败了答案，但这是我的尝试

SELECT t.transaction_id, t.fruit_id, t.person_id, t.quantity_bought, grouped_fruit.total_quantity_bought
FROM transactions t
JOIN (
    SELECT fruit_id, SUM(quantity_bought) total_quantity_bought
    FROM transactions
    GROUP BY fruit_id
) grouped_fruit ON t.fruit_id = grouped_fruit.fruit_id

如何将 sum() 的结果与 select 查询连接起来

How to join a sum()'s result with a select query

mysql

sum