我不确定我是否需要一个递归查询,但这就是我正在尝试做的(一个调用 B 查询的 A 查询,它调用一个 A 查询......递归地)。
这是我的最小完整可验证代码:
我有一个这样的表(MySQL v5.7,InnoDB):
CREATE TABLE transactions
(
id INT PRIMARY KEY AUTO_INCREMENT,
code VARCHAR(10),
date DATETIME,
mode ENUM('Buy', 'Sell', 'Count', 'Return'),
quantity INT,
price DECIMAL(10,2),
price_currency ENUM('ARS', 'USD'),
usd_to_ars DECIMAL(10,2),
return_id INT NULL DEFAULT NULL
)
然后我用一些项目填充它:
INSERT INTO transactions (code, date, mode, quantity, price, price_currency, usd_to_ars)
VALUES
("a", "20180101", 'Buy', 4, 10, 'ARS', 3.7),
("a", "20180102", 'Buy', 9, 8, 'ARS', 5.8),
("a", "20180103", 'Sell', -3, 0, 'USD', 0),
("b", "20180104", 'Buy', 5, 5, 'USD', 8.9),
("a", "20180105", 'Buy', 2, 7, 'USD', 3.4),
("b", "20180106", 'Buy', 1, 8, 'ARS', 9),
("a", "20180107", 'Sell', -8, 0, 'USD', 4.4),
("a", "20180108", 'Buy', 9, 9, 'ARS', 3.2);
INSERT INTO transactions (code, date, mode, quantity, price, price_currency, usd_to_ars, return_id)
VALUES ("a", "20180109", 'Return', 6, 2, 'ARS', 2, 2);
最后我执行这段代码:
SELECT *
FROM
(SELECT
id, date, code, mode, quantity, price, price_currency, usd_to_ars, return_id,
@acm := @acm + quantity as stock,
@avr := (@avr * (@acm - quantity) +
if(quantity > 0, quantity *
if(mode = "Return", @avr,
if(price_currency = 'USD', price, price / usd_to_ars)
),
quantity * @avr)
) / @acm as average_price_usd
FROM
transactions t1,
(SELECT @acm := 0) x,
(SELECT @avr := 0) y) t2
ORDER BY id DESC
如您所见,它不会引发错误,它会返回一个表,它可以正常工作...但不是我想要的。
第四行:
@avr := (@avr * (@acm - quantity) + if(quantity > 0, quantity * if(mode = "Return", @avr, if(price_currency = 'USD', price, price / usd_to_ars)), quantity * @avr)) / @acm as average_price_usd
我想更改 @avr
参数:
If(mode = "Return", @avr, [...])
目前,如果 mode = "Return"
为 true
,则使用当前的 @avr
值,但我想使用记录的 @avr
值代替 id = return_id
。由于 @avr
是一个计算值,我必须执行子查询才能再次计算它......我想。问题是我不知道该怎么做。
因此,id = 9
的平均价格不应为 3.02...
,而应为 2.72...
。
那么,当 mode = "Return"
时,如何在另一个查询中执行此查询,以便获取记录 WHERE 的
?@avr
值subquery.id = return_id
如果可能的话,我是 SQL 方面的新手,所以我仍然不确定哪些是可能的,哪些是不可能的。如果我需要在表中创建一个列、一个 View 或一个函数,对我来说都没有关系,我想我可以处理它们中的任何一个。
最佳答案
简介:您的数据库 fiddle 表明您似乎正在使用 MySQL 8.0。
由于您已经有一个有效的查询并且您只需要处理特定的 mode = 'Return'
情况,一个解决方案可能是将您的查询转换为 CTE,并将其自连接到将返回记录与其原始购买相关联。
这适用于 this db fiddle :
WITH cte AS (
SELECT * FROM (SELECT
id, date, code, mode, quantity, price, price_currency, usd_to_ars, return_id,
@acm := @acm + quantity as stock,
@avr := (@avr * (@acm - quantity) +
if(quantity > 0, quantity *
if(mode = "Return", @avr,
if(price_currency = 'USD', price, price / usd_to_ars)
),
quantity * @avr)
) / @acm as average_price_usd
FROM
transactions t1,
(SELECT @acm := 0) x,
(SELECT @avr := 0) y) t2
)
SELECT
cte.id,
cte.date,
cte.code,
cte.mode,
cte.quantity,
cte.price,
cte.price_currency,
cte.usd_to_ars,
cte.return_id,
cte.stock,
COALESCE(cte2.average_price_usd, cte.average_price_usd) average_price_usd
FROM cte LEFT JOIN cte cte2 ON cte.mode = 'Return' AND cte2.id = cte.return_id
ORDER BY cte.id DESC
此外,您似乎也可以使用 MySQL 8.0 窗口函数 来实现您的目标。我不清楚计算 average_price_usd
的逻辑,但是这里有一个使用此技术计算股票的查询。您也许可以修改它以添加平均价格的逻辑(它应该转到上述查询的 CTE 部分):
SELECT
id,
code,
date,
mode,
quantity,
price,
price_currency,
usd_to_ars,
return_id,
SUM(quantity) OVER (ORDER BY date) stock,
NULL average_price_usd
FROM transactions
ORDER BY date desc
编辑:这是 MySQL < 8.0 的(丑陋的)解决方法,使用两个子查询(参见 this db fiddle):
SELECT
cte.id,
cte.date,
cte.code,
cte.mode,
cte.quantity,
cte.price,
cte.price_currency,
cte.usd_to_ars,
cte.return_id,
cte.stock,
COALESCE(cte2.average_price_usd, cte.average_price_usd) average_price_usd
FROM
(
SELECT
id, date, code, mode, quantity, price, price_currency, usd_to_ars, return_id,
@acm := @acm + quantity as stock,
@avr := (@avr * (@acm - quantity) +
if(quantity > 0, quantity *
if(mode = "Return", @avr,
if(price_currency = 'USD', price, price / usd_to_ars)
),
quantity * @avr)
) / @acm as average_price_usd
FROM
transactions t1,
(SELECT @acm := 0) x,
(SELECT @avr := 0) y
) cte
LEFT JOIN (
SELECT
id, date, code, mode, quantity, price, price_currency, usd_to_ars, return_id,
@acm2 := @acm2 + quantity as stock,
@avr2 := (@avr2 * (@acm2 - quantity) +
if(quantity > 0, quantity *
if(mode = "Return", @avr2,
if(price_currency = 'USD', price, price / usd_to_ars)
),
quantity * @avr2)
) / @acm2 as average_price_usd
FROM
transactions t1,
(SELECT @acm2 := 0) x,
(SELECT @avr2 := 0) y
) cte2 ON cte.mode = 'Return' AND cte2.id = cte.return_id
ORDER BY cte.id DESC
关于mysql - 如何在查询中执行递归查询或子查询?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54269594/