Question

我有一个用于管理商店销售的数据库。我正在尝试编写一个查询，以获取每张发票的总费用和支付的总金额。

所有付款都记录在'invoice_payments'表中。发票的总成本是job_price +所有stock_unit项目（产品）。

'发票'表：

invoice_id  job_price
000071      100.00

'stock_unit'表：

unit_id     price       invoice_id
000261      10      000071
000262      10      000071
000263      10      000071

'invoice_payments'表：

payment_id  invoice_id  amount
000074      000071      100.00
000075      000071      30.00

查询：

SELECT
    invoice.invoice_id, 
    SUM(stock_unit.`sold_price` + invoice.job_price) AS price,
    SUM(`invoice_payments`.`amount`) AS paid

FROM invoice

LEFT JOIN stock_unit
    ON stock_unit.invoice_id = invoice.invoice_id

LEFT JOIN invoice_payments
    ON invoice_payments.invoice_id = invoice.invoice_id

GROUP BY invoice.invoice_id

当只有1个连接时，查询按预期工作，但是当添加2时，它们会相互影响。

在我添加'invoice_payments'之后：

invoice_id  paid
000071      130.00

后：

invoice_id  price       paid
000071      660.00      390.00

它应该是什么：

invoice_id  price       paid
000071      130.00      130.00

SQL小提琴无法正常工作，但我已在此处设置了一个架构，以便您可以更好地了解我的表：http://sqlfiddle.com/#!9/091a35

我最好有3个单独的查询并加入吗？

Answer 1

你应该选择e分离的结果而不是计算结果od两者是同一个加入的查询

SELECT t1.invoce_id, t1.price, t2.paid FROM 

( 
SELECT     
     invoice.invoice_id AS invoce_id, 
    SUM(stock_unit.`sold_price` + invoice.job_price) AS price
FROM invoice
LEFT JOIN stock_unit
    ON stock_unit.invoice_id = invoice.invoice_id
GROUP BY invoice.invoice_id ) AS t1

LEFT JOIN 

(
SELECT 
    invoice.invoice_id AS invoce_id, 
    SUM(`invoice_payments`.`amount`) AS paid
FROM invoice    
LEFT JOIN invoice_payments
    ON invoice_payments.invoice_id = invoice.invoice_id
GROUP BY invoice.invoice_id ) AS t2 

ON t1.invoce_id = t2.invoce_id

Answer 2

感谢Codecademy的Samantha解决这个问题！

第一个联接为每个发票返回多行。金额取决于您的付款金额。

执行第二次加入时，它尝试加入第一张发票的多个结果，这意味着它在一张发票上多次匹配。

要解决此问题，我们使用子查询对第一个连接进行分组，然后将精简结果加入到下一个查询中。

示例：

SELECT sub.invoice_id, sub.price, SUM(invoice_payments.amount) AS paid FROM (SELECT invoice.invoice_id AS invoice_id, SUM(stock_unit.sold_price) + invoice.job_price AS price FROM invoice LEFT JOIN stock_unit ON stock_unit.invoice_id = invoice.invoice_id GROUP BY invoice.invoice_id) AS sub LEFT JOIN invoice_payments ON invoice_payments.invoice_id = sub.invoice_id GROUP BY sub.invoice_id;

这也有效：

SELECT invoice_payments.invoice_id, p.price, SUM(invoice_payments.amount) AS paid FROM invoice_payments JOIN (select stock_unit.invoice_id, SUM(sold_price) + invoice.job_price AS price FROM stock_unit join invoice ON invoice.invoice_id=stock_unit.invoice_id) AS p ON p.invoice_id = invoice_payments.invoice_id GROUP BY invoice_id;

MySQL多个连接导致两个列都发生变化

2 个答案: