fri*_*nux 5 mysql sql algorithm syntax
我尝试从包含一些JOINS的查询中的列中SUM值.
例:
SELECT
p.id AS product_id,
SUM(out_details.out_details_quantity) AS stock_bought_last_month,
SUM(order_details.order_quantity) AS stock_already_commanded
FROM product AS p
INNER JOIN out_details ON out_details.product_id=p.id
INNER JOIN order_details ON order_details.product_id=p.id
WHERE p.id=9507
GROUP BY out_details.out_details_pk, order_details.id;
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我得到这个结果:
+------------+-------------------------+-------------------------+
| product_id | stock_bought_last_month | stock_already_commanded |
+------------+-------------------------+-------------------------+
| 9507 | 22 | 15 |
| 9507 | 22 | 10 |
| 9507 | 10 | 15 |
| 9507 | 10 | 10 |
| 9507 | 5 | 15 |
| 9507 | 5 | 10 |
+------------+-------------------------+-------------------------+
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现在,我想要求值,但当然有重复.我还必须按product_id进行分组:
SELECT
p.id AS product_id,
SUM(out_details.out_details_quantity) AS stock_bought_last_month,
SUM(order_details.order_quantity) AS stock_already_commanded
FROM product AS p
INNER JOIN out_details ON out_details.product_id=p.id
INNER JOIN order_details ON order_details.product_id=p.id
WHERE p.id=9507
GROUP BY p.id;
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结果:
+------------+-------------------------+-------------------------+
| product_id | stock_bought_last_month | stock_already_commanded |
+------------+-------------------------+-------------------------+
| 9507 | 74 | 75 |
+------------+-------------------------+-------------------------+
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想要的结果是:
+------------+-------------------------+-------------------------+
| product_id | stock_bought_last_month | stock_already_commanded |
+------------+-------------------------+-------------------------+
| 9507 | 37 | 25 |
+------------+-------------------------+-------------------------+
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我如何忽略重复?当然,行数可以改变!
Select P.Id
, Coalesce(DetailTotals.Total,0) As stock_bought_last_month
, Coalesce(OrderTotals.Total,0) As stock_already_commanded
From product As P
Left Join (
Select O1.product_id, Sum(O1.out_details_quantity) As Total
From out_details As O1
Group By O1.product_id
) As DetailTotals
On DetailTotals.product_id = P.id
Left Join (
Select O2.product_id, Sum(O2.order_quantity) As Total
From order_details As O2
Group By O2.product_id
) As OrderTotals
On OrderTotals.product_id = P.id
Where P.Id = 9507
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