Question

下面的查询目前按项目显示总数，但我想要的是计算采购订单中所有项目的小计。

由于

输出应为：

POID Item ItemQTY ItemPrice  ItemTotal   SubTotal
1      A       1     15.00       15.00      80.50
1      B       1     25.50       25.50      80.50
1      C       2     20.00       40.00      80.50
2      X       6      5.00       30.00      50.00
2      Y       2     10.00       20.00      50.00

关系：purchase_order 1 - N purchase_order_items

SELECT
purchase_order.id AS POID,

purchase_order_items.description AS Item,
purchase_order_items.quantity AS ItemQTY,
purchase_order_items.price AS ItemPrice,
(purchase_order_items.quantity*purchase_order_items.price) AS ItemTotal

/* Here, Subtotal should be calculated and displayed */

FROM purchase_order
INNER JOIN purchase_order_items ON purchase_order.id = purchase_order_items.fk_purchase_order

我查看了MySQL finding subtotals和Subtotals and SQL，但无法应用于我的查询。

Answer 1

您可以使用WITH ROLLUP功能获取小计：

SELECT
purchase_order.id AS POID,

purchase_order_items.description AS Item,
purchase_order_items.quantity AS ItemQTY,
purchase_order_items.price AS ItemPrice,
SUM(purchase_order_items.quantity*purchase_order_items.price) AS ItemTotal
FROM purchase_order
INNER JOIN purchase_order_items ON purchase_order.id = purchase_order_items.fk_purchase_order
GROUP BY POID, Item WITH ROLLUP

这将创建一个结果集，其中PO小计为Item = NULL，总计为POID = NULL。这些小计和总计将位于这些行的ItemTotal列中。

Answer 2

尝试

SELECT i.fk_purchase_order POID,
       description         Item,
       quantity            ItemQTY,
       price               ItemPrice,
       quantity * price    ItemTotal,
       s.subtotal          SubTotal
  FROM purchase_order_items i JOIN
(
  SELECT fk_purchase_order, SUM(quantity * price) subtotal
    FROM purchase_order_items
   GROUP BY fk_purchase_order
) s ON i.fk_purchase_order = s.fk_purchase_order

输出：

| POID | ITEM | ITEMQTY | ITEMPRICE | ITEMTOTAL | SUBTOTAL |
------------------------------------------------------------
|    1 |    A |       1 |        15 |        15 |     80.5 |
|    1 |    B |       1 |      25.5 |      25.5 |     80.5 |
|    1 |    C |       2 |        20 |        40 |     80.5 |
|    2 |    X |       6 |         5 |        30 |       50 |
|    2 |    Y |       2 |        10 |        20 |       50 |

这是 SQLFiddle 演示

Answer 3

害怕我认为最好的解决方案是与子选择交叉加入： -

SELECT
purchase_order.id AS POID,
purchase_order_items.description AS Item,
purchase_order_items.quantity AS ItemQTY,
purchase_order_items.price AS ItemPrice,
(purchase_order_items.quantity*purchase_order_items.price) AS ItemTotal,
Sub1.FullTotal
FROM purchase_order
INNER JOIN purchase_order_items ON purchase_order.id = purchase_order_items.fk_purchase_order
CROSS JOIN 
(
    SELECT  SUM(purchase_order_items.quantity*purchase_order_items.price) AS FullTotal
    FROM purchase_order
    INNER JOIN purchase_order_items ON purchase_order.id = purchase_order_items.fk_purchase_order
) Sub1

将其恢复为分组采购订单ID

SELECT
purchase_order.id AS POID,
purchase_order_items.description AS Item,
purchase_order_items.quantity AS ItemQTY,
purchase_order_items.price AS ItemPrice,
(purchase_order_items.quantity*purchase_order_items.price) AS ItemTotal,
Sub1.FullTotal
FROM purchase_order
INNER JOIN purchase_order_items ON purchase_order.id = purchase_order_items.fk_purchase_order
INNER JOIN 
(
    SELECT purchase_order.id, SUM(purchase_order_items.quantity*purchase_order_items.price GROUP BY purchase_order.id) AS FullTotal
    FROM purchase_order
    INNER JOIN purchase_order_items ON purchase_order.id = purchase_order_items.fk_purchase_order
) Sub1
ON purchase_order.id = Sub1.id

查询中所有项目的小计

3 个答案: