【发布时间】:2013-06-13 09:33:45
【问题描述】:
我正在尝试显示我正在建造的商店的运费,数据库中有三个表 1 用于服务,即皇家邮政、承运人......,一个用于乐队,即。 UK、Europe、Worldwide1 等。一个用于收费(数量 = 重量)
我有一个包含三个表的数据库,当它们连接时形成以下表格
+------------------+-----+-----------+-------+---------+---------------+----------+-------+-------------+
| name | qty | serviceID | basis | bandID | initial_charge | chargeID | price | total_price |
+------------------+-----+-----------+-------+---------+---------------+----------+-------+-------------+
| Collect in store | 0 | 3 | | 1 | 3 | 0.00 | 2 | 0.00 |
| Royal mail | 0 | 1 | 2 | 4 | 2.00 | 3 | 0.00 | 2.00 |
| Royal mail | 1 | 1 | 2 | 4 | 2.00 | 4 | 1.00 | 3.00 |
| APC | 0 | 2 | 1 | 1 | 0.00 | 6 | 5.95 | 5.95 |
+------------------+-----+-----------+-------+---------+---------------+----------+-------+-------------+
基本上我想做的是(如您所见)皇家邮政有两个条目,因为连接表中有多个条目。我想做的是显示两个皇家邮件条目中最高的条目(我最初尝试按 service_id 分组),同时还维护具有不同服务 ID 的其他两个服务
任何帮助都会很棒,因为这让我发疯。我觉得我已经尝试了所有组合!
在下面的示例中,物品的数量(重量)为 3kg
SELECT
`service`.`name`,
`charge`.`qty`,
`service`.`serviceID`,
`band`.`bandID`,
`band`.`initial_charge`,
`charge`.`chargeID`,
`charge`.`price`,
`band`.`initial_charge` + `charge`.`price` AS `total_price`
FROM
`delivery_band` AS `band`
LEFT JOIN
`delivery_charge` AS `charge`
ON
`charge`.`bandID` = `band`.`bandID`
AND
`charge`.`qty` < '3'
LEFT JOIN
`delivery_service` AS `service`
ON
`service`.`serviceID` = `band`.`serviceID`
WHERE
FIND_IN_SET( '225', `band`.`accepted_countries` )
AND
(
`band`.`min_qty` >= '3'
OR
`band`.`min_qty` = '0'
)
AND
(
`band`.`max_qty` <= '3'
OR
`band`.`max_qty` = '0'
)
delivery_service
+-----------+------------------+
| serviceID | name |
+-----------+------------------+
| 1 | Royal mail |
| 2 | APC |
| 3 | Collect in store |
+-----------+------------------+
delivery_band
+--------+-----------+-----------------+----------------+---------+---------+-------------------------------------------------------+
| bandID | serviceID | name | initial_charge | min_qty | max_qty | accepted_countries |
+--------+-----------+-----------------+----------------+---------+---------+-------------------------------------------------------+
| 1 | 2 | UK Mainland | 0.00 | 0 | 0 | 225 |
| 2 | 2 | UK Offshore | 14.00 | 0 | 0 | 240 |
| 3 | 3 | Bradford Store | 0.00 | 0 | 0 | 225 |
| 4 | 1 | UK | 2.00 | 0 | 0 | 225 |
| 5 | 2 | World wide | 15.00 | 0 | 0 | 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20... |
| 6 | 1 | World wide Mail | 5.00 | 0 | 0 | 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20... |
+--------+-----------+-----------------+----------------+---------+---------+-------------------------------------------------------+
delivery_charge
+----------+--------+-----+-------+
| chargeID | bandID | qty | price |
+----------+--------+-----+-------+
| 1 | 2 | 0 | 5.00 |
| 2 | 3 | 0 | 0.00 |
| 3 | 4 | 0 | 0.00 |
| 4 | 4 | 1 | 1.00 |
| 5 | 4 | 5 | 3.00 |
| 6 | 1 | 0 | 5.95 |
| 7 | 1 | 10 | 10.95 |
| 8 | 2 | 10 | 14.00 |
| 9 | 5 | 0 | 0.00 |
| 10 | 5 | 3 | 5.00 |
| 11 | 5 | 6 | 10.00 |
| 12 | 5 | 9 | 15.00 |
| 13 | 6 | 0 | 0.00 |
| 14 | 6 | 2 | 5.00 |
| 15 | 6 | 4 | 10.00 |
| 16 | 6 | 6 | 15.00 |
+----------+--------+-----+-------+
当我尝试将费用表添加为子查询然后限制该查询时,它给了我所有费用表字段的 NULL
如果我尝试以下查询:
SELECT
`service`.`name`,
`charge`.`qty`,
`service`.`serviceID`,
`band`.`bandID`,
`band`.`initial_charge`,
`charge`.`chargeID`,
MAX( `charge`.`price` ) AS `price`,
`band`.`initial_charge` + `charge`.`price` AS `total_price`
FROM
`delivery_band` AS `band`
LEFT JOIN
`delivery_charge` AS `charge`
ON
`charge`.`bandID` = `band`.`bandID`
AND
`charge`.`qty` < '3'
LEFT JOIN
`delivery_service` AS `service`
ON
`service`.`serviceID` = `band`.`serviceID`
WHERE
FIND_IN_SET( '225', `band`.`accepted_countries` )
AND
(
`band`.`min_qty` >= '3'
OR
`band`.`min_qty` = '0'
)
AND
(
`band`.`max_qty` <= '3'
OR
`band`.`max_qty` = '0'
)
GROUP BY
`service`.`serviceID`
我得到这个返回:
+------------------+-----+-----------+--------+----------------+----------+-------+-------------+
| name | qty | serviceID | bandID | initial_charge | chargeID | price | total_price |
+------------------+-----+-----------+--------+----------------+----------+-------+-------------+
| Royal mail | 0 | 1 | 4 | 2.00 | 3 | 1.00 | 2.00 |
| APC | 0 | 2 | 1 | 0.00 | 6 | 5.95 | 5.95 |
| Collect in store | 0 | 3 | 3 | 0.00 | 2 | 0.00 | 0.00 |
+------------------+-----+-----------+--------+----------------+----------+-------+-------------+
这在原则上看起来不错,直到您意识到chargeID = 3 的价格为 0.00,但表格显示的价格为 1.00,因此这些值似乎已解除关联
【问题讨论】:
-
sqlfiddle.com/#!2/9665c4/4 - 你能看出问题所在吗? - 谢谢你的链接顺便说一句 - 不知道这存在!
-
小提琴是为了让人们可以在得出结果的同时玩弄您的数据。您最初的问题没有明确说明您不仅想要最高价格,而且还想要 id 与之配套。我想现在应该更清楚了。
-
很抱歉 :( 我猜这是在我的脑海里,但没有传达出来
标签: mysql join group-by sql-order-by