我正在尝试显示我正在建设的商店的运费，数据库1中有三个表用于服务，即皇家邮政，运营商......，一个用于乐队即。英国，欧洲，Worldwide1等..和一个收费（数量=重量）
我有一个包含三个表的数据库，当从以下
加入时+------------------+-----+-----------+-------+---------+---------------+----------+-------+-------------+
| 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 |
+------------------+-----+-----------+-------+---------+---------------+----------+-------+-------------+
基本上我想要做的是（正如你所看到的）Royal Mail有两个条目，因为在连接表中有多个条目。我想要做的是显示两个皇家邮件条目中最高的（我最初尝试按service_id分组），同时还维护另外两个服务ID不同的服务
任何帮助都会很棒，因为这让我很生气。我觉得我已尝试过各种组合！
在下面的示例中，项目的数量（重量）为3千克
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'
)
+-----------+------------------+
| serviceID | name |
+-----------+------------------+
| 1 | Royal mail |
| 2 | APC |
| 3 | Collect in store |
+-----------+------------------+
+--------+-----------+-----------------+----------------+---------+---------+-------------------------------------------------------+
| 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... |
+--------+-----------+-----------------+----------------+---------+---------+-------------------------------------------------------+
+----------+--------+-----+-------+
| 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，它为所有费用表字段提供了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`
我得到了这个：
+------------------+-----+-----------+--------+----------------+----------+-------+-------------+
| 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 |
+------------------+-----+-----------+--------+----------------+----------+-------+-------------+
原则上看起来很好，直到你意识到chargeID = 3的价格为0.00而且该表显示的是1.00的价格，因此价值似乎已经解除了分离
答案 0 :(得分：0)
我想要做的是显示两个皇家邮件条目中最高的
您可以使用MAX获取给定列的最大值，例如
SELECT … MAX(charge.price) … FROM …
如果您绝对需要其他列（例如charge.chargeID）匹配，那么事情会变得非常复杂很多。因此，请确保您确实需要它。有关此类查询背后的一般概念的详细信息，请仔细查看Select one value from a group based on order from other columns。通过this answer调整@RichardTheKiwi，我想出了以下查询：
SELECT s.name,
c.qty,
s.serviceID,
b.bandID,
b.initial_charge,
c.chargeID,
c.price,
b.initial_charge + c.price AS total_price
FROM delivery_band AS b,
delivery_service AS s,
(SELECT chargeID, price, qty,
@rowctr := IF(bandId = @lastBand, @rowctr+1, 1) AS rowNumber,
@lastBand := bandId AS bandId
FROM (SELECT @rowctr:=0, @lastBand:=null) init,
delivery_charge
WHERE qty < 3
ORDER BY bandId, price DESC
) AS c
WHERE FIND_IN_SET(225, b.accepted_countries)
AND (b.min_qty >= 3 OR B.min_qty = 0)
AND (b.max_qty <= 3 OR B.max_qty = 0)
AND s.serviceID = b.serviceID
AND c.bandID = b.bandID
AND c.rowNumber = 1
请参阅this fiddle了解相应的输出。请注意，我只进行内部查询，而不是左查询，因为这对于有问题的查询来说似乎已经足够了，并且使事情更具可读性，因此您可以专注于重要部分，即涉及rowNumber的部分。这个想法是子查询为相同波段的项生成行号，将它们重置为下一个波段。当您只选择rowNumber为1的行时，您只能获得最高价格，以及与该相关联的所有其他列。