我有一个表，我的表结构是这样的：
unique_id vendor_name price1 price2 price3 code
1 Vendor 1 0.0012 0.0014 0.0054 125
2 Vendor 2 0.0015 0.0016 0.0050 125
3 Vendor 3 0.0011 0.0019 0.0088 125
4 Vendor 1 0.0025 0.0024 0.0034 126
5 Vendor 2 0.0043 0.0019 0.0065 126
6 Vendor 3 0.0019 0.0085 0.0082 126
我必须按代码获取每个价格列组的最低价格。我的预期输出如下：
Code price1 price2 price3 vendor for price1 vendor for price 2 vendor for price 3
125 0.0011 0.0014 0.0050 Vendor3 Vendor1 Vendor 2
126 0.0019 0.0019 0.0034 Vendor3 Vendor2 Vendor 1
那么获取这样的记录的MySQL查询是什么？而且我还必须进行查询以从表中获取最大值和第二个最高值，并且可能有任意数量的行包含单个代码。
我的数据位于此SQL Fiddle。
在第二个最高值的情况下，输出应为：
Code price1 price2 price3 vendor for price1 vendor for price 2 vendor for price 3
125 0.0012 0.0016 0.0054 Vendor1 Vendor2 Vendor 1
126 0.0025 0.0024 0.0065 Vendor1 Vendor1 Vendor 2
答案 0 :(得分：2)
以下是如何做到这一点
SELECT
vender_prices.code,
l.price1,
r.price2,
m.price3,
l.vendor_name `Vender1`,
r.vendor_name `Vender2`,
m.vendor_name `Vender3`
FROM vender_prices
LEFT JOIN (SELECT
code, vendor_name, vender_prices.price1
FROM vender_prices
INNER JOIN (SELECT MIN(price1) AS price1 FROM vender_prices GROUP BY vender_prices.code) AS l
ON l.price1 = vender_prices.price1
GROUP BY vender_prices.code
) as l ON vender_prices.code = l.code
LEFT JOIN (SELECT
code, vendor_name, vender_prices.price2
FROM vender_prices
INNER JOIN (SELECT MIN(price2) AS price2 FROM vender_prices GROUP BY vender_prices.code) AS l
ON l.price2 = vender_prices.price2
GROUP BY vender_prices.code
) as r ON vender_prices.code = r.code
LEFT JOIN (SELECT
code, vendor_name, vender_prices.price3
FROM vender_prices
INNER JOIN (SELECT MIN(price3) AS price3 FROM vender_prices GROUP BY vender_prices.code) AS l
ON l.price3 = vender_prices.price3
GROUP BY vender_prices.code
) as m ON vender_prices.code = m.code
GROUP BY vender_prices.code
输出
| CODE | PRICE1 | PRICE2 | PRICE3 | VENDER1 | VENDER2 | VENDER3 |
--------------------------------------------------------------------
| 125 | 0.0011 | 0.0014 | 0.0050 | Vendor 3 | Vendor 1 | Vendor 2 |
| 126 | 0.0019 | 0.0019 | 0.0034 | Vendor 3 | Vendor 2 | Vendor 1 |
答案 1 :(得分：1)
SELECT
data.*,
v1.vendor_name 'vendor for price1',
v2.vendor_name 'vendor for price2',
v3.vendor_name 'vendor for price3'
FROM
(
SELECT
Code,
MIN(price1) price1,
MIN(price2) price2,
MIN(price3) price3
FROM
tbl
GROUP BY Code
) data
LEFT JOIN
(
SELECT
MIN(vendor_name) vendor_name,
Code
FROM
tbl
WHERE
price1 =
(
SELECT
MIN(price1)
FROM
tbl t
WHERE
t.Code = tbl.Code
)
GROUP BY Code
) v1 ON data.Code = v1.Code
LEFT JOIN
(
SELECT
MIN(vendor_name) vendor_name
Code
FROM
tbl
WHERE
price2 =
(
SELECT
MIN(price2)
FROM
tbl t
WHERE
t.Code = tbl.Code
)
GROUP BY Code
) v2 ON data.Code = v2.Code
LEFT JOIN
(
SELECT
MIN(vendor_name) vendor_name
Code
FROM
tbl
WHERE
price3 =
(
SELECT
MIN(price3)
FROM
tbl t
WHERE
t.Code = tbl.Code
)
GROUP BY Code
) v3 ON data.Code = v3.Code
即使查询本身看起来很大，但连接重复了3次。
更新我更新了删除LIMIT并添加MIN(vendor_name) vendor_name而不是vendor_name的查询。
输出
| CODE | PRICE1 | PRICE2 | PRICE3 | VENDOR FOR PRICE1 | VENDOR FOR PRICE2 | VENDOR FOR PRICE3 |
-----------------------------------------------------------------------------------------------
| 125 | 0.0011 | 0.0014 | 0.0050 | Vendor 3 | Vendor 1 | Vendor 2 |
| 126 | 0.0019 | 0.0019 | 0.0034 | Vendor 3 | Vendor 2 | Vendor 1 |
答案 2 :(得分：0)
试试这个：
select table1.vendor_name as vendor_name_for_price1,table2.vendor_name as vendor_name_for_price2, table3.vendor_name as vendor_name_for_price3
from
table table1, table table2, table table3,
(Select code, min(price1),min(price2),min(price3) from table group by code ) TMP1
where
table1.price1=min(price1) and table1.code=TMP1.code
and
table2.price2=min(price2) and table2.code=TMP2.code
and
table3.price3=min(price3) and table3.code=TMP2.code ;
答案 3 :(得分：0)
这是一个很好的做法，我先尝试一种粗暴的方法。
select 1 as wh,code,min(price1) as price,vendor_name
from(
select code,price1,vendor_name
from ForgeRock
order by code,price1
)t1
group by code
union all
select 2 as wh,code,min(price2) as price,vendor_name
from(
select code,price2,vendor_name
from ForgeRock
order by code,price2
)t1
group by code
union all
select 3 as wh,code,min(price3) as price,vendor_name
from(
select code,price3,vendor_name
from ForgeRock
order by code,price3
)t1
group by code
这是一种非常简单易行的方法，无需加入， 我认为这足以使用了。 我添加了一列wh，你可以根据需要调整结果。 但是，使用上面的代码，你无法获得第二低的结果。 有几种方法可以达到这个目标， 我建议你尝试制作像：
这样的分区CREATE TABLE ForgeRock
(`unique_id` tinyint, `vendor_name` varchar(8), `price1` float(5,4)
, `price2` float(5,4), `price3` float(5,4),`code` tinyint)
PARTITION BY KEY(code)
PARTITIONS 2;
INSERT INTO ForgeRock
(`unique_id`, `vendor_name`, `price1`, `price2`, `price3`, `code`)
VALUES
(1,'Vendor 1',0.0012,0.0014,0.0054,125),
(2,'Vendor 2',0.0015,0.0016,0.0050,125),
(3,'Vendor 3',0.0011,0.0019,0.0088,125),
(4,'Vendor 1',0.0025,0.0024,0.0034,126),
(5,'Vendor 2',0.0043,0.0019,0.0065,126),
(6,'Vendor 3',0.0019,0.0085,0.0082,126);
这有点棘手，但如果code中没有这么多不同的值，那么非常有用。
你可以得到结果：
select * from
(select code,price1 as price,vendor_name
from ForgeRock partition (p0)
order by price1
Limit 1,1
)t1
union all
select * from
(select code,price1 as price,vendor_name
from ForgeRock partition (p1)
order by price1
Limit 1,1
)t1
union all
select * from
(select code,price2 as price,vendor_name
from ForgeRock partition (p0)
order by price2
Limit 1,1
)t1
union all
select * from
(select code,price2 as price,vendor_name
from ForgeRock partition (p1)
order by price2
Limit 1,1
)t1
union all
select * from
(select code,price3 as price,vendor_name
from ForgeRock partition (p0)
order by price3
Limit 1,1
)t1
union all
select * from
(select code,price3 as price,vendor_name
from ForgeRock partition (p1)
order by price3
Limit 1,1
)t1
虽然我认为你可以明显改善我的答案， 我会把这份工作留给你。