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我正在寻找一个快速的 MySQL 查询，它返回与所有价格类别 (和) 相比价格最低 ( ) 或最高 ( ) 的产品的 id。minmaxa, b, cd

我有一个chocolate_stock包含多个价格类别的产品表。从特定类别 (或或) 中获得最低 ( )或min最高 ( ) 价格非常容易。maxabcd

id |      name       | price_a | price_b | price_c | price_d |
--------------------------------------------------------------
1  |   Chips Ahoy    |   250   |   530   |   720   |   120 
--------------------------------------------------------------
2  | Chocolate Chunk |   250   |   90    |  32.92  |   110   
--------------------------------------------------------------
3  |      Oreo       |   103   |  44.52  |   250   |   850   
--------------------------------------------------------------

价格类别是decimal(10,2)。这是一个从类别中返回最高价格但不返回 id 的示例：

$t = 'chocolate_stock';
$arrIds = array(1, 3);

$strQuery = "SELECT id, 
             MAX(price_a) AS price_a, 
             MAX(price_b) AS price_b, 
             MAX(price_c) AS price_c, 
             MAX(price_d) AS price_d 
             FROM $t WHERE id IN(". implode(',', array_map('intval', $arrIds)) .")";

检索此信息的最快方法是什么？

4

5 回答 5

1

如果您将希望输出的样子制成表格可能会有所帮助，但我认为您缺少的部分是 HAVING 子句。

首先 - 试试这个

select min(id), max(price_a) from $t having price_a = max(price_a)

然后尝试

select min(id), min(price_a) from $t having price_a = min(price_a)
union
select min(id), max(price_a) from $t having price_a = max(price_a)
于 2013-05-28T13:41:06.787 回答
1

此查询执行您想要的操作：

(select t.*
 from $t t
 where . . .
 order by price_a desc
 limit 1) union all
 (select t.*
 from $t t
 where . . .
 order by price_b desc
 limit 1) union all
(select t.*
 from $t t
 where . . .
 order by price_c desc
 limit 1) union all
(select t.*
 from $t t
 where . . .
 order by price_d desc
 limit 1)

如果你有一个索引，id它应该表现得相当好。

这种方法需要四次遍历表（尽管索引id应该大大减少）。以下方法只需要通过表一次：

select MAX(price_a),
       substring_index(group_concat(id order by price_a desc), ',', 1),
       max(price_b),
       substring_index(group_concat(id order by price_b desc), ',', 1),
       max(price_c),
       substring_index(group_concat(id order by price_c desc), ',', 1),
       max(price_d),
       substring_index(group_concat(id order by price_d desc), ',', 1)
from $t
where . . .

它使用group_concat()andsubstring_index()来获取每个列的最大 id。

于 2013-05-28T13:43:50.057 回答
1

您要做的第一件事是规范化您的数据，为了便于以后查询，我将创建以下视图：

CREATE VIEW NormalT
AS
    SELECT  ID, Name, 'Price_a' AS Type, Price_a AS Price
    FROM    T
    UNION ALL
    SELECT  ID, Name, 'Price_b' AS Type, Price_b AS Price
    FROM    T
    UNION ALL
    SELECT  ID, Name, 'Price_c' AS Type, Price_c AS Price
    FROM    T
    UNION ALL
    SELECT  ID, Name, 'Price_d' AS Type, Price_d AS Price
    FROM    T;

然后我不确定你想要的格式，如果你想要每个价格的最小值和最大值，你可以使用这个：

SELECT  mt.Type2,
        mt.Type,
        mt.Price,
        t.ID,
        t.Name
FROM    (   SELECT  Type, MIN(Price) AS Price, 'MIN' AS Type2
            FROM    NormalT
            GROUP BY Type
            UNION ALL
            SELECT  Type, MAX(Price) AS Price, 'MAX' AS Type2
            FROM    NormalT
            GROUP BY Type
        ) mt
        INNER JOIN NormalT T
            ON mt.Type = T.Type
            AND mt.Price = t.Price
ORDER BY mt.Type2, mt.Type, t.ID;

这将从您的示例数据中输出以下内容：

TYPE2   TYPE        PRICE   ID  NAME
MAX     Price_a     250     1   Chips Ahoy
MAX     Price_a     250     2   Chocolate Chunk
MAX     Price_b     530     1   Chips Ahoy
MAX     Price_c     720     1   Chips Ahoy
MAX     Price_d     850     3   Oreo
MIN     Price_a     103     3   Oreo
MIN     Price_b     44.52   3   Oreo
MIN     Price_c     32.92   2   Chocolate Chunk
MIN     Price_d     110     2   Chocolate Chunk

但是，如果它只是所有价格（a、b、c 和 d）的最小值和最大值，那么您可以使用以下命令：

SELECT  mt.Type2,
        t.Type,
        mt.Price,
        t.ID,
        t.Name
FROM    (   SELECT  MIN(Price) AS Price, 'MIN' AS Type2
            FROM    NormalT
            UNION ALL
            SELECT  MAX(Price) AS Price, 'MAX' AS Type2
            FROM    NormalT
        ) mt
        INNER JOIN NormalT T
            ON mt.Price = t.Price;

这将输出：

TYPE2   TYPE    PRICE       ID  NAME
MIN     Price_c     32.92   2   Chocolate Chunk
MAX     Price_d     850     3   Oreo

SQL Fiddle 上的示例

于 2013-05-28T13:44:59.610 回答
1

试试这个，它正在模拟分析，因为 MYSQL 默认没有它们：

SELECT id, 
             ( select MAX(price_a) from $t t2 where  t2.id = t1.id ) AS price_a, 
             ( select MAX(price_b) from $t t2 where  t2.id = t1.id ) AS price_b, 
             ( select MAX(price_c) from $t t2 where  t2.id = t1.id ) AS price_c, 
             ( select MAX(price_d) from $t t2 where  t2.id = t1.id ) AS price_d 
             FROM $t t1 WHERE id IN(". implode(',', array_map('intval', $arrIds)) .")

来源：http ://www.oreillynet.com/pub/a/mysql/2007/03/29/emulating-analytic-aka-ranking-functions-with-mysql.html?page=3

于 2013-05-28T14:35:16.830 回答
-1

你没有得到 id 因为， MAX 返回一个值。但 id 并非如此。您可以使用单独的查询，例如

SELECT id,MAX(price_a) FROM $t WHERE id IN (". implode(',', array_map('intval', $arrIds)).")";
SELECT id,MAX(price_b) FROM $t WHERE id IN (". implode(',', array_map('intval', $arrIds)).")";

ETC

于 2013-05-28T13:03:40.137 回答