MYSQL连接/索引优化

Question

我有一个查询，试图找到包含一组给定包装的所有购物车。 对于每个包裹，我都会加入对应的Cartitem表一次，因为我只对包含所有给定包裹的推车感兴趣。

当我到达15个以上的程序包（联接）时，查询性能迅速下降。

我在相应的外部列上有两个索引，并且知道mysql仅使用其中之一。 当我在两列（cartitem_package_id，cartitem_cart_id）上添加索引时，它可以工作，但这是解决这种情况的唯一方法吗？ 我想知道为什么MYSQL突然卡在这种情况下，并且可能是mysql内部问题，因为我看不到此定义和查询有任何更深层次的问题？ 这可能是查询优化器的问题，我是否可以做一些事情（例如添加方括号）来支持或强制执行特定的查询？ 还是有人在使用另一个查询来采用其他方法？

查询看起来像这样：

             SELECT cart_id
             FROM cart
             INNER JOIN cartitem as c1 ON cart_id=c1.cartitem_cart_id AND c1.cartitem_package_id= 7  
             INNER JOIN cartitem as c2 ON cart_id=c2.cartitem_cart_id AND c2.cartitem_package_id= 8  
             INNER JOIN cartitem as c3 ON cart_id=c3.cartitem_cart_id AND c3.cartitem_package_id= 9  
             INNER JOIN cartitem as c4 ON cart_id=c4.cartitem_cart_id AND c4.cartitem_package_id= 10  
             INNER JOIN cartitem as c5 ON cart_id=c5.cartitem_cart_id AND c5.cartitem_package_id= 11  
             INNER JOIN cartitem as c6 ON cart_id=c6.cartitem_cart_id AND c6.cartitem_package_id= 12  
             INNER JOIN cartitem as c7 ON cart_id=c7.cartitem_cart_id AND c7.cartitem_package_id= 13  
             INNER JOIN cartitem as c8 ON cart_id=c8.cartitem_cart_id AND c8.cartitem_package_id= 14  
             INNER JOIN cartitem as c9 ON cart_id=c9.cartitem_cart_id AND c9.cartitem_package_id= 15  
             INNER JOIN cartitem as c10 ON cart_id=c10.cartitem_cart_id AND c10.cartitem_package_id= 16  
             INNER JOIN cartitem as c11 ON cart_id=c11.cartitem_cart_id AND c11.cartitem_package_id= 17  
             INNER JOIN cartitem as c12 ON cart_id=c12.cartitem_cart_id AND c12.cartitem_package_id= 18  
             INNER JOIN cartitem as c13 ON cart_id=c13.cartitem_cart_id AND c13.cartitem_package_id= 19  
             INNER JOIN cartitem as c14 ON cart_id=c14.cartitem_cart_id AND c14.cartitem_package_id= 20  
             INNER JOIN cartitem as c15 ON cart_id=c15.cartitem_cart_id AND c15.cartitem_package_id= 21  
             INNER JOIN cartitem as c16 ON cart_id=c16.cartitem_cart_id AND c16.cartitem_package_id= 22  
             INNER JOIN cartitem as c17 ON cart_id=c17.cartitem_cart_id AND c17.cartitem_package_id= 23 

输出：

No result.

考虑以下示例结构：

CREATE TABLE IF NOT EXISTS `cart` (
  `cart_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `cart_state` smallint(20) DEFAULT NULL,
  PRIMARY KEY (`cart_id`)
) ENGINE=MyISAM  DEFAULT CHARSET=utf8 AUTO_INCREMENT=80 ;



INSERT INTO `cart` (`cart_id`, `cart_state`) VALUES
(1, 0),(2, 5),(3, 0),(4, 0),(5, 0),(6, 0),(7, 0),(8, 0),(9, 0),(10, 0),(11, 0),(12, 0),(13, 0),(14, 5),(15, 5),(16, 10),(17, 0),(18, 10),(19, 40),(20, 10),(21, 5),(22, 0),(23, 10),(24, 10),(25, 0),(26, 10),(27, 5),(28, 5),(29, 0),(30, 5),(31, 0),(32, 0),(33, 0),(34, 0),(35, 0),(36, 0),(37, 0),(38, 0),(39, 0),(40, 0),(41, 0),(42, 0),(43, 0),(44, 0),(45, 40),(46, 0),(47, 0),(48, 1),(49, 0),(50, 5),(51, 0),(52, 0),(53, 5),(54, 5),(55, 0),(56, 0),(57, 10),(58, 0),(59, 0),(60, 5),(61, 0),(62, 0),(63, 10),(64, 0),(65, 5),(66, 5),(67, 10),(68, 10),(69, 0),(70, 0),(71, 10),(72, 0),(73, 10),(74, 0),(75, 10),(76, 0),(77, 10),(78, 0),(79, 10);


CREATE TABLE IF NOT EXISTS `cartitem` (
  `cartitem_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `cartitem_package_id` int(10) unsigned DEFAULT NULL,
  `cartitem_cart_id` int(10) unsigned DEFAULT NULL,
  `cartitem_price` decimal(7,2) NOT NULL DEFAULT '0.00',
  PRIMARY KEY (`cartitem_id`),
  KEY `cartitem_package_id` (`cartitem_package_id`),
  KEY `cartitem_cart_id` (`cartitem_cart_id`)
) ENGINE=MyISAM  DEFAULT CHARSET=utf8 AUTO_INCREMENT=89 ;


INSERT INTO `cartitem` (`cartitem_id`, `cartitem_package_id`, `cartitem_cart_id`, `cartitem_price`) VALUES
(1, 4, 2, 200.00),(2, 7, 3, 30.00),(3, 14, 9, 255.00),(4, 14, 9, 255.00),(5, 22, 9, 120.00),(6, 22, 9, 120.00),(7, 13, 13, 300.00),(8, 13, 13, 300.00),(9, 7, 14, 450.00),(10, 13, 14, 250.00),(11, 17, 14, 150.00),(12, 7, 15, 450.00),(13, 13, 15, 250.00),(14, 18, 15, 127.50),(15, 7, 16, 450.00),(16, 17, 16, 150.00),(17, 7, 18, 450.00),(18, 7, 19, 450.00),(19, 17, 19, 150.00),(20, 21, 19, 25.00),(21, 13, 20, 300.00),(22, 7, 21, 550.00),(23, 19, 21, 105.00),(24, 22, 21, 120.00),(25, 17, 22, 150.00),(26, 7, 23, 550.00),(27, 11, 24, 245.00),(31, 7, 26, 450.00),(32, 21, 26, 25.00),(33, 21, 26, 25.00),(34, 22, 26, 120.00),(35, 23, 26, 120.00),(36, 10, 27, 382.50),(37, 22, 27, 120.00),(38, 13, 27, 250.00),(39, 10, 28, 297.50),(43, 7, 29, 550.00),(41, 20, 28, 82.50),(42, 22, 28, 120.00),(44, 7, 30, 550.00),(46, 22, 30, 120.00),(47, 23, 30, 120.00),(48, 21, 18, 25.00),(49, 21, 19, 25.00),(50, 17, 37, 150.00),(51, 17, 37, 150.00),(52, 21, 37, 25.00),(53, 21, 37, 25.00),(54, 4, 45, 1.20),(55, 6, 45, 0.00),(56, 7, 47, 450.00),(57, 4, 50, 200.00),(58, 13, 52, 250.00),(59, 13, 19, 300.00),(60, 9, 19, 0.00),(61, 17, 53, 150.00),(62, 7, 53, 450.00),(63, 22, 18, 120.00),(64, 7, 16, 450.00),(65, 7, 54, 450.00),(66, 7, 57, 450.00),(67, 17, 57, 150.00),(68, 7, 56, 450.00),(69, 17, 59, 150.00),(70, 7, 60, 450.00),(71, 17, 61, 150.00),(72, 17, 63, 150.00),(73, 21, 65, 25.00),(74, 7, 66, 450.00),(75, 7, 67, 450.00),(76, 11, 68, 385.00),(77, 7, 71, 450.00),(78, 11, 73, 385.00),(79, 13, 73, 300.00),(80, 4, 75, 200.00),(82, 7, 73, 30.00),(83, 18, 73, 127.50),(84, 23, 73, 120.00),(85, 7, 73, 30.00),(86, 10, 77, 382.50),(87, 7, 79, 550.00),(88, 17, 79, 150.00);

给定的查询是可能的边缘情况，在此示例中没有结果。

             SELECT cart_id
         FROM cart
         INNER JOIN cartitem as c1 ON cart_id=c1.cartitem_cart_id AND c1.cartitem_package_id= 7
         INNER JOIN cartitem as c3 ON cart_id=c3.cartitem_cart_id AND c3.cartitem_package_id= 9  
         INNER JOIN cartitem as c4 ON cart_id=c4.cartitem_cart_id AND c4.cartitem_package_id= 13  
         INNER JOIN cartitem as c5 ON cart_id=c5.cartitem_cart_id AND c5.cartitem_package_id= 17  
         INNER JOIN cartitem as c6 ON cart_id=c6.cartitem_cart_id AND c6.cartitem_package_id= 21

输出：

cart_id
-------------
19
19

在这种情况下，查询应返回包含连接到package（7,9,13,17,21）的商品的所有购物车。

Answer 1

我对您的问题的处理方法是：

SELECT
    cart_id
FROM
    cart
INNER JOIN
    cartitem
ON
    cart_id = cartitem_cart_id
WHERE
    cartitem_package_id IN (7,9,13,17,21)      -- items that got to be in the cart
GROUP BY
    cart_id
HAVING
    count(distinct cartitem_package_id) = 5    -- number of different packages
;

演示与您的数据

说明

原理是先过滤所需值列表，此处是您的软件包。 现在计算每个购物车的不同包裹数（ GROUP BY cart_id ）。 如果此计数与过滤器列表中的值数匹配，则每个单个包装都必须在此购物车中。

如果从子选择中获得了这些值，则可以用子选择替换IN子句的值列表。

您应该看到这种方法应该很容易适应类似的需求。