[英]why does this query fail in MySQL 5.1.56?
以下查询在MySQL 5.1.56中失败:
SELECT
shop_id, products.product_id AS
product_id, brand, title, price, image, image_width, image_height
FROM products, users LEFT JOIN
(
SELECT fav5.product_id AS product_id, SUM(CASE
WHEN fav5.current = 1 AND fav5.closeted = 1 THEN 1
WHEN fav5.current = 1 AND fav5.closeted = 0 THEN -1
ELSE 0
END) AS favorites_count
FROM favorites fav5
GROUP BY fav5.product_id
) AS fav6 ON products.product_id=fav6.product_id
WHERE products.product_id= 46876 AND users.user_id!=products.product_id
错误是
#1054 - Unknown column 'products.product_id' in 'on clause'
没有用户表的修改不会失败:
SELECT
shop_id, products.product_id AS
product_id, brand, title, price, image, image_width, image_height
FROM products LEFT JOIN
(
SELECT fav5.product_id AS product_id, SUM(CASE
WHEN fav5.current = 1 AND fav5.closeted = 1 THEN 1
WHEN fav5.current = 1 AND fav5.closeted = 0 THEN -1
ELSE 0
END) AS favorites_count
FROM favorites fav5
GROUP BY fav5.product_id
) AS fav6 ON products.product_id=fav6.product_id
WHERE products.product_id= 46876
在MySQL 5.0.67中,查询均不会失败。 (我从5.0.67导出数据库,然后导入5.1.56,因此结构应相同。)
产品表确实有一个int(10)类型的product_id列。 收藏夹表还具有类型为int(10)的product_id列。 到底是怎么回事?
在MySQL中,JOIN处理运算符的优先级在5.1中已更改。 在5.0.12更改后 ,人们从5.0 MySQL LEFT JOIN升级是一个常见问题-如何重写查询
这是您的原始查询,对其进行了重新格式化,并添加了两个括号:
SELECT shop_id, products.product_id AS
product_id, brand, title, price, image, image_width, image_height
FROM products,
( -- Parenthesis added
users LEFT JOIN
(
SELECT fav5.product_id AS product_id, SUM(CASE
WHEN fav5.current = 1 AND fav5.closeted = 1 THEN 1
WHEN fav5.current = 1 AND fav5.closeted = 0 THEN -1
ELSE 0
END) AS favorites_count
FROM favorites fav5
GROUP BY fav5.product_id
) AS fav6 ON products.product_id=fav6.product_id
) -- Parenthesis added
WHERE products.product_id= 46876 AND users.user_id!=products.product_id
括号表示SQL解析器如何解释查询,并且括号中没有products
表。
混合使用旧样式和新样式(例如,自SQL-92起)是一个坏主意。
采用:
SELECT shop_id, products.product_id AS
product_id, brand, title, price, image, image_width, image_height
FROM products JOIN users ON users.user_id != products.product_id
LEFT JOIN
(SELECT fav5.product_id AS product_id,
SUM(CASE WHEN fav5.current = 1 AND fav5.closeted = 1 THEN 1
WHEN fav5.current = 1 AND fav5.closeted = 0 THEN -1
ELSE 0
END) AS favorites_count
FROM favorites fav5
GROUP BY fav5.product_id
) AS fav6 ON products.product_id=fav6.product_id
WHERE products.product_id = 46876
!=
连接将会很慢(实际上是笛卡尔积)。
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