mysql - SQL:选择其他表中不存在的位置

Question

我在每个表中都有相同的名称，但它仍然从配料表中选择它，即使确切的名称在 ingredients_master 表中。有什么想法吗？

SELECT
    `ingredients`.`id`,
    `ingredients`.`name` 
FROM
    `ingredients` 
WHERE
    `ingredients`.`name` LIKE '%makers%' ESCAPE '!' 
    OR `ingredients`.`name` LIKE '%jack%' ESCAPE '!' 
    OR `ingredients`.`name` LIKE '%hennesey%' ESCAPE '!' 
    OR `ingredients`.`name` LIKE '%bourbon%' ESCAPE '!' 
    OR `ingredients`.`name` LIKE '%RYE%' ESCAPE '!' 
    AND NOT EXISTS ( 
        SELECT * FROM ingredients_master WHERE ingredients.NAME = ingredients_master.NAME 
    ) 
GROUP BY
    `ingredients`.`name`

Answer 1

您的问题与您使用 AND 和 OR 运算符的方式有关:AND 的优先级高于 OR，所以这个:

WHERE
    `ingredients`.`name` LIKE '%makers%' ESCAPE '!' 
    OR `ingredients`.`name` LIKE '%jack%' ESCAPE '!' 
    OR `ingredients`.`name` LIKE '%hennesey%' ESCAPE '!' 
    OR `ingredients`.`name` LIKE '%bourbon%' ESCAPE '!' 
    OR `ingredients`.`name` LIKE '%RYE%' ESCAPE '!' 
    AND NOT EXISTS ( 
        SELECT * FROM ingredients_master WHERE ingredients.NAME = ingredients_master.NAME 
    ) 

实际上等同于:

WHERE
    `ingredients`.`name` LIKE '%makers%' ESCAPE '!' 
    OR `ingredients`.`name` LIKE '%jack%' ESCAPE '!' 
    OR `ingredients`.`name` LIKE '%hennesey%' ESCAPE '!' 
    OR `ingredients`.`name` LIKE '%bourbon%' ESCAPE '!' 
    OR (
        `ingredients`.`name` LIKE '%RYE%' ESCAPE '!' 
        AND NOT EXISTS ( 
            SELECT * FROM ingredients_master WHERE ingredients.NAME = ingredients_master.NAME 
        ) 
    )

因此，如果第一个 OR 条件中的任何一个为真，则结果为真，与 NOT EXIST 条件的结果无关。

你可能想要:

WHERE
    (
        `ingredients`.`name` LIKE '%makers%' ESCAPE '!' 
        OR `ingredients`.`name` LIKE '%jack%' ESCAPE '!' 
        OR `ingredients`.`name` LIKE '%hennesey%' ESCAPE '!' 
        OR `ingredients`.`name` LIKE '%bourbon%' ESCAPE '!' 
        OR `ingredients`.`name` LIKE '%RYE%' ESCAPE '!' 
    )
    AND NOT EXISTS ( 
        SELECT * FROM ingredients_master WHERE ingredients.NAME = ingredients_master.NAME 
    )