Question

我知道有很多“类似”的问题，但我无法弄清楚如何使这项工作。我希望对其他人来说很简单！

我有一个如下所示的查询：

SELECT
    d.name DishName,
    alt.name AllergyName
FROM meals_dish d
JOIN meals_dishingredient di ON di.dish_id = d.id
LEFT JOIN meals_food_allergytypes fat ON fat.food_id = di.food_id
LEFT JOIN meals_allergytypes alt ON alt.id = fat.allergytypes_id
GROUP BY d.id, alt.id
ORDER BY d.name

这给了我类似于以下内容的表格：

Dish Name               |    AllergyName
-----------------------------------------
Almond Pudding          |     NULL
Almond Pudding          |     Egg
Apple Cake              |     Milk
Apple Cake              |     Lactose
Apple Cake              |     Egg
Apple Cake              |     NULL
Avocado Chocolate Cream |     Milk
Avocado Chocolate Cream |     Lactose
Avocado Chocolate Cream |     NULL
Avocado Chocolate Cream |     Peanuts
Avocado Chocolate Cream |     Tree Nuts
Avocado Chocolate Cream |     Soy
Avocado Egg             |     NULL
Avocado Egg             |     Egg
Bacon Wrapped Asparagus |     NULL
Baked Cauliflower       |     NULL
Baked Cauliflower       |     Milk
Baked Cauliflower       |     Lactose
Baked Salmon            |     NULL
Baked Salmon            |     Milk
Baked Salmon            |     Lactose

现在，我想做的是列出所有菜肴的清单，其中特定菜肴的任何过敏名称都不是“鸡蛋”等。

例如，如果我只想获得那些不能吃鸡蛋的菜，那么我会运行这个查询，并应该取回Avocado Chocolate Cream, Bacon Wrapped Asparagus, Baked Cauliflower and Baked Salmon的菜肴。

我试图设置一些WHERE NOT IN和其他类似的，但我继续只从菜中取出某种成分而不是不选择菜。例如，在我尝试的查询中，使用'Egg'示例，Apple Cake仍然会返回，但是'Egg'的特定行现在不会显示。

Answer 1

首先，考虑如何确定哪些菜肴中含有“蛋”。

SELECT
   FoodAllergyTbl.dish_id 
FROM
    `meals_food_allergytypes` AS FoodAllergyTbl
INNER JOIN
    `meals_allergytypes` AS AllergyTbl
     ON FoodAllergyTbl.allergy_types_id = AllergyTbl.id
WHERE
    AllergyTbl.name = 'Egg';

然后，您可以选择上面列表中没有id的菜肴列表。

SELECT
    name
FROM
    `meal_dish`
WHERE
    `id` NOT IN (
    SELECT
       FoodAllergyTbl.dish_id 
    FROM
        `meals_food_allergytypes` AS FoodAllergyTbl
    INNER JOIN
        `meals_allergytypes` AS AllergyTbl
         ON FoodAllergyTbl.allergy_types_id = AllergyTbl.id
    WHERE
        AllergyTbl.name = 'Egg'
);

Answer 2

你可以用这种方式用于Dish Name，例如Egg：

  select name 
  from meals_dish
  where  name not in (
  SELECT distinct d.name     
  FROM meals_dish d
  JOIN meals_dishingredient di ON di.dish_id = d.id
  LEFT JOIN meals_food_allergytypes fat ON fat.food_id = di.food_id
  LEFT JOIN meals_allergytypes alt ON alt.id = fat.allergytypes_id 
             AND alt .AllergyName = 'Egg'
  )

MySQL加入场景

2 个答案: