Question

我有以下表格：

餐具

id
name

客户

id
name

成分

id
name

Dishes_Ingredients （关于菜肴和食材的表格）

id
dish_id
ingredient_id

Customers_Allergic_Ingredients （客户对某些成分过敏）

id
customer_id
ingredient_id

Customers_Intolerance_Ingredients （客户不能容忍某些成分）

id
customer_id
ingredient_id

我需要从数据库中获取以下信息：对于给定的customer_id，我想要检索客户不过敏且不耐受的所有菜肴，使用Laravel查询生成器。

这是我到目前为止所尝试的：

$dishes = DB::table('dishes')
           ->join('dishes_ingredients', 'dishes.id', '=', 'dishes_ingredients.dish_id')
           ->join('customers_allergic_ingredients', 'dishes_ingredients.ingredient_id', '<>', 'customers_allergic_ingredients.ingredient_id')
           ->join('customers_intolerance_ingredients', 'dishes_ingredients.ingredient_id', '<>', 'customers_intolerance_ingredients.ingredient_id')
           ->where('customers.id', 1)
           ->select('dishes.id', 'dish_translations.name')
           ->get();

Answer 1

我不知道Laravel，但如果你只想解决问题：

SELECT *
FROM Dishes d
WHERE d.id NOT IN
    (
        SELECT DISTINCT dish_id
        FROM Dishes_Ingridients
        WHERE ingredient_id IN
            (SELECT DISTINCT ingredient_id FROM Customers_Allergic_Ingredients cai WHERE cai.customer_id = ?)
            OR
            ingredient_id IN
            (SELECT DISTINCT ingredient_id FROM Customers_Intolerance_Ingredients cii WHERE cii.customer_id = ?)
    )

Answer 2

尝试使用 Laravel Eloquent：Relationships

这将节省您的时间和时间。你的代码会更简单......

https://laravel.com/docs/5.2/eloquent-relationships

Answer 3

  $dishes = DB::table("dishes AS a") 
                   ->select(array("a.*" ))
                   ->join("dishes_ingredients AS b", "a.id",  "b.dish_id") 
                   ->join("Customers_Allergic_Ingredients AS c", "b.id", "=", "c.ingredient_id")
                   ->where("c.customer_id", "!=", $customer_id) 
                   ->groupBy("a.id")
                   ->get();

这将为您提供所有使用客户不会过敏的成分制作的菜肴

Laravel复杂查询

3 个答案: