1

食品表で野菜だけを購入する顧客を選択しようとしています.しかし、クエリ以外では使用できません.どうすればよいですか? 

    create table Order (
        email varchar(30),
        nOrder integer,
        primary key (email, nOrder),
        foreign key (email) references Client(email)
    );

    create table RegisterOrder (
        email varchar(30),
        nOrder integer,
        nameA varchar(30),
        primary key (email,nOrder,nameA),
        foreign key (email,nOrder) references Order(email, nOrder),
        foreign key (nameA) references Dishe(nameA)
    );

create table Dishe(
    nameA varchar(30),
    primary key (nameA),
    foreign key (nameA) references Food(nameA));

    create table Food ( 
      nameA varchar(30), 
      type varchar(30),
      primary key (nameA)
    );
4

2 に答える 2

2

この例を考えてみましょう...

 SELECT * FROM recipes;
 +-----------+---------------------------------+
 | recipe_id | recipe                          |
 +-----------+---------------------------------+
 |         6 | Beans & Macaroni                |
 |         9 | Beans on Jacket Potato          |
 |         3 | Beans on Toast                  |
 |        10 | Cheese & Beans on Jacket Potato |
 |         4 | Cheese & Beans on Toast         |
 |         2 | Cheese on Toast                 |
 |         1 | Macaroni & Cheese               |
 |        12 | Peanut Butter on Toast          |
 |         5 | Toast & Jam                     |
 +-----------+---------------------------------+

 SELECT * FROM recipe_ingredient;
 +-----------+---------------+
 | recipe_id | ingredient_id |
 +-----------+---------------+
 |         1 |             1 |
 |         1 |             2 |
 |         2 |             2 |
 |         2 |             4 |
 |         3 |             3 |
 |         3 |             4 |
 |         4 |             2 |
 |         4 |             3 |
 |         4 |             4 |
 |         5 |             4 |
 |         5 |             5 |
 |         6 |             1 |
 |         6 |             3 |
 |         9 |             3 |
 |         9 |             6 |
 |        10 |             2 |
 |        10 |             3 |
 |        10 |             6 |
 |        12 |             4 |
 |        12 |             7 |
 +-----------+---------------+

 SELECT * FROM ingredients;
 +---------------+---------------+
 | ingredient_id | ingredient    |
 +---------------+---------------+
 |             3 | Beans         |
 |             2 | Cheese        |
 |             6 | Jacket Potato |
 |             5 | Jam           |
 |             1 | Macaroni      |
 |             7 | Peanut Butter |
 |             4 | Toast         |
 +---------------+---------------+

以下は、すべてのレシピとそれらを構築するために必要な材料のリストを返します...

 SELECT r.*
      , i.*
   FROM recipes r
   JOIN recipe_ingredient ri
     ON ri.recipe_id = r.recipe_id
   JOIN ingredients i
     ON i.ingredient_id = ri.ingredient_id;
 +-----------+---------------------------------+---------------+---------------+
 | recipe_id | recipe                          | ingredient_id | ingredient    |
 +-----------+---------------------------------+---------------+---------------+
 |         6 | Beans & Macaroni                |             1 | Macaroni      |
 |         6 | Beans & Macaroni                |             3 | Beans         |
 |         9 | Beans on Jacket Potato          |             3 | Beans         |
 |         9 | Beans on Jacket Potato          |             6 | Jacket Potato |
 |         3 | Beans on Toast                  |             3 | Beans         |
 |         3 | Beans on Toast                  |             4 | Toast         |
 |        10 | Cheese & Beans on Jacket Potato |             2 | Cheese        |
 |        10 | Cheese & Beans on Jacket Potato |             3 | Beans         |
 |        10 | Cheese & Beans on Jacket Potato |             6 | Jacket Potato |
 |         4 | Cheese & Beans on Toast         |             2 | Cheese        |
 |         4 | Cheese & Beans on Toast         |             3 | Beans         |
 |         4 | Cheese & Beans on Toast         |             4 | Toast         |
 |         2 | Cheese on Toast                 |             2 | Cheese        |
 |         2 | Cheese on Toast                 |             4 | Toast         |
 |         1 | Macaroni & Cheese               |             1 | Macaroni      |
 |         1 | Macaroni & Cheese               |             2 | Cheese        |
 |        12 | Peanut Butter on Toast          |             4 | Toast         |
 |        12 | Peanut Butter on Toast          |             7 | Peanut Butter |
 |         5 | Toast & Jam                     |             4 | Toast         |
 |         5 | Toast & Jam                     |             5 | Jam           |
 +-----------+---------------------------------+---------------+---------------+

ここで、チーズ、ビーンズ、トーストが入ったパントリーがあるとします。それらの材料のみを使用して、何を構築できますか?

 SELECT r.*
      , SUM(CASE WHEN ingredient IN ('Cheese','Beans','Toast') THEN 1 ELSE 0 END) x
      , COUNT(*) y
   FROM recipes r
   JOIN recipe_ingredient ri ON ri.recipe_id = r.recipe_id
   JOIN ingredients i ON i.ingredient_id = ri.ingredient_id
  GROUP
     BY r.recipe_id;
 +-----------+---------------------------------+------+---+
 | recipe_id | recipe                          | x    | y |
 +-----------+---------------------------------+------+---+
 |         1 | Macaroni & Cheese               |    1 | 2 |
 |         2 | Cheese on Toast                 |    2 | 2 | <-- 
 |         3 | Beans on Toast                  |    2 | 2 | <--
 |         4 | Cheese & Beans on Toast         |    3 | 3 | <-- *
 |         5 | Toast & Jam                     |    1 | 2 |
 |         6 | Beans & Macaroni                |    1 | 2 |
 |         9 | Beans on Jacket Potato          |    1 | 2 |
 |        10 | Cheese & Beans on Jacket Potato |    2 | 3 |
 |        12 | Peanut Butter on Toast          |    1 | 2 |
 +-----------+---------------------------------+------+---+

 x = y : recipes use only those ingredients found in the pantry.

 x = y = total no of ingredients in pantry : recipes using EXACTLY the ingredients found in the pantry

これは次のように書き換えることができます...

 SELECT r.*
   FROM recipes r
   JOIN recipe_ingredient ri ON ri.recipe_id = r.recipe_id
   JOIN ingredients i ON i.ingredient_id = ri.ingredient_id
  GROUP
     BY r.recipe_id
 HAVING  SUM(CASE WHEN ingredient IN ('Cheese','Beans','Toast') THEN 1 ELSE 0 END) = COUNT(*);
于 2013-11-12T14:04:05.967 に答える
0 
select ro.email
from registerorder ro 
inner join food f on f.namea = ro.namea
group by ro.email
having sum(f.type <> 'vegtables') = 0
于 2013-11-12T13:14:35.400 に答える