选择带有浇头 SQL 的比萨饼答案

【问题标题】：SELECT pizzas with toppings SQL选择带有浇头 SQL 的比萨饼
【发布时间】：2021-11-20 16:52:12
【问题描述】：

谁能告诉我如何展示所有的比萨饼及其浇头，而不用每次浇头多次显示同一个比萨饼？

name	name
Cajun Combo	bacon crumble
Cajun Combo	ham
Cajun Combo	spicy pepperoni
Cajun Combo	Cajun spice
Cajun Combo	Fresh garlic
Cajun Combo	garlic sauce topping
Bistro	black pepper
Bistro	cheddar cheese
Bistro	chili flakes
Bistro	pepperoni
Bistro	red onion
BBQ King	cream cheese
BBQ King	jalapeno
BBQ King	pepper cheese
BBQ King	pepperoni
BBQ King	pulled pork
BBQ King	BBQ topping
Fiesta	cream cheese
Fiesta	Fresh garlic
Fiesta	mushrooms
Fiesta	oregano
Fiesta	pepperoni
Fiesta	pineapple
Italiana	balsamic glaze
Italiana	cream cheese
Italiana	dates
Italiana	pepperoni
Italiana	semi dried tomatoes
Italiana	spinach
Champion	bacon crumble
Champion	black pepper
Champion	Fresh garlic
Champion	ham
Champion	jalapeno
Champion	mushrooms
Champion	pepperoni
Champion	pineapple
Champion	red onion

    CREATE TABLE toppings (
      idTopping INTEGER PRIMARY KEY,
      name text
    );
    
    CREATE TABLE pizzas (
      idPizza INTEGER PRIMARY KEY,
      name text
    );
    
    CREATE TABLE pizzasAndToppings (
      idTopping INTEGER,
      idPizza INTEGER,
      FOREIGN KEY (idTopping)
        REFERENCES toppings (idTopping),
      FOREIGN KEY (idPizza)
        REFERENCES pizzas (idPizza)
    );
    
    
    
    INSERT INTO pizzasAndToppings (idTopping, idPizza) VALUES (5, 1), (7, 1), (12, 1), (27, 1), (29, 1), (30, 1);
    
    INSERT INTO pizzasAndToppings (idTopping, idPizza) VALUES (26, 2), (1, 2), (28, 2), (9, 2), (21, 2);
    
    INSERT INTO pizzasAndToppings (idTopping, idPizza) VALUES (2, 3), (18, 3), (4, 3), (9, 3), (11, 3), (25, 3);
    
    INSERT INTO pizzasAndToppings (idTopping, idPizza) VALUES (2, 4), (29, 4), (19, 4), (31, 4), (9, 4), (20, 4);
    
    INSERT INTO pizzasAndToppings (idTopping, idPizza) VALUES (24, 5), (2, 5), (15, 5), (9, 5), (22, 5), (23, 5);
    
    INSERT INTO pizzasAndToppings (idTopping, idPizza) VALUES (5, 6), (26, 6), (29, 6), (7, 6), (18, 6), (19, 6), (9, 6), (20 ,6), (21, 6);
    
    INSERT INTO toppings (name) VALUES ('cheddar cheese'), ('cream cheese'), ('Havarti'), ('pepper cheese'),
    ('bacon crumble'), ('chicken fajitas'), ('ham'), ('minced beef'), ('pepperoni'), ('prime bacon slices'), ('pulled pork'),
    ('spicy pepperoni'), ('Vegan chicken'), ('black olives'), ('dates'), ('fresh chili'), ('green pepper'), ('jalapeno'), ('mushrooms'),
    ('pineapple'), ('red onion'), ('semi dried tomatoes'), ('spinach'), ('balsamic glaze'), ('BBQ topping'), ('black pepper'),
    ('Cajun spice'), ('chili flakes'), ('Fresh garlic'), ('garlic sauce topping'), ('oregano');
    
    INSERT INTO pizzas (name) VALUES ('Cajun Combo'), ('Bistro'), ('BBQ King'), ('Fiesta'), ('Italiana'), ('Champion');
   
/* SELECT */
    SELECT pizzas.name, toppings.name FROM pizzas INNER JOIN pizzasAndToppings ON pizzasAndToppings.idPizza = pizzas.idPizza INNER JOIN toppings ON toppings.idTopping = pizzasAndToppings.idTopping;

【问题讨论】：

使用 GROUP_CONCAT：dbfiddle.uk/…
对此的预期输出是什么？您尝试过什么来解决问题？
@NicoHaase 使用 Mureinik 答案解决了

标签： sql sqlite select

【解决方案1】：

你可以使用group_concat聚合函数：

SELECT     pizzas.name AS pizza_name, 
           GROUP_CONCAT(toppings.name) AS toppings_names
FROM       pizzas
INNER JOIN pizzasAndToppings ON pizzasAndToppings.idPizza = pizzas.idPizza
INNER JOIN toppings ON toppings.idTopping = pizzasAndToppings.idTopping
GROUP BY   pizzas.name

【讨论】：