将一列分成多列

Question

我正在尝试从我们的数据库中检索数据，并且我加入了另一个表，我需要从该列中获取数据，但被分成多个列。

到目前为止我有这个代码：

SELECT 
    restaurants.name AS 'Name',
    extras_time_slots.restaurant_id AS 'DC ID',
    SUM(capacity) AS 'Seats opened', 
    meal_group_name AS 'Category',
    CASE
        WHEN extras_events.status = 0 THEN 'Offline'
        WHEN extras_events.status = 1 THEN 'Submitted'
        WHEN extras_events.status = 2 THEN 'Online'
        ELSE 'Unknown'
    END AS 'Status',
    CASE
        WHEN restaurants.region_id = 29 THEN 'Beijing'
        WHEN restaurants.region_id = 31 THEN 'Shanghai'
        WHEN restaurants.region_id = 63 THEN 'Guangzhou'
        WHEN restaurants.region_id = 83 THEN 'Shenzhen'
        WHEN restaurants.region_id = 86 THEN 'Hangzhou'
        WHEN restaurants.region_id = 88 THEN 'Suzhou'
        WHEN restaurants.region_id = 91 THEN 'Chengdu'
        ELSE 'Unknown'
        END AS 'City',
    (
        SELECT extras_event_meals.price
        FROM extras_event_meals
        WHERE meal_type LIKE 'lunch'
        AND extras_events_id 
    ) AS 'Lunch price',
    (
        SELECT extras_event_meals.price
        FROM extras_event_meals
        WHERE meal_type LIKE 'dinner'
        AND extras_events_id 
    ) AS 'Dinner price',
    (
        SELECT extras_event_meals.price
        FROM extras_event_meals
        WHERE meal_type LIKE 'brunch'
        AND extras_events_id 
    ) AS 'Brunch price'
FROM extras_time_slots
JOIN restaurants
    ON extras_time_slots.restaurant_id = restaurants.id
JOIN extras_events
    ON extras_events.restaurant_id = restaurants.id AND extras_events.project = extras_time_slots.project
JOIN extras_event_meals
    ON extras_event_meals.extras_event_id = extras_events.id
WHERE extras_time_slots.project = 'rwcn_autumn_2018'
GROUP BY restaurants.id;

因此，extras_events_meal_price 表为一家餐厅提供了多个具有类型和价格的“餐点”。

以下是列：

id
extras_events_id
meal_type
price
created_at
updated_at

一家餐厅可以有多个餐点类型和价格，但它们在同一列中，因此在最后一个 SELECT 子查询中，我试图将它们分成 3 列：早午餐价格、午餐价格、晚餐价格。

我一直得到的错误是：

子查询返回多于 1 行。

Answer 1

您从未在子选择语句中完成查询。 extras_events_id 应该等于 extras_events.id。如果不包括它，您只需检查它是否为true，这将返回表中“午餐”的每个meal_type：

SELECT extras_event_meals.price
FROM extras_event_meals
WHERE meal_type LIKE 'lunch'
AND extras_event_id = extras_events.id