如何将 mysql 计数与选择查询一起使用？

Question

我有一个这样的 mysql 选择查询：

select r.restaurant_id, r.restaurant_name, r.city_id, c.name
from restaurants r
inner join cities c on c.id = r.city_id;  

这是上面查询的结果：

+---------------+----------------------+---------+-------------+
| restaurant_id | restaurant_name      | city_id | name        |
+---------------+----------------------+---------+-------------+
|             7 | Somasiri Bake House  |       5 | Mumbai      |
|             8 | Indian    Bake House |       7 | Chennai     |
|             9 | KFC Rest             |       5 | Mumbai      |    
|            10 | Indian t             |       5 | Mumbai      |
+---------------+----------------------+---------+-------------+

现在我想显示所有可用城市以及一个城市现有的餐馆数量。

例如：孟买（3）、钦奈（1）等

我在下面用 mysql COUN() 尝试过，但它对我不起作用。

SELECT c.name, count(r.city_id) AS count
FROM cities c
INNER JOIN restaurants r ON c.id = r.city_id;

谁能告诉我这有什么问题？

希望有人可以帮助我。 谢谢你。

Answer 1

SELECT c.name, COALESCE(count(r.city_id), 0) AS count
FROM cities c
LEFT JOIN restaurants r ON c.id = r.city_id
GROUP BY c.id

Answer 2

如果您不需要餐厅数据，请使用简单的分组方式：

select c.name, count(r.city_id) as available
from cities c
left join restaurants r on c.id = r.city_id
group by r.city_id

见SQLFiddle。

---

或者，如果您也需要餐厅数据，请先从城市中选择，然后左连接到其他表，这样仍然会返回没有餐厅的城市。向计算每个城市频率的子查询添加左连接：

select
  r.restaurant_id,
  r.restaurant_name,
  c.id,
  c.name,
  coalesce(available, 0) available
from cities c
left join restaurants r on c.id = r.city_id
left join (select city_id, count(*) available from restaurants group by 1) a
  on a.city_id = r.city_id

见SQLFiddle。

Answer 3

这称为分组或聚合查询，您需要告诉它如何对元素进行分组。

只需添加

GROUP BY r.restaurant_id, r.restaurant_name, r.city_id, c.name

最后，在最后一个分号之前。