Question

我有一个像这样的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;

这是上述查询的结果：

+---------------+----------------------+---------+-------------+
| 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的城市，其中包含一个城市的餐馆数量。

例如：孟买（3），金奈（1）等等

我尝试使用mysql COUN()，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;

有谁可以告诉我这有什么问题？

希望有人可以帮助我。谢谢。

Answer 1

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。

添加

result["nodes"].flatten!.uniq!
result

最后，在你的最后一次分号之前。

Answer 2

如果您不想要餐馆数据，请使用简单的小组：

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

请参阅SQLFiddle。

或者，如果您也想要餐馆数据，请先从城市中选择，然后再加入其他餐桌，以便仍然可以返回没有餐厅的城市。将左连接添加到子查询中，该子查询计算每个城市的频率：

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

请参阅SQLFiddle。

Answer 3

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

我如何使用选择查询的mysql计数？

3 个答案: