Question

我试图获取每个城市中没有旅馆的数量。

我有2张桌子。第一张桌子是hotel_master

lval *z = malloc(sizeof(lval));
z->value.sexpr1 = malloc(sizeof(sexpr));
z->value.sexpr1->cell= NULL;

第二个表是city_master

 hotel_master
 id | hotel_name | city_code

是否可以从查询中获取计数？

首先，我已经尝试过此查询，或者我需要使用do while循环？

 city_master
 id | city_code | city_name

请帮助我。

Answer 1

SELECT COUNT(h.city_code) as total_hotel, c.city_name
FROM hotel_master h
INNER JOIN city_master c ON h.city_code = c.city_code
GROUP BY h.city_code

在COUNT中的字段上使用hotel_master。它将根据城市获得旅馆的数量。

Answer 2

您可以尝试以下方法：

SELECT COUNT(*) 
  FROM hotel_master 
  JOIN city_master 
    ON hotel_master.city_code = city_master.city_code 
  GROUP 
    BY hotel_master.city_code

Answer 3

只需使用COUNT()，GROUP BY就可以做到

SELECT
  count(hotel_master.hotel_name),
  city_master.city_name
FROM hotel_master
INNER JOIN city_master ON hotel_master.city_code = city_master.city_code
GROUP BY city_master.city_name

Answer 4

需要在查询中使用GROUP BY子句来获取每个城市的旅馆数量。

SELECT city_master.city_name, count(hotel_master.id) hotetcount 
  FROM hotel_master 
  JOIN city_master 
    ON hotel_master.city_code = city_master.city_code 
  GROUP 
    BY city_master.city_code

PHP计数每个城市的酒店数量

4 个答案: