Question

我有一个问题要解决：

-列出没有城市人口超过1百万的国家/地区的名称。

（我认为这将意味着COUNT（name）应该提取等于0的名称。我不确定如何用该语句表达短语。这是我的代码，其中有两个表。

<mysql>SELECT ct.code, ct.fullname, COUNT(name) FROM city AS cy JOIN country AS ct ON cy.country = ct.code 
WHERE cy.population < 1000000 
GROUP BY ct.code 
ORDER BY ct.code ASC; 

+------+---------------------------------------+-------------+
| code | fullname                              | COUNT(name) |
+------+---------------------------------------+-------------+
| ABW  | Aruba                                 |           1 |
| AFG  | Afghanistan                           |           3 |
| AGO  | Angola                                |           4 |
| AIA  | Anguilla                              |           2 |
| ALB  | Albania                               |           1 |
| AND  | Andorra                               |           1 |
| ANT  | Netherlands Antilles                  |           1 |
| ARE  | United Arab Emirates                  |           5 |

Answer 1

Count()将只计算人口少于1000000的国家/地区中的所有城市。Where子句基本上仅返回那些符合您给定条件的城市（人口少于1000000 ）。不一定表示一个国家的所有城市都在1000000以下。
您需要在Group By中加上Having子句，以便检查一个国家的所有城市，以使所有城市都没有population >= 1000000

尝试以下查询：

SELECT ct.code, 
       ct.fullname 
FROM city AS cy 
JOIN country AS ct ON cy.country = ct.code 
GROUP BY ct.code, ct.fullname 
HAVING SUM(cy.population > 1000000) = 0 
ORDER BY ct.code ASC;

Answer 2

SELECT cc.name
from city c, country cc
where CountryCode = code
group by cc.name
having MAX(c.population) < 1000000;

不确定如何表达COUNT（）语句

2 个答案: