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mysql - 不确定如何表达 COUNT() 语句

转载 作者:行者123 更新时间:2023-11-29 05:52:33 25 4
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我有一个问题要解决:

-列出没有城市人口超过 1,000,000 的国家的名称。

(我认为这意味着 COUNT(name) 应该提取等于零的名称。我不确定如何表达该语句。这是我的代码,有两个表连接。

<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 |

最佳答案

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

关于mysql - 不确定如何表达 COUNT() 语句,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52802693/

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