php – SQL：将条件置于聚合函数的结果上

这件事很好：

SELECT c.id, c.name, c.ascii_name, COUNT(*) AS nr
    FROM cities c 
    INNER JOIN jobs j ON (j.city_id = c.id ) 
    WHERE j.is_active = 1 
    GROUP BY c.name
limit 100

但是,当我想把条件放在新列上时,它表示未找到列

SELECT c.id, c.name, c.ascii_name, COUNT(*) AS nr
    FROM cities c 
    INNER JOIN jobs j ON (j.city_id = c.id ) 
    WHERE j.is_active = 1 and nr > 100
    GROUP BY c.name
limit 100

解决方法:

您应该在HAVING子句中将条件放在nr上,如下所示：

SELECT c.id, c.name, c.ascii_name, COUNT(*) AS nr
    FROM cities c 
    INNER JOIN jobs j ON (j.city_id = c.id ) 
    WHERE j.is_active = 1
    GROUP BY c.name
    HAVING nr > 100
limit 100

这是因为nr是聚合函数(COUNT(*))的结果,因此在应用WHERE过滤器时不可用.

编辑：在某些数据库服务器中,对nr的引用不起作用;你也可以使用HAVING COUNT(*)> 100.