SQL:将条件置于聚合函数的结果上

ehm*_*d11 4 php mysql

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
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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
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Spi*_*man 9

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
Run Code Online (Sandbox Code Playgroud)

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; 你也可以用HAVING COUNT(*) > 100.