mysql - MySQL group by 和 Where

Question

我想在下面的 sql 中添加一个 where 子句，以便名为“howmany”的列只显示大于 1 的结果，但是当我尝试时我不断收到错误消息。

 select monthname(birthdate) as 'Month', Count(*) as howmany
 from tblBday
 group by month(birthdate)
 ;

Answer 1

您不能在WHERE子句中使用别名，并且由于要使用聚合函数，因此您需要一个HAVING子句：

select monthname(birthdate) as `Month`, Count(*) as howmany
from tblBday
group by month(birthdate)
having Count(*) > 1

如果要在WHERE子句中使用别名，则可以将查询设为子查询：

select `month`, howmany
from
(
    select monthname(birthdate) as `Month`, Count(*) as howmany
    from tblBday
    group by month(birthdate)
) src
where howmany > 1

Answer 2

这是因为您不能在 WHERE 子句中使用别名

此外，您不能在 WHERE 子句中执行聚合函数

你需要这样的东西

SELECT monthname(birthdate) as Month, Count(*) AS howmany
FROM tblBday
GROUP BY month(birthdate)
HAVING Count(*) > 1
 ;

这是关于 HAVING 的教程

Answer 3

 select monthname(birthdate) as 'Month', Count(*) as howmany
                                ^--   ^--

你不能像那样引用别名。单引号 ( ') 将事物转换为字符串 - 字符串不是别名，也不能是别名。删除引号，或使用反引号：

select monthname(birthdate) as Month, Count(*) as howmany

Answer 4

尝试

select monthname(birthdate) as 'Month', Count(*) as howmany
from tblBday
group by month(birthdate)
having howmany > 1
;

Answer 5

 select monthname(birthdate) as 'Month', Count(*) as howmany
 from tblBday
 group by monthname(birthdate)
Having count(*) > 1
;