psql中不存在该列

Question

我正在尝试过滤出错误率超过1％的网站的日子。 我已经有一个表来显示各个日期及其各自的错误率，但是当我尝试包括“ where”或“ having”子句以过滤比率低于.01的日期时，查询将停止工作并显示即使我之前声明了几个字符，我的列也不存在。 这是代码：

select date(time) as day, 
    (trunc(cast(count(*) filter (where status similar to '%404%') as decimal) / count(*), 5)) as col1 
    from log
    where col1 > 0.01 
    group by day 
    order by col1 desc;

这是我得到的错误

ERROR:  column "col1" does not exist
LINE 4: where col1 > 0.01 

谢谢！！

Answer 1

col1是聚合的结果。 Postgres允许group by列的别名，但不能having 。 因此，将条件移至having子句：

select date(time) as day, 
      (trunc(cast(count(*) filter (where status similar to '%404%') as decimal) / count(*), 5)) as col1 
from log
group by day 
having (trunc(cast(count(*) filter (where status similar to '%404%') as decimal) / count(*), 5)) > 0.01 
order by col1 desc;

尽管filter确实很花哨，但我认为此版本的逻辑更简单：

      trunc(cast(avg( (status similar to '%404%')::decimal), 5) as col1 

也可以更容易地将它放入having子句中。

Answer 2

问题是，除非对查询进行分层，否则无法在WHERE子句中引用列别名col1 。

重复条件选项：

select date(time) as day, 
       (trunc(cast(count(*) filter (where status similar to '%404%') as decimal) / count(*), 5)) as col1 
from log
group by day 
having (trunc(cast(count(*) filter (where status similar to '%404%') as decimal) / count(*), 5)) > 0.01
order by col1 desc;

派生表选项：

select day, col1
from (select date(time) as day, 
             (trunc(cast(count(*) filter (where status similar to '%404%') as decimal) / count(*), 5)) as col1 
      from log
      group by day) as derivedTable
where col1 > 0.01