psql中不存在該列

Question

我正在嘗試過濾出錯誤率超過1％的網站的日子。 我已經有一個表來顯示各個日期及其各自的錯誤率，但是當我嘗試包括“ where”或“ having”子句以過濾比率低於.01的日期時，查詢將停止工作並顯示即使我之前聲明了幾個字符，我的列也不存在。 這是代碼：

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;

這是我得到的錯誤

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

謝謝！！

Answer 1

col1是聚合的結果。 Postgres允許group by列的別名，但不能having 。 因此，將條件移至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;

盡管filter確實很花哨，但我認為此版本的邏輯更簡單：

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

也可以更容易地將它放入having子句中。

Answer 2

問題是，除非對查詢進行分層，否則無法在WHERE子句中引用列別名col1 。

重復條件選項：

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;

派生表選項：

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