如何从一组具有派生自现有列的列的联接表中正确选择

Question

这里是新的 sql 用户，我正在尝试对

postgresql

中实现的现有视图进行更改。该视图计算一个列，

computed_column

，它是一个函数 2 个变量（我将在第一个片段中将其称为 #1 和 #2），它可以从现有的列中派生出来，我不确定它是什么做到这一点的最佳方法是不要在

computed_column

选择中做大量案例工作。

视图如下：

create or replace view my_view as
select
  -- some other columns not included here
  (case when t.flag = 'on' THEN (#1) ELSE #2 END) / (#1 - #2) as computed_column
from table1 t1
left join table2 t2 on ....
left join table2 t3 on ....

我尝试了这样的事情（请注意，

derived_column1

是上面的#1，同样是#2）

create or replace view my_view as
select
  -- some other columns not included here
  (case when t1.proto = 'p' then t2.c2 else t2.c3) as derived_column1,
  (case when t1.proto = 's' then t2.c4 else t2.c5) as derived_column2,
  (case when t1.flag = 'on' then (derived_column1) ELSE derived_column2 END) / (derived_column1 - derived_column2) as computed_column
from table1 t1
left join table2 t2 on ....
left join table2 t3 on ....

但这不起作用，因为我显然无法引用同一 select 子句中定义的列别名，所以我决定做一个子查询，但它似乎相当冗长，因为我有一长串列被选中，我必须将其放入外部选择和内部子查询选择中。有更好的方法吗？

Answer 1

尝试使用子查询

create or replace view my_view as
select *
  (case when t1.flag = 'on' then (derived_column1) ELSE derived_column2 END) / (derived_column1 - derived_column2) as computed_column
from(
  select 
    -- some other columns not included here
    (case when t1.proto = 'p' then t2.c2 else t2.c3) as derived_column1,
    (case when t1.proto = 's' then t2.c4 else t2.c5) as derived_column2,
  from table1 t1
 )t1
left join table2 t2 on ....
left join table2 t3 on ....