如何比较postgresql中新计算列的值

Question

我正在运行此查询

  select 
    case when "Payment Collection Fee" = 0  then
    round(("Comm (Incl. S.Tax)" - ((("Marketing Fee" *16)/100)+"Marketing Fee"+0+"Courier Fee")),2) 
    when "Web Sale Price" < 0 then round(("Comm (Incl. S.Tax)" - ((("Marketing Fee" *16)/100)+"Marketing Fee"-20+"Courier Fee")),2) 
    else
       round(("Comm (Incl. S.Tax)" - ((("Marketing Fee" *16)/100)+"Marketing Fee"+20+"Courier Fee")),2) 
    end as diff

  from 
    meta.sd_payment_error

现在我想添加条件

  where diff > 10 

但是它给出错误diff列不存在

我如何为上述查询添加条件？

Answer 1

使用派生表：

select *
from
(
select 
    case when "Payment Collection Fee" = 0  then
    round(("Comm (Incl. S.Tax)" - ((("Marketing Fee" *16)/100)+"Marketing Fee"+0+"Courier Fee")),2) 
    when "Web Sale Price" < 0 then round(("Comm (Incl. S.Tax)" - ((("Marketing Fee" *16)/100)+"Marketing Fee"-20+"Courier Fee")),2) 
    else
       round(("Comm (Incl. S.Tax)" - ((("Marketing Fee" *16)/100)+"Marketing Fee"+20+"Courier Fee")),2) 
    end as diff

  from 
    meta.sd_payment_error
) as dt
where dt.diff > 10

Answer 2

您可以使用Common Table Expression：

#include <stdio.h>

void main()
{
    int c, nl;
    nl = 0;
    while ((c = getchar()) != EOF )
        if (c == '\n')
            ++nl;
    printf("%d \n", nl);

    return 0;
}