Postgresql 如何不在select中计算两次值？

psql（9.6.1，服务器9.5.5）

员工

     Column     |            Type             |                            Modifiers                            | Storage  | Stats target | Description 
----------------+-----------------------------+-----------------------------------------------------------------+----------+--------------+----    ---------
 employee_id    | integer                     | not null default nextval('employees_employee_id_seq'::regclass) | plain    |              | 
 first_name     | character varying(20)       |                                                                 | extended |              | 
 last_name      | character varying(25)       | not null                                                        | extended |              | 
 email          | character varying(25)       | not null                                                        | extended |              | 
 phone_number   | character varying(20)       |                                                                 | extended |              | 
 hire_date      | timestamp without time zone | not null                                                        | plain    |              | 
 job_id         | character varying(10)       | not null                                                        | extended |              | 
 salary         | numeric(8,2)                |                                                                 | main     |              | 
 commission_pct | numeric(2,2)                |                                                                 | main     |              | 
 manager_id     | integer                     |                                                                 | plain    |              | 
 department_id  | integer 

我需要提取员工编号、姓氏、工资、加薪15.5%（以整数表示）以及新旧工资之间的差异

我是这样做的：

select employee_id,
    last_name,
    salary,
    round(salary * 1.155, 0) as "New Salary",
    round(salary * 1.155, 0) - salary as "Increase"
from employees;

select 2 as val_a, val_a - 4; --not working

select t.*,
    New_Salary - salary as Increase
from (
    select employee_id,
        last_name,
        salary,
        round(salary * 1.155, 0) as New_Salary,
    from employees
    ) t;

使我烦恼的是，我已经计算了两次新的工资

我尝试在同一选择中使用alias。试验如下：

select employee_id,
    last_name,
    salary,
    round(salary * 1.155, 0) as "New Salary",
    round(salary * 1.155, 0) - salary as "Increase"
from employees;

select 2 as val_a, val_a - 4; --not working

select t.*,
    New_Salary - salary as Increase
from (
    select employee_id,
        last_name,
        salary,
        round(salary * 1.155, 0) as New_Salary,
    from employees
    ) t;

---

嗯，我的解决方案输出了可接受的结果。但是，难道没有更好的解决方案吗？

如果您担心性能，那么这种计算实际上什么都不是。一些优化器甚至可以在内部重用计算

如果必须自己执行，可以使用如下子查询：

select employee_id,
    last_name,
    salary,
    round(salary * 1.155, 0) as "New Salary",
    round(salary * 1.155, 0) - salary as "Increase"
from employees;

select 2 as val_a, val_a - 4; --not working

select t.*,
    New_Salary - salary as Increase
from (
    select employee_id,
        last_name,
        salary,
        round(salary * 1.155, 0) as New_Salary,
    from employees
    ) t;

---

如果您“挑剔”不计算两次相同的值，则可以使用子查询编写：

SELECT
     *, "New Salary" - salary as "Increase"
FROM
(
    SELECT
        employee_id,
        last_name,
        salary,
        round(salary * 1.155, 0) as "New Salary"
    FROM 
        employees
) AS s0 ;

实际上，运行几次时的差异是可以忽略的：

---

dbfiddle

让我担心的主要是代码的可读性和风格。如果我的解决方案难看，或者你的建议风格更好？我觉得你的问题很好。尽管我建议使用标准别名，使用字母、数字和下划线，这样就不必使用双引号。到处都是双引号肯定会损害可读性。如果你担心性能，那就别担心。