有没有办法在查询中使用SELECT的结果作为列标识符？ （PostgreSQL的）

Question

我需要做一个这样的查询

select * from calendar where (select to_char(now(), 'day')) = true; 

但这是无效和失败，ERROR: failed to find conversion function from unknown to boolean。

我尝试编写查询，今日运行时会归结到

select * from calendar where thursday = true; 

但明天，应该是

select * from calendar where friday = true; 

表有这个模式

mbta=# \d calendar 
      Table "public.calendar" 
    Column |   Type   | Modifiers 
------------+------------------------+----------- 
service_id | character varying(255) | not null 
monday  | boolean    | 
tuesday | boolean    | 
wednesday | boolean    | 
thursday | boolean    | 
friday  | boolean    | 
saturday | boolean    | 
sunday  | boolean    | 
start_date | integer    | 
end_date | integer    | 

如何正确编写此查询？

Answer 1

这一个丑陋的架构......一堆的替代品：

1. 替换“周一，周二......”由一个整型字段的字段，被解释为一个位掩码 - 或使用bit-string data type

2. 将它们替换为包含整数（星期几）的单个字段。

3. 非规范化为一个额外的表，带有一个day_of_week字段和一个FK到日历表。

Answer 2

select * from calendar where (to_char(now(), 'day') != 'Monday' || monday) && (to_char(now(), 'day') != 'Tuesday' || tuesday) && … 

鉴于新的模式，我认为这样的事情是你最好的选择。

Answer 3

是的，有一个解决方案。显然，你不能使用子选择的结果来代替列，但是你可以重新排列关系来适应这种查询。首先，构建调换各个列到一个列子查询

SELECT calendar.*, 'monday' AS weekday, monday AS dayvalue FROM calendar 
UNION ALL 
SELECT calendar.*, 'tuesday' AS weekday, tuesday AS dayvalue FROM calendar 
UNION ALL 
SELECT calendar.*, 'wednesday' AS weekday, wednesday AS dayvalue FROM calendar 
UNION ALL 
SELECT calendar.*, 'thursday' AS weekday, thursday AS dayvalue FROM calendar 
UNION ALL 
SELECT calendar.*, 'friday' AS weekday, friday AS dayvalue FROM calendar 
UNION ALL 
SELECT calendar.*, 'saturday' AS weekday, saturday AS dayvalue FROM calendar 
UNION ALL 
SELECT calendar.*, 'sunday' AS weekday, sunday AS dayvalue FROM calendar 

然后你可以用这都归结为一个子查询，并挑选出恰好与右平日行：

SELECT * FROM (
    SELECT calendar.*, 'monday' AS weekday, monday AS dayvalue FROM calendar 
    UNION ALL 
    SELECT calendar.*, 'tuesday' AS weekday, tuesday AS dayvalue FROM calendar 
    UNION ALL 
    ...   -- You get the idea. 
    UNION ALL 
    SELECT calendar.*, 'sunday' AS weekday, sunday AS dayvalue FROM calendar 
) AS ss WHERE to_char(now(), 'day') = ss.weekday AND dayvalue = true;