有沒有辦法在查詢中使用SELECT的結果作爲列標識符？ （PostgreSQL的）

Question

我需要做一個這樣的查詢

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

但這是無效和失敗，ERROR: failed to find conversion function from unknown to boolean。

我嘗試編寫查詢，今日運行時會歸結到

select * from calendar where thursday = true; 

但明天，應該是

select * from calendar where friday = true; 

表有這個模式

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    | 

如何正確編寫此查詢？

Answer 1

這一個醜陋的架構......一堆的替代品：

1. 替換「週一，週二......」由一個整型字段的字段，被解釋爲一個位掩碼 - 或使用bit-string data type

2. 將它們替換爲包含整數（星期幾）的單個字段。

3. 非規範化爲一個額外的表，帶有一個day_of_week字段和一個FK到日曆表。

Answer 2

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

鑑於新的模式，我認爲這樣的事情是你最好的選擇。

Answer 3

是的，有一個解決方案。顯然，你不能使用子選擇的結果來代替列，但是你可以重新排列關係來適應這種查詢。首先，構建調換各個列到一個列子查詢

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 

然後你可以用這都歸結爲一個子查詢，並挑選出恰好與右平日行：

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;