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[英]What is the most concise way of pulling database objects from a db into a list?
[英]In Postgres: What is the most concise way to use `IS DISTINCT FROM` on multiple values?
我正在嘗試查找具有枚舉列的所有行,該列要么是 NULL,要么與給定的一組枚舉值不同。 我可以用很多IS DINSTINCT FROM
調用來做到這一點,但它真的很冗長,我寧願使用NOT IN()
語法,但NULL
會拋棄它。
這是我想在 SQL 中做的一個例子,來自這個小提琴: http://sqlfiddle.com/#!17/dfae4d/8
CREATE TYPE mood AS ENUM ('sad', 'ok', 'happy');
CREATE TABLE people (
name text,
current_mood mood,
current_mood_as_text varchar
);
insert into people values('Mr Happy', 'happy', 'happy');
insert into people values('Mr Sad', 'sad', 'sad');
insert into people values('Mr Null', NULL, NULL);
-- This doesn't return MrUnknown because of the NULL value:
select * from people where current_mood NOT IN ('happy');
-- This works great, and includes Mr Null:
select * from people where current_mood IS DISTINCT FROM 'happy';
-- But If I want to start comparing to multiple moods, it gets more verbose fast:
SELECT * FROM people
WHERE current_mood IS DISTINCT FROM 'happy' AND
current_mood IS DISTINCT FROM 'sad';
-- You can't write this, but it's kinda what I want:
-- SELECT * FROM people
-- WHERE current_mood IS DISTINCT FROM ('happy', 'sad');
-- For the non enum column, I could do this to make my intention and code clear and concise
SELECT * FROM people
WHERE coalesce(current_mood_as_text, '') NOT IN ('happy', 'sad');
-- But if I write this, I get an error because '' is not a valid enum value
-- SELECT * FROM people
-- WHERE coalesce(current_mood, '') NOT IN ('happy', 'sad');
還有另一種方法可以使這種多重比較更簡潔嗎?
使用coalesce()
的解決方案:
select *
from people
where coalesce(current_mood not in ('happy', 'sad'), true)
還有一種方式:
SELECT * FROM people
WHERE NOT EXISTS (
SELECT FROM (VALUES ('happy'::mood), ('sad')) v(m)
WHERE people.current_mood = v.m
);
Cleanest(IMO) 方法是在 CTE 中使用VALUES()
並對該表表達式執行NOT EXISTS()
:
WITH m(m) AS ( VALUES( 'happy'::mood) )
SELECT *
FROM people p
WHERE NOT EXISTS (
SELECT * FROM m
WHERE m.m = p.current_mood
);
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