繁体   English   中英

在 Postgres 中:在多个值上使用“IS DISTINCT FROM”的最简洁方法是什么?

[英]In Postgres: What is the most concise way to use `IS DISTINCT FROM` on multiple values?

我正在尝试查找具有枚举列的所有行,该列要么是 NULL,要么与给定的一组枚举值不同。 我可以用很多IS DINSTINCT FROM调用来做到这一点,但它真的很冗长,我宁愿使用NOT IN()语法,但NULL会抛弃它。

这是我想在 SQL 中做的一个例子,来自这个小提琴: 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');

还有另一种方法可以使这种多重比较更简洁吗?

使用coalesce()的解决方案:

select * 
from people 
where coalesce(current_mood not in ('happy', 'sad'), true)

SQLFiddle。

还有一种方式:


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()并对该表表达式执行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
        );

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM