让我们对已经说过的内容有一个直观的印象。例如，你有一个带有一些任务的表:

SELECT a.id,a.status,a.description FROM minicloud_tasks as a ORDER BY random();

 id |   status   |   description    
----+------------+------------------
  4 | processing | work on postgres
  6 | deleted    | need some rest
  3 | pending    | garden party
  5 | completed  | work on html

并且您希望按状态对任务列表进行排序。 status是一个字符串值列表:

(processing, pending,  completed, deleted)

诀窍是给每个状态值一个整数，并以数字顺序排列列表:

SELECT a.id,a.status,a.description FROM minicloud_tasks AS a
  JOIN (
    VALUES ('processing', 1), ('pending', 2), ('completed', 3), ('deleted', 4)
  ) AS b (status, id) ON (a.status = b.status)
  ORDER BY b.id ASC;

这就导致:

 id |   status   |   description    
----+------------+------------------
  4 | processing | work on postgres
  3 | pending    | garden party
  5 | completed  | work on html
  6 | deleted    | need some rest

信贷@user80168

无序列，只适用于8.4:

select * from comments c
join 
(
    select id, row_number() over() as id_sorter  
    from (select unnest(ARRAY[1,3,2,4]) as id) as y
) x on x.id = c.id
order by x.id_sorter

要做到这一点，我认为您可能应该有一个额外的“ORDER”表，它定义了id到ORDER的映射(有效地执行您对自己的问题的回答)，然后您可以将其用作选择上的附加列，然后对其进行排序。

通过这种方式，您可以显式地描述您希望在数据库中的顺序，以及它应该在哪里。

在Postgres 9.4中，这可以做得更短一些:

select c.*
from comments c
join (
  select *
  from unnest(array[43,47,42]) with ordinality
) as x (id, ordering) on c.id = x.id
order by x.ordering;

或者在没有派生表的情况下更加紧凑:

select c.*
from comments c
  join unnest(array[43,47,42]) with ordinality as x (id, ordering) 
    on c.id = x.id
order by x.ordering

无需手动为每个值分配/维护位置。

在Postgres 9.6中，可以使用array_position():

with x (id_list) as (
  values (array[42,48,43])
)
select c.*
from comments c, x
where id = any (x.id_list)
order by array_position(x.id_list, c.id);

使用CTE时，值列表只需要指定一次。如果这并不重要，也可以写成:

select c.*
from comments c
where id in (42,48,43)
order by array_position(array[42,48,43], c.id);

在Postgresql:

select *
from comments
where id in (1,3,2,4)
order by position(id::text in '1,3,2,4')

相比使用序列的版本略有改进，我认为:

CREATE OR REPLACE FUNCTION in_sort(anyarray, out id anyelement, out ordinal int)
LANGUAGE SQL AS
$$
    SELECT $1[i], i FROM generate_series(array_lower($1,1),array_upper($1,1)) i;
$$;

SELECT 
    * 
FROM 
    comments c
    INNER JOIN (SELECT * FROM in_sort(ARRAY[1,3,2,4])) AS in_sort
        USING (id)
ORDER BY in_sort.ordinal;